24 Essential SQL Interview Questions*

What does UNION do? What is the difference between UNION and UNION ALL?

UNION merges the contents of two structurally-compatible tables into a single combined table. The difference between UNION and UNION ALL is that UNION will omit duplicate records whereas UNION ALL will include duplicate records.

It is important to note that the performance of UNION ALL will typically be better than UNION, since UNION requires the server to do the additional work of removing any duplicates. So, in cases where is is certain that there will not be any duplicates, or where having duplicates is not a problem, use of UNION ALL would be recommended for performance reasons.

Fields marked with an asterisk (*) are required
Comment submitted succesfully. Thank you.
We are going to review the comment and get back to you as soon as possible.

List and explain the different types of JOIN clauses supported in ANSI-standard SQL.

ANSI-standard SQL specifies five types of JOIN clauses as follows:

  • INNER JOIN (a.k.a. “simple join”): Returns all rows for which there is at least one match in BOTH tables. This is the default type of join if no specific JOIN type is specified.

  • LEFT JOIN (or LEFT OUTER JOIN): Returns all rows from the left table, and the matched rows from the right table; i.e., the results will contain all records from the left table, even if the JOIN condition doesn’t find any matching records in the right table. This means that if the ON clause doesn’t match any records in the right table, the JOIN will still return a row in the result for that record in the left table, but with NULL in each column from the right table.

  • RIGHT JOIN (or RIGHT OUTER JOIN): Returns all rows from the right table, and the matched rows from the left table. This is the exact opposite of a LEFT JOIN; i.e., the results will contain all records from the right table, even if the JOIN condition doesn’t find any matching records in the left table. This means that if the ON clause doesn’t match any records in the left table, the JOIN will still return a row in the result for that record in the right table, but with NULL in each column from the left table.

  • FULL JOIN (or FULL OUTER JOIN): Returns all rows for which there is a match in EITHER of the tables. Conceptually, a FULL JOIN combines the effect of applying both a LEFT JOIN and a RIGHT JOIN; i.e., its result set is equivalent to performing a UNION of the results of left and right outer queries.

  • CROSS JOIN: Returns all records where each row from the first table is combined with each row from the second table (i.e., returns the Cartesian product of the sets of rows from the joined tables). Note that a CROSS JOIN can either be specified using the CROSS JOIN syntax (“explicit join notation”) or (b) listing the tables in the FROM clause separated by commas without using a WHERE clause to supply join criteria (“implicit join notation”).

Fields marked with an asterisk (*) are required
Comment submitted succesfully. Thank you.
We are going to review the comment and get back to you as soon as possible.

Consider the following two query results:

SELECT count(*) AS total FROM orders;

+-------+
| total |
+-------+
|  100  |
+-------+

SELECT count(*) AS cust_123_total FROM orders WHERE customer_id = '123';

+----------------+
| cust_123_total |
+----------------+
|       15       |
+----------------+

Given the above query results, what will be the result of the query below?

SELECT count(*) AS cust_not_123_total FROM orders WHERE customer_id <> '123'

The obvious answer is 85 (i.e, 100 - 15). However, that is not necessarily correct. Specifically, any records with a customer_id of NULL will not be included in either count (i.e., they won’t be included in cust_123_total, nor will they be included in cust_not_123_total). For example, if exactly one of the 100 customers has a NULL customer_id, the result of the last query will be:

+--------- ----------+
| cust_not_123_total |
+--------------------+
|         84         |
+--------------------+
Fields marked with an asterisk (*) are required
Comment submitted succesfully. Thank you.
We are going to review the comment and get back to you as soon as possible.

Find top SQL talent today. Toptal can match you with the best developers to finish your project.

View Toptal’s SQL Developers

What will be the result of the query below? Explain your answer and provide a version that behaves correctly.

select case when null = null then 'Yup' else 'Nope' end as Result;

This query will actually yield “Nope”, seeming to imply that null is not equal to itself! The reason for this is that the proper way to compare a value to null in SQL is with the is operator, not with =.

