21 Essential PHP Interview Questions *
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echo
and print
are largely the same in PHP. Both are used to output data to the screen.
The only differences are as follows:
-
echo
does not return a value whereasprint
does return a value of 1 (this enablesprint
to be used in expressions). -
echo
can accept multiple parameters (although such usage is rare) while print can only take a single argument.
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.
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:
- 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. - 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"
.
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.
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) }
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).
Please note that as of PHP 7.2, this code produces an error.
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 $text
will be the string “John D” (i.e., “John”, followed by six spaces, followed by “D”) and strlen($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 five. 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 five-character string, PHP instead is very “accommodating” and instead allows this and sets all intermediate characters to blanks.
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.
The number 1
.
Note, in PHP 7.2 this throws a warning that a non-numeric value is encountered on the line and does not do the conversion to the int
, so it echoes PHP
instead.
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.
Traits are a mechanism that provides some of the reuse advantages of multiple inheritance in languages like PHP where multiple inheritance is not supported. Traits enable developers to reuse combinations of methods from different class hierarchies.
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.
-
header()
is used to redirect from one page to another:header("Location: index.php");
-
header()
is used to send an HTTP status code:header("HTTP/1.0 this Not Found");
-
header()
is used to send a raw HTTP header:header('Content-Type: application/json');
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.
$arr=array("0"=>"Even","1"=>"Odd");
$check=13;
echo "Your number is: ".$arr[$check%2];
What is the difference between include_once()
and require_once()
, which one would you use in circumstances where you need to connect to a database, and why?
include_once()
or include
allows a file to be included, and in cases where the file is missing or has the wrong name, we receive an error message and execution will still continue regardless.
On the other hand, require_once()
or require
would be suitable in cases where a file needs to be included once and if it is missing or has a wrong name then we receive a fatal error and the execution of the program stops.
require_once
or require
is a suitable method in cases where a database connection file is involved and helps alleviate the possibility of multiple instances of the same file being included several times.
A session stores the value on the server and cookies store the value in the user’s browser.
$str = 'drinking giving jogging 喝 喝 passing 制图 giving 跑步 吃'; // Example input
Highlight all the Chinese characters in red and return the string.
$str = 'drinking giving jogging 喝 喝 passing 制图 giving 跑步 吃';
$string = explode(' ', $str);
$chi = array_filter( explode('_', preg_replace(array('/[^\p{Han}?]/u', '/(\s)+/'), array('_', '$1'), $str)));
$value = array ();
foreach ($string as $s) {
if (in_array($s, $chi)) {
$value[] = ''.$s.'';
} else {
$value[] = $s;
}
}
return (implode(' ', $value));
$number_class = $number == 0 ? 'blue' : ($number > 0 ? 'green' : 'red');
Here we are assigning different strings to the number_class
variable based on a numeric value ($number
).
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 $str2
in $str1
(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
.
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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