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[edit] Last updated: Fri, 10 Feb 2012

list

(PHP 4, PHP 5)

list — Assign variables as if they were an array

Description

array list ( mixed $varname [, mixed $... ] )

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 = mysql_query("SELECT id, name, salary FROM employees", $conn);
while (list($id, $name, $salary) = mysql_fetch_row($result)) {
    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.


Although the warning claims of undefined behaviour, it's reasonably straightforward what happens.



<?php

    $a = array(1, 2, 3);

    list($a, $b, $c) = $a;

    var_dump($a, $b, $c);

    // int(1)

    // int(2)

    // int(3)

?>



Since list() works from right to left on the target variables, $a is replaced last, thus keeping the remaining array element values around long enough to populate $b and $c.



<?php

    $a = array(1, 2, 3);

    list($c, $b, $a) = $a;

    var_dump($a, $b, $c);

    // int(3)

    // NULL

    // NULL

?>



$a is populated first (remember, right to left) with the value 3 from itself. Since $a is no longer an array, list() is forced to substitute NULL values for $b and $c.



This appears to be perfectly sensible, though there may be edge cases and other strange behaviour I'm unaware of.

Ultimater at gmail dot com 08-May-2011 03:37


Note that list(...) isn't limited to scalar variables.

It can even create an associative array:



<?php

$html=<<<EOT

<table>

<tr class="row-odd">

<td><span class="username">Foo</span></td>

<td><span class="userid">30185</span></td>

</tr><tr class="row-even">

<td><span class="username">Bar</span></td>

<td><span class="userid">3093</span></td>

</tr>

</table>

EOT;



interface RegExps

{

    const PROFILE_CONTENT='/<tr[^>]*>.*?<span class="username">(.*?)<\/span>[

        ]?.*?<span class="userid">(.*?)<\/span>.*?<\/tr>/msi';

}



preg_match_all(RegExps::PROFILE_CONTENT,$html, $matches,PREG_SET_ORDER);

$profiles=array();

foreach($matches as $match)

{

    $profile=array();

    list(,$profile['username'],$profile['userid'])=$match;

    //$profile=array_reverse($profile);

    $profiles[]=$profile;

}

echo '<pre>'.print_r($profiles,true).'</pre>';

?>



Just be careful, as the manual warns us, regarding the right-most parameter being handled first.

If this becomes a problem, I'd suggest using array_reverse as shown in my code.

boukeversteegh at gmail dot com 06-Apr-2011 12:49


If your array is shorter than the number of arguments in list(), you will get an "undefined index" notice.



You can solve it by making sure the array is long enough:



<?php

   $array = Array( "one", "two" );



   # This will give a Notice: undefined index [2]:

   list( $one, $two, $three ) = $array;



   # This won't:

   list( $one, $two, $three ) = $array + Array( null, null, null );



   # If you know count($array) will be at least 1, you could skip the first index:

   list( $one, $two, $three ) = $array + Array( 1 => null, null );



   # You could of course also use other default values:

   list( $one, $two, $three ) = $array + Array( "one", "two", "three" );

?>

lili at nikha dot org 04-Apr-2011 04:02


Keep it simple! 

For associative arrays, my replacement for list() is this:

<?php

foreach ($associative_array as $key => $value) { $$key = $value; }

?>



Example:

<?php

$petnames = array();

$petnames['dog'] = 'Paul';

$petnames['cat'] = 'Lili';



foreach ($petnames as $name => $value) { $$name = $value; }



echo 'my pets are '.$dog.' and '.$cat;

?>



Will give you:

my pets are Paul and Lili

develop at dieploegers dot de 31-Mar-2010 03:37


Remember, that list starts from index 0. You can skip an index if you just leave the column blank like this:





<?php


list(,$a,$b,$c) = array(1,2,3,4);


?>





You CAN'T (at least not in 5.3.1, what I have tested) set the column to null:





<?php


list(null,$a,$b,$c) = array(1,2,3,4);


?>





This will fail.