Accordingly, the correct version of the above query that yields the expected result (i.e., “Yup”) would be as follows:

select case when null is null then 'Yup' else 'Nope' end as Result;
Fields marked with an asterisk (*) are required
Comment submitted succesfully. Thank you.
We are going to review the comment and get back to you as soon as possible.

Given the following tables:

sql> SELECT * FROM runners;
+----+--------------+
| id | name         |
+----+--------------+
|  1 | John Doe     |
|  2 | Jane Doe     |
|  3 | Alice Jones  |
|  4 | Bobby Louis  |
|  5 | Lisa Romero  |
+----+--------------+

sql> SELECT * FROM races;
+----+----------------+-----------+
| id | event          | winner_id |
+----+----------------+-----------+
|  1 | 100 meter dash |  2        |
|  2 | 500 meter dash |  3        |
|  3 | cross-country  |  2        |
|  4 | triathalon     |  NULL     |
+----+----------------+-----------+

What will be the result of the query below?

SELECT * FROM runners WHERE id NOT IN (SELECT winner_id FROM races)

Explain your answer and also provide an alternative version of this query that will avoid the issue that it exposes.

Surprisingly, given the sample data provided, the result of this query will be an empty set. The reason for this is as follows: If the set being evaluated by the SQL NOT IN condition contains any values that are null, then the outer query here will return an empty set, even if there are many runner ids that match winner_ids in the races table.

Knowing this, a query that avoids this issue would be as follows:

SELECT * FROM runners WHERE id NOT IN (SELECT winner_id FROM races WHERE winner_id IS NOT null)
Fields marked with an asterisk (*) are required
Comment submitted succesfully. Thank you.
We are going to review the comment and get back to you as soon as possible.

Given two tables created and populated as follows:

CREATE TABLE dbo.envelope(id int, user_id int);
CREATE TABLE dbo.docs(idnum int, pageseq int, doctext varchar(100));

INSERT INTO dbo.envelope VALUES
  (1,1),
  (2,2),
  (3,3);

INSERT INTO dbo.docs(idnum,pageseq) VALUES
  (1,5),
  (2,6),
  (null,0);

What will the result be from the following query:

UPDATE docs SET doctext=pageseq FROM docs INNER JOIN envelope ON envelope.id=docs.idnum
WHERE EXISTS (
  SELECT 1 FROM dbo.docs
  WHERE id=envelope.id
);

Explain your answer.

The result of the query will be as follows:

idnum  pageseq  doctext
1      5        5
2      6        6
NULL   0        NULL

The EXISTS clause in the above query is a red herring. It will always be true since ID is not a member of dbo.docs. As such, it will refer to the envelope table comparing itself to itself!

The idnum value of NULL will not be set since the join of NULL will not return a result when attempting a match with any value of envelope.

Fields marked with an asterisk (*) are required
Comment submitted succesfully. Thank you.
We are going to review the comment and get back to you as soon as possible.

What is wrong with this SQL query? Correct it so it executes properly.

SELECT Id, YEAR(BillingDate) AS BillingYear 
FROM Invoices
WHERE BillingYear >= 2010;

The expression BillingYear in the WHERE clause is invalid. Even though it is defined as an alias in the SELECT phrase, which appears before the WHERE phrase, the logical processing order of the phrases of the statement is different from the written order. Most programmers are accustomed to code statements being processed generally top-to-bottom or left-to-right, but T-SQL processes phrases in a different order.

The correct query should be:

SELECT Id, YEAR(BillingDate) AS BillingYear
FROM Invoices
WHERE YEAR(BillingDate) >= 2010;
Fields marked with an asterisk (*) are required
Comment submitted succesfully. Thank you.
We are going to review the comment and get back to you as soon as possible.

Given these contents of the Customers table:

Id	Name			ReferredBy
1	John Doe		NULL
2	Jane Smith		NULL
3	Anne Jenkins		2
4	Eric Branford		NULL
5	Pat Richards		1
6	Alice Barnes		2

Here is a query written to return the list of customers not referred by Jane Smith:

SELECT Name FROM Customers WHERE ReferredBy <> 2;

What will be the result of the query? Why? What would be a better way to write it?

