14 Essential PHP Interview Questions*

Consider the following code:

$str1 = 'yabadabadoo';
$str2 = 'yaba';
if (strpos($str1,$str2)) {
    echo "\"" . $str1 . "\" contains \"" . $str2 . "\"";
} else {
    echo "\"" . $str1 . "\" does not contain \"" . $str2 . "\"";
}

The output will be:

"yabadabadoo" does not contain "yaba"

Why? How can this code be fixed to work correctly?

The problem here is that strpos() returns the starting position index of $str1 in $str2 (if found), otherwise it returns false. So in this example, strpos() returns 0 (which is then coerced to false when referenced in the if statement). That’s why the code doesn’t work properly.

The correct solution would be to explicitly compare the value returned by strpos() to false as follows:

$str1 = 'yabadabadoo';
$str2 = 'yaba';
if (strpos($str1,$str2) !== false) {
    echo "\"" . $str1 . "\" contains \"" . $str2 . "\"";
} else {
    echo "\"" . $str1 . "\" does not contain \"" . $str2 . "\"";
}

Note that we used the !== operator, not just the != operator. If we use !=, we’ll be back to the problem that 0 is coerced to false when referenced in a boolean expression, so 0 != false will evaluate to false.

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What will be the output of the code below and why?

$x = 5;
echo $x;
echo "<br />";
echo $x+++$x++;
echo "<br />";
echo $x;
echo "<br />";
echo $x---$x--;
echo "<br />";
echo $x;

The output will be as follows:

5
11
7
1
5

Here’s are the two key facts that explain why:

  1. The term $x++ says to use the current value of $x and then increment it. Similarly, the term $x-- says to use the current value of $x and then decrement it.
  2. The increment operator (++) has higher precedence then the sum operator (+) in order of operations.

With these points in mind, we can understand that $x+++$x++ is evaluated as follows: The first reference to $x is when its value is still 5 (i.e., before it is incremented) and the second reference to $x is then when its value is 6 (i.e., before it is again incremented), so the operation is 5 + 6 which yields 11. After this operation, the value of $x is 7 since it has been incremented twice.

Similarly, we can understand that $x---$x-- is evaluated as follows: The first reference to $x is when its value is still 7 (i.e., before it is decremented) and the second reference to $x is then when its value is 6 (i.e., before it is again decremented), so the operation is 7 - 6 which yields 1. After this operation, the value of $x is back to its original value of 5, since it has been incremented twice and then decremented twice.

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What will be the values of $a and $b after the code below is executed? Explain your answer.

$a = '1';
$b = &$a;
$b = "2$b";

Both $a and $b will be equal to the string "21" after the above code is executed.

Here’s why:

The statement $b = &$a; sets $b equal to a reference to $a (as opposed to setting $b to the then-current value of $a). Thereafter, as long as $b remains a reference to $a, anything done to $a will affect $b and vice versa.

So when we subsequently execute the statement $b = "2$b", $b is set equal to the string "2" followed by the then-current value of $b (which is the same as $a) which is 1, so this results in $b being set equal to the string "21" (i.e., the concatenation of "2" and "1"). And, since $b is a reference to $a, this has the same affect on the value of $a, so both end up equal to "21".

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What will be the output of each of the statements below and why?

var_dump(0123 == 123);
var_dump('0123' == 123);
var_dump('0123' === 123);

var_dump(0123 == 123) will output bool(false) because the leading 0 in 0123 tells the PHP interpreter to treat the value as octal (rather than decimal) value, and 123 octal is equal to 83 decimal, so the values are not equal.

var_dump('0123' == 123) will output bool(true) since the string 0123 will automatically be coerced to an integer when being compared with an integer value. Interestingly, when this conversion is performed, the leading 0 is ignored and the value is treated as a decimal (rather than octal) value, so the values are bother 123 (decimal) and are therefore equal.

var_dump('0123' === 123) outputs bool(false) since it performs a more strict comparison and does not do the automatic type coercion of the string to an integer.

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What is the problem with the code below? What will it output? How can it be fixed?

$referenceTable = array();
$referenceTable['val1'] = array(1, 2);
$referenceTable['val2'] = 3;
$referenceTable['val3'] = array(4, 5);

$testArray = array();

$testArray = array_merge($testArray, $referenceTable['val1']);
var_dump($testArray);
$testArray = array_merge($testArray, $referenceTable['val2']);
var_dump($testArray);
$testArray = array_merge($testArray, $referenceTable['val3']);
var_dump($testArray);

The output will be as follows:

array(2) { [0]=> int(1) [1]=> int(2) }
NULL
NULL

You may also see two warnings generated, similar to the following:

Warning: array_merge(): Argument #2 is not an array
Warning: array_merge(): Argument #1 is not an array

The issue here is that, if either the first or second argument to array_merge() is not an array, the return value will be NULL. For example, although one might reasonably expect that a call such as array_merge($someValidArray, NULL) would simply return $someValidArray, it instead returns NULL! (And to make matters worse, this is not documented well at all in the PHP documentation.)