Anonymous 12-Mar-2010 09:29


Quick little function that is similar to list but for objects.





<?php


    function listObj() {


        $stack = debug_backtrace();


        if (isset($stack[0]['args'])) {


            $i = 0;


            $args = $stack[0]['args'];


            foreach ($args[0] as $key => $value)


                $args[++$i] = $value;


        }


    }





    class obj {public $var = "test"; public $vars = "test2"; function obj() {}}


    listObj(new obj, &$var, &$var2);


    echo $var, $var2;


?>

claude dot pache at gmail dot com 13-May-2009 03:26


A simple way to swap variables (correction of a note of mario dot mueller dot work at gmail dot com below):

<?php

list($var1, $var2) = array($var2, $var1); // swaps the values of $var1 and $var2

?>

Note that this is not equivalent to:

<?php

$var2 = $var1; $var1 = $var2; // $var1 and $var2 get both the old value of $var1

?>

as one could fear. Indeed, the array is constructed with the values of $var1 and $var2 (and not with the variables $var1 and $var2 themselves) before the assignment is carried out.



Similarly, it is possible to bypass the problem pointed by sasha in the previous note by providing an expression rather than a variable on the right-hand side of the assignment operator:

<?php

$var = array ("test" ,"blah");

list ($a,$var) = $var + array();

echo $a ; // prints "test", not "b"

echo $var ; // prints "blah"

?>

tristan in oregon 08-Apr-2008 07:44


Here's yet another way to make a list()-like construct for associative arrays. This one has the advantage that it doesn't depend on the order of the keys, it only extracts the keys that you specify, and only extracts them into the current scope instead of the global scope (which you can still do, but at least here you have the option).



<?php

    $arr  = array("foo" => 1, "bar" => 2, "baz" => 3);

    $keys = array("baz");



//  $foo = 10;

    $bar = 20;

    $baz = 30;



    extract(array_intersect_key($arr, $keys));



    var_dump($foo);

    var_dump($bar);

    var_dump($baz);

?>



Should print

NULL

int(20)

int(3)



If your version of PHP doesn't have array_intersect_key() yet (below 5.1 I think), it's easy to write a limited feature replacement for this purpose.



<?php

function my_array_intersect_key ($assoc, $keys)

{

    $intersection = array();

    foreach ($assoc as $key => $val)

        if (in_array($key, $keys))

            $intersection[$key] = $val;



    return $intersection;

}

?>

kevin at vanzonneveld dot net 06-Feb-2008 07:12


Another way to do it associative (if your array isn't numeric), is to just use array_values like this:



<?php

$os = array();

$os["main"] = "Linux";

$os["distro"] = "Ubuntu";

$os["version"] = "7.10";



list($main, $distro, $version) = array_values($os);

?>

danieljames3 at g mail 19-Jan-2008 05:51


With regard to the note written by ergalvan at bitam dot com:



You must take note that list() assigns variables starting from the rightmost one (as stated in the warning). That makes $record having the value "value4" and then $var1, $var2 and $var3 take their values from the "new" $record variable.



It's clear that the behavior stated in the warning wasn't followed by version 5.0.4 (and perhaps previous versions?)



----------



I'm still seeing this behavior in PHP 5.2.5.  Hopefully someone can comment on why it's been changed.

Hayley Watson 04-Nov-2007 12:36


In the code by tenz699 at hotmail dot com, the list() construct is taking values from the result of the each() function, not from the associative array; the example is therefore spurious.



each() returns an array of four elements, indexed in the order 1, 'value', 0, 'key'. As noted in the documentation, the associative keys are ignored, and the numerically-indexed values are assigned in key order.