Although there are 4 customers not referred by Jane Smith (including Jane Smith herself), the query will only return one: Pat Richards. All the customers who were referred by nobody at all (and therefore have NULL in their ReferredBy column) don’t show up. But certainly those customers weren’t referred by Jane Smith, and certainly NULL is not equal to 2, so why didn’t they show up?

SQL Server uses three-valued logic, which can be troublesome for programmers accustomed to the more satisfying two-valued logic (TRUE or FALSE) most programming languages use. In most languages, if you were presented with two predicates: ReferredBy = 2 and ReferredBy <> 2, you would expect one of them to be true and one of them to be false, given the same value of ReferredBy. In SQL Server, however, if ReferredBy is NULL, neither of them are true and neither of them are false. Anything compared to NULL evaluates to the third value in three-valued logic: UNKNOWN.

The query should be written:

SELECT Name FROM Customers WHERE ReferredBy IS NULL OR ReferredBy <> 2

Watch out for the following, though!

SELECT Name FROM Customers WHERE ReferredBy = NULL OR ReferredBy <> 2

This will return the same faulty set as the original. Why? We already covered that: Anything compared to NULL evaluates to the third value in the three-valued logic: UNKNOWN. That “anything” includes NULL itself! That’s why SQL Server provides the IS NULL and IS NOT NULL operators to specifically check for NULL. Those particular operators will always evaluate to true or false.

Even if a candidate doesn’t have a great amount of experience with SQL Server, diving into the intricacies of three-valued logic in general can give a good indication of whether they have the ability learn it quickly or whether they will struggle with it.

Fields marked with an asterisk (*) are required
Comment submitted succesfully. Thank you.
We are going to review the comment and get back to you as soon as possible.

Considering the database schema displayed in the SQLServer-style diagram below, write a SQL query to return a list of all the invoices. For each invoice, show the Invoice ID, the billing date, the customer’s name, and the name of the customer who referred that customer (if any). The list should be ordered by billing date.

SELECT i.Id, i.BillingDate, c.Name, r.Name AS ReferredByName
FROM Invoices i
 JOIN Customers c ON i.CustomerId = c.Id
 LEFT JOIN Customers r ON c.ReferredBy = r.Id
ORDER BY i.BillingDate;

This question simply tests the candidate’s ability take a plain-English requirement and write a corresponding SQL query. There is nothing tricky in this one, it just covers the basics:

  • Did the candidate remember to use a LEFT JOIN instead of an inner JOIN when joining the customer table for the referring customer name? If not, any invoices by customers not referred by somebody will be left out altogether.

  • Did the candidate alias the tables in the JOIN? Most experienced T-SQL programmers always do this, because repeating the full table name each time it needs to be referenced gets tedious quickly. In this case, the query would actually break if at least the Customer table wasn’t aliased, because it is referenced twice in different contexts (once as the table which contains the name of the invoiced customer, and once as the table which contains the name of the referring customer).

  • Did the candidate disambiguate the Id and Name columns in the SELECT? Again, this is something most experienced programmers do automatically, whether or not there would be a conflict. And again, in this case there would be a conflict, so the query would break if the candidate neglected to do so.

Note that this query will not return Invoices that do not have an associated Customer. This may be the correct behavior for most cases (e.g., it is guaranteed that every Invoice is associated with a Customer, or unmatched Invoices are not of interest). However, in order to guarantee that all Invoices are returned no matter what, the Invoices table should be joined with Customers using LEFT JOIN:

SELECT i.Id, i.BillingDate, c.Name, r.Name AS ReferredByName
FROM Invoices i
 LEFT JOIN Customers c ON i.CustomerId = c.Id
 LEFT JOIN Customers r ON c.ReferredBy = r.Id
ORDER BY i.BillingDate;
Fields marked with an asterisk (*) are required
Comment submitted succesfully. Thank you.
We are going to review the comment and get back to you as soon as possible.

Assume a schema of Emp ( Id, Name, DeptId ) , Dept ( Id, Name).

If there are 10 records in the Emp table and 5 records in the Dept table, how many rows will be displayed in the result of the following SQL query:

Select * From Emp, Dept

Explain your answer.