As a result, the call to $testArray = array_merge($testArray, $referenceTable['val2']) evaluates to $testArray = array_merge($testArray, 3) and, since 3 is not of type array, this call to array_merge() returns NULL, which in turn ends up setting $testArray equal to NULL. Then, when we get to the next call to array_merge(), $testArray is now NULL so array_merge() again returns NULL. (This also explains why the first warning complains about argument #2 and the second warning complains about argument #1.)

The fix for this is straightforward. If we simply typecast the second argument to an array, we will get the desired results. The corrected array_merge() calls would therefore be as follows:

$testArray = array_merge($testArray, (array)$referenceTable['val1']);
var_dump($testArray);
$testArray = array_merge($testArray, (array)$referenceTable['val2']);
var_dump($testArray);
$testArray = array_merge($testArray, (array)$referenceTable['val3']);
var_dump($testArray);

which will yield the following output (and no warnings):

array(2) { [0]=> int(1) [1]=> int(2) } 
array(3) { [0]=> int(1) [1]=> int(2) [2]=> int(3) } 
array(5) { [0]=> int(1) [1]=> int(2) [2]=> int(3) [3]=> int(4) [4]=> int(5) }
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What will this code output and why?

$x = true and false;
var_dump($x);

Surprisingly to many, the above code will output bool(true) seeming to imply that the and operator is behaving instead as an or.

The issue here is that the = operator takes precedence over the and operator in order of operations, so the statement $x = true and false ends up being functionally equivalent to:

$x = true;       // sets $x equal to true
true and false;  // results in false, but has no affect on anything

This is, incidentally, a great example of why using parentheses to clearly specify your intent is generally a good practice, in any language. For example, if the above statement $x = true and false were replaced with $x = (true and false), then $x would be set to false as expected.

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What will $x be equal to after the statement $x = 3 + "15%" + "$25"?

The correct answer is 18.

Here’s why:

PHP supports automatic type conversion based on the context in which a variable or value is being used.

If you perform an arithmetic operation on an expression that contains a string, that string will be interpreted as the appropriate numeric type for the purposes of evaluating the expression. So, if the string begins with one or more numeric characters, the remainder of the string (if any) will be ignored and the numeric value is interpreted as the appropriate numeric type. On the other hand, if the string begins with a non-numeric character, then it will evaluate to zero.

With that understanding, we can see that "15%" evaluates to the numeric value 15 and "$25" evaluates to the numeric value zero, which explains why the result of the statement $x = 3 + "15%" + "$25" is 18 (i.e., 3 + 15 + 0).

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After the code below is executed, what will be the value of $text and what will strlen($text) return? Explain your answer.

$text = 'John ';
$text[10] = 'Doe';

After the above code is executed, the value of $text will be the string “John D” (i.e., “John”, followed by 5 spaces, followed by “D”) and strlen($text) will return 11.

There are two things going on here.

First of all, since $text is a string, setting a single element of $text simply sets that single character to the value specified. The statement $text[10] = 'Doe' therefore sets that single character to 'D' (i.e., the first character in the string "Doe", since an element of a string can only be a single character).

Secondly, $text[10] = 'Doe' says to set the 11th character of the string (remember that indices are zero-based) to 'D'. Prior to that statement, though, the length of the string $text ("John ") was only 5. Whereas compilers or interpreters in other languages might barf (with something akin to an out-of-bounds-index error) when you then attempt to set the 11th character of a 5 character string, PHP instead is very “accommodating” and instead allows this and sets all intermediate characters to blanks.

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PHP_INT_MAX is a PHP constant that corresponds to the largest supported integer value (value is based on the version of PHP being run and the platform it is running on).

Assume that var_dump(PHP_INT_MAX) will yield int(9223372036854775807).

In that case, what will be the result of var_dump(PHP_INT_MAX + 1)? Also, what will be the result of var_dump((int)(PHP_INT_MAX + 1))?

NOTE: It’s not important to supply the exact value when answering the question, but rather to explain what will happen and why.

The result of var_dump(PHP_INT_MAX + 1) will be displayed as a double (in the case of this specific example, it will display double(9.2233720368548E+18)). The key here is for the candidate to know that PHP handles large integers by converting them to doubles (which can store larger values).