<?php

$array = array('foo'=>'bar');

$t = each($array);

print_r($t);

list($a,$b,$c,$d) = $t;

var_dump($a);

var_dump($b);

var_dump($c);

var_dump($d);

?>



Output:

Array

(

    [1] => bar

    [value] => bar

    [0] => foo

    [key] => foo

)

string(3) "foo"

string(3) "bar"

NULL

NULL

tenz699 at hotmail dot com 18-Sep-2007 10:50


PhP manual's NOTE says: list() only works on numerical arrays and assumes the numerical indices start at 0.



I'm finding it do works for associative arrays too,as below:



<?

$tenzin = array ("1" => "one", "2" => "two","3"=>"three");

while(list($keys,$values) = each($tenzin))

echo($keys." ".$values."<br>");

?>



gives O/P

1 one  

2 two 

3 three



tsarma

mick at wireframe dot com 08-Aug-2007 12:08


It's worth noting that, as expected, list() does not have to have as many variables (and/or empty skips) as there are elements in the array. PHP will disregard all elements that there are no variables for. So:



<?php

$Array_Letters = array('A', 'B', 'C', 'D', 'E', 'F');



list($Letter_1, $Letter_2) = $Array_Letters;



echo $Letter_1 . $Letter_2;

?>



Will output: AB



Mick

tobylewis at logogriph dot com 08-May-2007 03:55


The list construct assigns elements from a numbered array starting from element zero.  It does not assign elements from associative arrays.  So



$arr = array();

$arr[1] = 'x';

list($a, $b) = $arr;

var_dump($a); //outputs NULL because there is no element [0]

var_dump($b); //outputs 'x'



and 



$arr = array('red'=>'stop','green'=>'go');

list($a, $b) = $arr;

var_dump($a); //outputs NULL

var_dump($b); //outputs NULL



If there are not enough elements in the array for the variables in the list the excess variables are assigned NULL.



If there are more elements in the array than variables in the list, the extra array elements are ignored without error.



Also the warning above about order of assignment is confusing until you get used to php arrays.  The order in which array elements are stored is the order in which elements are assigned to the array.  So even in a numbered array if you assign $may_arr[2] before you assign $my_array[0] then element [2] will be in the array before [0].  This becomes apparent when using commands like, push, shift or foreach which work with the stored order of the elements.  So the warning only applies when the variables in the list are themselves array elements which have not already been assigned to their array.

ergalvan at bitam dot com 04-May-2006 11:29


With regard to the note written by dolan at teamsapient dot com:



You must take note that list() assigns variables starting from the rightmost one (as stated in the warning). That makes $record having the value "value4" and then $var1, $var2 and $var3 take their values from the "new" $record variable.



It's clear that the behavior stated in the warning wasn't followed by version 5.0.4 (and perhaps previous versions?)

dolan at teamsapient dot com 06-Apr-2006 11:08


I noticed w/ version 5.1.2, the behavior of list() has changed (this occurred at some point between version 5.0.4 and 5.1.2).  When re-using a variable name in list() that list() is being assigned to, instead of the values being assigned all at once, the reused variable gets overwritten before all the values are read.



Here's an example:

** disclaimer: obviously this is sloppy code, but I want to point out the behavior change (in case anyone else comes across similar code) **



<?

$data = array();

$data[] = array("value1", "value2", "value3", "value4");

$data[] = array("value1", "value2", "value3", "value4");

$data[] = array("value1", "value2", "value3", "value4");

$data[] = array("value1", "value2", "value3", "value4");



foreach($data as $record)

{

    list($var1, $var2, $var3, $record) = $record;

    echo "var 1: $var1, var 2: $var2, var 3: $var3, record: $record\\n";