The query will result in 50 rows as a “cartesian product” or “cross join”, which is the default whenever the ‘where’ clause is omitted.

Fields marked with an asterisk (*) are required
Comment submitted succesfully. Thank you.
We are going to review the comment and get back to you as soon as possible.

Given a table SALARIES, such as the one below, that has m = male and f = female values. Swap all f and m values (i.e., change all f values to m and vice versa) with a single update query and no intermediate temp table.

Id  Name  Sex  Salary
1   A     m    2500
2   B     f    1500
3   C     m    5500
4   D     f    500
UPDATE SALARIES SET sex = CASE sex WHEN 'm' THEN 'f' ELSE 'm' END
Fields marked with an asterisk (*) are required
Comment submitted succesfully. Thank you.
We are going to review the comment and get back to you as soon as possible.

Given two tables created as follows

create table test_a(id numeric);

create table test_b(id numeric);

insert into test_a(id) values
  (10),
  (20),
  (30),
  (40),
  (50);

insert into test_b(id) values
  (10),
  (30),
  (50);

Write a query to fetch values in table test_a that are and not in test_b without using the NOT keyword.

In SQL Server, PostgreSQL, and SQLite, this can be done using the except keyword as follows:

select * from test_a
except
select * from test_b;

In Oracle, the minus keyword is used instead.

MySQL does not support the except function, so it is necessary to use not in.

Fields marked with an asterisk (*) are required
Comment submitted succesfully. Thank you.
We are going to review the comment and get back to you as soon as possible.

Given a table TBL with a field Nmbr that has rows with the following values:

1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 1, 0, 1, 0, 1

Write a query to add 2 where Nmbr is 0 and add 3 where Nmbr is 1.

This can be done as follows:

update TBL set Nmbr = case when Nmbr > 0 then Nmbr+3 else Nmbr+2 end;
Fields marked with an asterisk (*) are required
Comment submitted succesfully. Thank you.
We are going to review the comment and get back to you as soon as possible.

Write a SQL query to find the 10th highest employee salary from an Employee table. Explain your answer.

(Note: You may assume that there are at least 10 records in the Employee table.)

This can be done as follows:

SELECT TOP (1) Salary FROM
(
    SELECT DISTINCT TOP (10) Salary FROM Employee ORDER BY Salary DESC
) AS Emp ORDER BY Salary

This works as follows:

First, the SELECT DISTINCT TOP (10) Salary FROM Employee ORDER BY Salary DESC query will select the top 10 salaried employees in the table. However, those salaries will be listed in descending order. That was necessary for the first query to work, but now picking the top 1 from that list will give you the highest salary not the the 10th highest salary.

Therefore, the second query reorders the 10 records in ascending order (which the default sort order) and then selects the top record (which will now be the lowest of those 10 salaries).

Not all databases support the TOP keyword. For example, MySQL and PostreSQL use the LIMIT keyword, as follows:

SELECT Salary FROM
(
    SELECT DISTINCT Salary FROM Employee ORDER BY Salary DESC LIMIT 10
) AS Emp ORDER BY Salary LIMIT 1;
Fields marked with an asterisk (*) are required
Comment submitted succesfully. Thank you.
We are going to review the comment and get back to you as soon as possible.

Write a SQL query using UNION ALL (not UNION) that uses the WHERE clause to eliminate duplicates. Why might you want to do this?

You can avoid duplicates using UNION ALL and still run much faster than UNION DISTINCT (which is actually same as UNION) by running a query like this:

SELECT * FROM mytable WHERE a=X UNION ALL SELECT * FROM mytable WHERE b=Y AND a!=X

The key is the AND a!=X part. This gives you the benefits of the UNION (a.k.a., UNION DISTINCT) command, while avoiding much of its performance hit.

Fields marked with an asterisk (*) are required
Comment submitted succesfully. Thank you.
We are going to review the comment and get back to you as soon as possible.