And interestingly, the result of var_dump((int)(PHP_INT_MAX + 1)) will be displayed as a negative number (in the case of this specific example, it will display int(-9223372036854775808)). Again, the key here is for the candidate to know that the value will be displayed as a negative number, not to know the precise value.

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How would you sort an array of strings to their natural case-insensitive order, while maintaing their original index association?

For example, the following array:

array(
	'0' => 'z1',
	'1' => 'Z10',
	'2' => 'z12',
	'3' => 'Z2',
	'4' => 'z3',
)

After sorting, should become:

array(
	'0' => 'z1',
	'3' => 'Z2',
	'4' => 'z3',
	'1' => 'Z10',
	'2' => 'z12',
)

The trick to solving this problem is to use three special flags with the standard asort() library function:

asort($arr, SORT_STRING|SORT_FLAG_CASE|SORT_NATURAL)

The function asort() is a variant of the standard function sort() that preserves the index association. The three flags used above SORT_STRING, SORT_FLAG_CASE and SORT_NATURAL forces the sort function to treat the items as strings, sort in a case-insensitive way and maintain natural order respectively.

Note: Using the natcasesort() function would not be a correct answer, since it would not maintain the original index association of the elements of the array.

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What is PEAR in php?

PEAR (PHP Extension and Application Repository) is a framework and repository for reusable PHP components. PEAR is a code repository containing all kinds of php code snippets and libraries.

PEAR also offers a command-line interface that can be used to automatically install packages.

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What are the differences between echo and print in PHP?

echo and print are largely the same in PHP. Both are used to output data to the screen.

The only differences are as follows:

  1. echo does not return a value whereas print does return a value of 1 (this enables print to be used in expressions).
  2. echo can accept multiple parameters (although such usage is rare) while print can only take a single argument.
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What will be the output of the PHP code snippet below?

$v = 1;
$m = 2;
$l = 3;

if( $l > $m > $v){
    echo "yes";
}else{
    echo "no";
}

Since 3 > 2 > 1, one might be fooled into thinking the output would be “yes”.

In fact, though, the output will be “no”.

Here’s why:

First, $l > $m will be evaluated which yields a boolean value of 1 or true. Comparing that boolean value to the integer value 1 (i.e., bool(1) > $v) will result in NULL, so the output will be “no”.

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Consider the following code:

$x = NULL;

if ('0xFF' == 255) {
    $x = (int)'0xFF';
}

What will be the value of $x after this code executes? Explain your answer.

Perhaps surprisingly, the answer is neither NULL nor 255. Rather, the answer is that $x will equal 0 (zero).

Why?

First, let’s consider whether '0xFF' == 255 will evaluate to true or false. When a hex string is loosely compared to an integer, it is converted to an integer. Internally, PHP uses is_numeric_string to detect that the string contains a hex value and converts it to an integer (since the other operand is an integer). So in this case, ‘0xFF’ is converted to its integer equivalent which is 255. Since 255 = 255, this condition evaluates to true. (Note that this only works for hex strings, not for octal or binary strings.)

But if that’s the case, shouldn’t the statement $x = (int)'0xFF'; execute and result in $x being set equal to 255?

Well, the statement does execute, but it results in $x being set equal to 0, not 255 (i.e., it is not set to the integer equivalent of ‘0xFF’). The reason is that the explicit type cast of the string to an integer uses convert_to_long (which works differently than the is_numeric_string function that was used in evaluating the conditional expression, as explained above). convert_to_long processes the string one character at a time from left to right and stops at the first non-numeric character that it reaches. In the case of ‘0xFF’, the first non-numeric character is ‘x’, so the only part of the string processed is the initial ‘0’. As a result, the value returned by (int)'0xFF' is 0, so when the code completes, $x will be equal to 0.

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* 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.
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Julien Loutre
United States
Julien is a former Facebook Team Leader and User Operations Analyst, as well as a Project Manager and Developer. He has extensive experience working with 2 of the most popular stacks: PHP/MySQL (over 10 years of experience) and Node.js/MongoDB (4 years of experience).
Ignacio Freiberg
Argentina
Ignacio is a software engineer and architect with more than 8 years of experience in many programming languages, libraries, and frameworks. He enjoys working in freelance mode to give him the opportunity to constantly develop new skills, and he is familiar with many different paradigms and technologies.
James Koshigoe
United States
James is an entrepreneur, programmer, and software engineer focused on web development. He has a vast amount of experience in startup and small business environments where full-stack skills, creative thinking, and high attention to detail are key and fast turnaround is paramount.
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