数组 函数
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list

(PHP 4, PHP 5)

list — 把数组中的值赋给一些变量

说明

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

像 array() 一样，这不是真正的函数，而是语言结构。list() 用一步操作给一组变量进行赋值。

Note:

list() 仅能用于数字索引的数组并假定数字索引从 0 开始。

Example #1 list() 例子

<?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";

?>

Example #2 使用 list() 的例子

<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>

Warning

list() 从最右边一个参数开始赋值。如果你用单纯的变量，不用担心这一点。但是如果你用了具有索引的数组，通常你期望得到的结果和在 list() 中写的一样是从左到右的，但实际上不是。是以相反顺序赋值的。

Example #3 在 list() 中使用数组索引

<?php
$info = array('coffee', 'brown', 'caffeine');
list($a[0], $a[1], $a[2]) = $info;
var_dump($a);
?>

产生如下输出（注意单元顺序和 list() 语法中所写的顺序的比较）： 

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

参见 each()，array() 和 extract()。

数组 函数
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PHP手册 - N: 把数组中的值赋给一些变量

用户评论:

dan dot lugg at gmail dot com (15-Jan-2012 01:43)

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 11:37)

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 08:49)

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 (05-Apr-2011 12:02)

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 11:37)

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 05:29)

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 11:26)

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 (09-Apr-2008 03:44)

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 03:12)

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 01:51)

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 08:36)

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 06:50)

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 08:08)

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 11:55)

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 07:29)

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 07:08)

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 04:49)

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 03:34)

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 07:05)

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 09:29)

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 04:29)

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 09:08)

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 03:28)

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 01:15)

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;
  ...
}