}

?>



OUTPUT on version 5.0.4:

var 1: value1, var 2: value2, var 3: value3, record: value4

var 1: value1, var 2: value2, var 3: value3, record: value4

var 1: value1, var 2: value2, var 3: value3, record: value4

var 1: value1, var 2: value2, var 3: value3, record: value4



OUTPUT on version 5.1.2:

var 1: v, var 2: a, var 3: l, record: value4

var 1: v, var 2: a, var 3: l, record: value4

var 1: v, var 2: a, var 3: l, record: value4

var 1: v, var 2: a, var 3: l, record: value4

mzizka at hotmail dot com 02-Jan-2006 08:49


Elements on the left-hand side that don't have a corresponding element on the right-hand side will be set to NULL. For example,



<?php

$y = 0;

list($x, $y) = array("x");

var_dump($x);

var_dump($y);

?>



Results in:



string(1) "x"

NULL

Nearsighted 25-Jul-2005 07:34


list, coupled with while, makes for a handy way to populate arrays.



while (list($repcnt[], $replnk[], $date[]) = mysql_fetch_row($seek0))

{

// insert what you want to do here.

}



PHP will automatically assign numerical values for the array because of the [] signs after the variable.



From here, you can access their row values by array numbers.



eg.



for ($i=0;$i<$rowcount;$i++)

{

echo "The title number $repcnt[$i] was written on $date[$i].";

}

webmaster at miningstocks dot com 31-May-2005 11:05


One way to use the list function with non-numerical keys is to use the array_values() function



<?php

$array = array ("value1" => "one", "value2" => "two");

list ($value1, $value2) = array_values($array);

?>

mortoray at ecircle-ag dot com 16-Feb-2005 01:29


There is no way to do reference assignment using the list function, therefore list assignment is will always be a copy assignment (which is of course not always what you want).



By example, and showing the workaround (which is to just not use list):



    function &pass_refs( &$a ) {

        return array( &$a );

    }



    $a = 1;

    list( $b ) = pass_refs( $a ); //*

    $a = 2;

    print( "$b" ); //prints 1



    $ret = pass_refs( $a );

    $b =& $ret[0];

    $a = 3;

    print( "$b" ); //prints 3



*This is where some syntax like the following would be desired:

   list( &$b ) = pass_refs( $a );

or maybe:

   list( $b ) =& pass_refs( $a );

jennevdmeer at zonnet dot nl 21-Oct-2004 08:29


This is a function simulair to that of 'list' it lists an array with the 'key' as variable name and then those variables contain the value of the key in the array.

This is a bit easier then list in my opinion since you dont have to list up all variable names and it just names them as the key.



<?php

 function lista($a) {

  foreach ($a as $k => $v) {

   $s = "global \$".$k;

   eval($s.";");

   $s = "\$".$k ." = \"". $v."\"";

   eval($s.";");

  }

 }

?>

HW 14-Aug-2004 01:08


The list() construct can be used within other list() constructs (so that it can be used to extract the elements of multidimensional arrays):

<?php

$matrix = array(array(1,2),

                array(3,4));



list(list($tl,$tr),list($bl,$br)) = $matrix;



echo "$tl $tr $bl $br";

?>

Outputs "1 2 3 4".

jeronimo at DELETE_THIS dot transartmedia dot com 28-Jan-2004 07:28


If you want to swap values between variables without using an intermediary, try using the list() and array() language constructs. For instance:



<?



// Initial values.

$biggest = 1;

$smallest = 10;



// Instead of using a temporary variable...

$temp = $biggest;

$biggest = $smallest;

$smallest = $temp;



// ...Just swap the values.

list($biggest, $smallest) = array($smallest, $biggest);



?>



This works with any number of variables; you're not limited to just two.

Cheers,

Jeronimo

rubein at earthlink dot net 28-Dec-2000 05:15


Note: If you have an array full of arrays, you can't use list() in conjunction to foreach() when traversing said array, e.g.





$someArray = array(


  array(1, "one"),


  array(2, "two"),


  array(3, "three")


);





foreach($somearray as list($num, $text)) { ... }








This, however will work





foreach($somearray as $subarray) {


  list($num, $text) = $subarray;


  ...


}

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