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list

(PHP 4, PHP 5)

list — 配列と同様の形式で、複数の変数への代入を行う

説明

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

array() と同様に、この関数は実際には関数ではなく言語の構成要素です。 list() は、単一の操作で一連の変数に値を代入するために使われます。

パラメータ

varname: 変数。

返り値

値を返しません。

例

例1 list() の例


<?php

$info = array('コーヒー', '茶色', 'カフェイン');

// すべての変数の取得
list($drink, $color, $power) = $info;
echo "$drink の色は $color で、$power が含まれています。\n";

// 一部の変数の取得
list($drink, , $power) = $info;
echo "$drink には $power が含まれています。\n";

// 三番目のみの取得
list( , , $power) = $info;
echo "$power 欲しい!\n";

// list() は文字列では動作しません
list($bar) = "abcde";
var_dump($bar); // NULL
?>

例2 list() の使用法の例


<table>
 <tr>
  <th>社員氏名</th>
  <th>給与</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>

例3 ネストした list() の使用法


<?php

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

var_dump($a, $b, $c);

?>

int(1)
int(2)
int(3)

例4 配列の添字を使用した list() の例


<?php

$info = array('coffee', 'brown', 'caffeine');

list($a[0], $a[1], $a[2]) = $info;

var_dump($a);

?>

次のような出力になります(list() の文法に書かれた順番と、要素の順番の違いに注意):

array(3) {
  [2]=>
  string(8) "caffeine"
  [1]=>
  string(5) "brown"
  [0]=>
  string(6) "coffee"
}

注意

警告

list()は、最も右のパラメータから値を代入します。プレーンな変数を使用している場合には、このことを気にする必要はありません。しかし、添字配列を使用している場合には、配列の添字の順番が list() に書いたものと同じく左から右となることを通常は期待しますが、そうはなりません。この配列の添字は逆の順番となります。

注意: list() は数値添字の配列のみを生成します。添字は 0 から始まります。

参考

natcasesort

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ksort

Last updated: Fri, 06 Jun 2008

add a note User Contributed Notes
list

tristan in oregon
09-Apr-2008 11:44


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
07-Feb-2008 12:12


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
20-Jan-2008 10:51


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
05-Nov-2007 05:36


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
19-Sep-2007 02:50


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
09-Aug-2007 04:08


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 07:55


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
05-May-2006 03:29


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
07-Apr-2006 03:08


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
03-Jan-2006 01:49


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 11:34


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
01-Jun-2005 03:05


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 06:29


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
22-Oct-2004 12:29


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
15-Aug-2004 05:08


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
29-Jan-2004 12:28


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
29-Dec-2000 10:15


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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Last updated: Fri, 06 Jun 2008