ksort
The example states the following:
<?php
// list() doesn't work with strings
list($bar) = "abcde";
var_dump($bar);
// output: NULL
?>
If the string is in a variable however, it seems using list() will treat the string as an array:
<?php
$string = "abcde";
list($foo) = $string;
var_dump($foo);
// output: string(1) "a"
?>
list
(PHP 4, PHP 5)
list — Assign variables as if they were an array
Description
Like array(), this is not really a function, but a language construct. list() is used to assign a list of variables in one operation.
Parameters
-
varname -
A variable.
Return Values
Returns the assigned array.
Examples
Example #1 list() examples
<?php
$info = array('coffee', 'brown', 'caffeine');
// Listing all the variables
list($drink, $color, $power) = $info;
echo "$drink is $color and $power makes it special.\n";
// Listing some of them
list($drink, , $power) = $info;
echo "$drink has $power.\n";
// Or let's skip to only the third one
list( , , $power) = $info;
echo "I need $power!\n";
// list() doesn't work with strings
list($bar) = "abcde";
var_dump($bar); // NULL
?>
Example #2 An example use of list()
<table>
<tr>
<th>Employee name</th>
<th>Salary</th>
</tr>
<?php
$result = $pdo->query("SELECT id, name, salary FROM employees");
while (list($id, $name, $salary) = $result->fetch(PDO::FETCH_NUM)) {
echo " <tr>\n" .
" <td><a href=\"info.php?id=$id\">$name</a></td>\n" .
" <td>$salary</td>\n" .
" </tr>\n";
}
?>
</table>
Example #3 Using nested list()
<?php
list($a, list($b, $c)) = array(1, array(2, 3));
var_dump($a, $b, $c);
?>
int(1) int(2) int(3)
Example #4 Using list() with array indices
<?php
$info = array('coffee', 'brown', 'caffeine');
list($a[0], $a[1], $a[2]) = $info;
var_dump($a);
?>
Gives the following output (note the order of the elements compared in which order they were written in the list() syntax):
array(3) {
[2]=>
string(8) "caffeine"
[1]=>
string(5) "brown"
[0]=>
string(6) "coffee"
}
Notes
Warning
list() assigns the values starting with the right-most parameter. If you are using plain variables, you don't have to worry about this. But if you are using arrays with indices you usually expect the order of the indices in the array the same you wrote in the list() from left to right; which it isn't. It's assigned in the reverse order.
Warning
Modification of the array during list() execution (e.g. using list($a, $b) = $b) results in undefined behavior.
Note:
list() only works on numerical arrays and assumes the numerical indices start at 0.
add a note
User Contributed Notes
list - [3 notes]
chris at chlab dot ch ¶
5 months ago
svennd ¶
4 months ago
The list() definition won't throw an error if your array is longer then defined list.
<?php
list($a, $b, $c) = array("a", "b", "c", "d");
var_dump($a); // a
var_dump($b); // b
var_dump($c); // c
?>
srikanth at networthindia dot com ¶
2 months ago
Note: list cannot assign array cast of object to variables straight away. first you need to convert the object to numeric indexed array.
ex:
list($a, $b, $d) = (array) $abc; // $abc is an object; this will not assign.
list($a, $b, $c) = array_values((array) $abc); // This will work.