Given the following tables:

SELECT * FROM users;

user_id  username
1        John Doe                                                                                            
2        Jane Don                                                                                            
3        Alice Jones                                                                                         
4        Lisa Romero

SELECT * FROM training_details;

user_training_id  user_id  training_id  training_date
1                 1        1            "2015-08-02"
2                 2        1            "2015-08-03"
3                 3        2            "2015-08-02"
4                 4        2            "2015-08-04"
5                 2        2            "2015-08-03"
6                 1        1            "2015-08-02"
7                 3        2            "2015-08-04"
8                 4        3            "2015-08-03"
9                 1        4            "2015-08-03"
10                3        1            "2015-08-02"
11                4        2            "2015-08-04"
12                3        2            "2015-08-02"
13                1        1            "2015-08-02"
14                4        3            "2015-08-03"

Write a query to to get the list of users who took the a training lesson more than once in the same day, grouped by user and training lesson, each ordered from the most recent lesson date to oldest date.

SELECT
      u.user_id,
      username,
      training_id,
      training_date,
      count( user_training_id ) AS count
  FROM users u JOIN training_details t ON t.user_id = u.user_id
  GROUP BY user_id,
           training_id,
           training_date
  HAVING count( user_training_id ) > 1
  ORDER BY training_date DESC;
user_id  username      training_id  training_date             count
4        Lisa Romero   2            August, 04 2015 00:00:00  2
4        Lisa Romero   3            August, 03 2015 00:00:00  2
1        John Doe      1            August, 02 2015 00:00:00  3
3        Alice Jones   2            August, 02 2015 00:00:00  2
Fields marked with an asterisk (*) are required
Comment submitted succesfully. Thank you.
We are going to review the comment and get back to you as soon as possible.

What is an execution plan? When would you use it? How would you view the execution plan?

An execution plan is basically a road map that graphically or textually shows the data retrieval methods chosen by the SQL server’s query optimizer for a stored procedure or ad hoc query. Execution plans are very useful for helping a developer understand and analyze the performance characteristics of a query or stored procedure, since the plan is used to execute the query or stored procedure.

In many SQL systems, a textual execution plan can be obtained using a keyword such as EXPLAIN, and visual representations can often be obtained as well. In Microsoft SQL Server, the Query Analyzer has an option called “Show Execution Plan” (located on the Query drop down menu). If this option is turned on, it will display query execution plans in a separate window when a query is run.

Fields marked with an asterisk (*) are required
Comment submitted succesfully. Thank you.
We are going to review the comment and get back to you as soon as possible.

List and explain each of the ACID properties that collectively guarantee that database transactions are processed reliably.

ACID (Atomicity, Consistency, Isolation, Durability) is a set of properties that guarantee that database transactions are processed reliably. They are defined as follows:

  • Atomicity. Atomicity requires that each transaction be “all or nothing”: if one part of the transaction fails, the entire transaction fails, and the database state is left unchanged. An atomic system must guarantee atomicity in each and every situation, including power failures, errors, and crashes.
  • Consistency. The consistency property ensures that any transaction will bring the database from one valid state to another. Any data written to the database must be valid according to all defined rules, including constraints, cascades, triggers, and any combination thereof.
  • Isolation. The isolation property ensures that the concurrent execution of transactions results in a system state that would be obtained if transactions were executed serially, i.e., one after the other. Providing isolation is the main goal of concurrency control. Depending on concurrency control method (i.e. if it uses strict - as opposed to relaxed - serializability), the effects of an incomplete transaction might not even be visible to another transaction.
  • Durability. Durability means that once a transaction has been committed, it will remain so, even in the event of power loss, crashes, or errors. In a relational database, for instance, once a group of SQL statements execute, the results need to be stored permanently (even if the database crashes immediately thereafter). To defend against power loss, transactions (or their effects) must be recorded in a non-volatile memory.
Fields marked with an asterisk (*) are required
Comment submitted succesfully. Thank you.
We are going to review the comment and get back to you as soon as possible.

What is a key difference between Truncate and Delete?

Truncate is used to delete table content and the action can not be rolled back, whereas Delete is used to delete one or more rows in the table and can be rolled back.

Fields marked with an asterisk (*) are required
Comment submitted succesfully. Thank you.
We are going to review the comment and get back to you as soon as possible.

Given a table dbo.users where the column user_id is a unique identifier, how can you efficiently select the first 100 odd user_id values from the table?

(Assume the table contains well over 100 records with odd user_id values.)

SELECT TOP 100 user_id FROM dbo.users WHERE user_id % 2 = 1 ORDER BY user_id

Fields marked with an asterisk (*) are required
Comment submitted succesfully. Thank you.
We are going to review the comment and get back to you as soon as possible.

How can you select all the even number records from a table? All the odd number records?

To select all the even number records from a table:

Select * from table where id % 2 = 0 

To select all the odd number records from a table:

Select * from table where id % 2 != 0
Fields marked with an asterisk (*) are required
Comment submitted succesfully. Thank you.
We are going to review the comment and get back to you as soon as possible.

What are the NVL and the NVL2 functions in SQL? How do they differ?

Both the NVL(exp1, exp2) and NVL2(exp1, exp2, exp3) functions check the value exp1 to see if it is null.

With the NVL(exp1, exp2) function, if exp1 is not null, then the value of exp1 is returned; otherwise, the value of exp2 is returned, but case to the same data type as that of exp1.

With the NVL2(exp1, exp2, exp3) function, if exp1 is not null, then exp2 is returned; otherwise, the value of exp3 is returned.

Fields marked with an asterisk (*) are required
Comment submitted succesfully. Thank you.
We are going to review the comment and get back to you as soon as possible.

What is the difference between the RANK() and DENSE_RANK() functions? Provide an example.

The only difference between the RANK() and DENSE_RANK() functions is in cases where there is a “tie”; i.e., in cases where multiple values in a set have the same ranking. In such cases, RANK() will assign non-consecutive “ranks” to the values in the set (resulting in gaps between the integer ranking values when there is a tie), whereas DENSE_RANK() will assign consecutive ranks to the values in the set (so there will be no gaps between the integer ranking values in the case of a tie).

For example, consider the set {25, 25, 50, 75, 75, 100}. For such a set, RANK() will return {1, 1, 3, 4, 4, 6} (note that the values 2 and 5 are skipped), whereas DENSE_RANK() will return {1,1,2,3,3,4}.

Fields marked with an asterisk (*) are required
Comment submitted succesfully. Thank you.
We are going to review the comment and get back to you as soon as possible.

What is the difference between the WHERE and HAVING clauses?

When GROUP BY is not used, the WHERE and HAVING clauses are essentially equivalent.

However, when GROUP BY is used:

  • The WHERE clause is used to filter records from a result. The filtering occurs before any groupings are made.
  • The HAVING clause is used to filter values from a group (i.e., to check conditions after aggregation into groups has been performed).
Fields marked with an asterisk (*) are required
Comment submitted succesfully. Thank you.
We are going to review the comment and get back to you as soon as possible.
* There is more to interviewing than tricky technical questions, so these are intended merely as a guide. Not every “A” candidate worth hiring will be able to answer them all, nor does answering them all guarantee an “A” candidate. At the end of the day, hiring remains an art, a science — and a lot of work.
Submit an interview question
Submitted questions and answers are subject to review and editing, and may or may not be selected for posting, at the sole discretion of Toptal, LLC.
All fields are required
Thanks for submitting your question.
Our editorial staff will review it shortly. Please note that submitted questions and answers are subject to review and editing, and may or may not be selected for posting, at the sole discretion of Toptal, LLC.
Gustav Stieger
Australia
Gustav is a top engineer and programmer with 20 years of experience with all levels of software and architecture. A creative abstract and theoretical thinker with the ability to turn ideas into re-usable modules and solutions, Gustav has worked at all levels of participation, from contractor to technical lead.
Mike Stankavich
United States
Mike is a senior data engineer and freelance architect experienced across the development stack. He has extensive cloud and infrastructure experience, with multiple certifications from Microsoft, ISC2, Powersoft, and more. He currently builds data back-ends for web apps at scale on the RDBMS or NoSQL platforms.
Jake Varghese
United States
As an architect, Jake understands the process of solving complex problems that plague enterprise and medium size businesses. He is a full-stack problem solver--he can build entire apps from scratch or he can optimize an existing app already in the field.