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Array Funktionen




(PHP 4, PHP 5)

listWeist Variablen zu, als wären sie ein Array


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

Wie array() ist auch dies keine wirkliche Funktion, sondern ein Sprachkonstrukt. list() wird verwendet, um eine Liste von Variablen innerhalb einer Operation zuzuweisen.


list() funktioniert nur bei numerischen Arrays und basiert auf der Annahme, dass die numerischen Indizes bei 0 beginnen.

Beispiel #1 list()


= array('Kaffee''braun''Koffein');

// Auflisten aller Variablen
list($drink$color$power) = $info;
"$drink ist $color und $power macht es zu etwas besonderem.\n";

// Ein paar davon auflisten
list($drink, , $power) = $info;
"$drink hat $power.\n";

// Oder nur die dritte ausgeben
list( , , $power) = $info;
"Ich brauche $power!\n";


Beispiel #2 list()



mysql_query ("SELECT id, name, salary FROM employees",$conn);
while (list (
$id$name$salary) = mysql_fetch_row ($result)) {
" <tr>\n".
"  <td><a href=\"info.php?id=$id\">$name</a></td>\n".
"  <td>$salary</td>\n".
" </tr>\n";




list() weist die Werte von rechts beginnend zu. Wenn Sie einfache Variablen benutzen, brauchen Sie sich nicht darum zu kümmern. Wenn Sie jedoch Arrays mit Indizes verwenden, erwarten Sie gewöhnlich die Reihenfolge der Indizes in dem Array genau so, wie Sie sie in list() geschrieben haben (von links nach rechts), was jedoch nicht der Fall ist. Es wird in der umgekehrten Reihenfolge zugewiesen.

Beispiel #3 list() mit Array Indizes verwenden


= array('coffee''brown''caffeine');

$a[0], $a[1], $a[2]) = $info;



Die Ausgabe (Beachten Sie die Reihenfolge der Elemente, verglichen mit der in list() eingetragenen Reihenfolge):

array(3) {
  string(8) "caffeine"
  string(5) "brown"
  string(6) "coffee"

Siehe auch each(), array() und extract().

20 BenutzerBeiträge:
- Beiträge aktualisieren...
develop at dieploegers dot de
31.03.2010 13:37
Remember, that list starts from index 0. You can skip an index if you just leave the column blank like this:

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:

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

This will fail.
12.03.2010 18:29
Quick little function that is similar to list but for objects.

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;

obj {public $var = "test"; public $vars = "test2"; function obj() {}}
listObj(new obj, &$var, &$var2);
$var, $var2;
claude dot pache at gmail dot com
13.05.2009 12:26
A simple way to swap variables (correction of a note of mario dot mueller dot work at gmail dot com below):
list($var1, $var2) = array($var2, $var1); // swaps the values of $var1 and $var2
Note that this is not equivalent to:
= $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:
= array ("test" ,"blah");
list (
$a,$var) = $var + array();
$a ; // prints "test", not "b"
echo $var ; // prints "blah"
tristan in oregon
9.04.2008 4:44
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).

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

//  $foo = 10;
$bar = 20;
$baz = 30;

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


Should print

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.

function my_array_intersect_key ($assoc, $keys)
$intersection = array();
    foreach (
$assoc as $key => $val)
        if (
in_array($key, $keys))
$intersection[$key] = $val;

kevin at vanzonneveld dot net
6.02.2008 16:12
Another way to do it associative (if your array isn't numeric), is to just use array_values like this:

= array();
$os["main"] = "Linux";
$os["distro"] = "Ubuntu";
$os["version"] = "7.10";

$main, $distro, $version) = array_values($os);
danieljames3 at g mail
20.01.2008 2:51
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
4.11.2007 21:36
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.

= array('foo'=>'bar');
$t = each($array);
$a,$b,$c,$d) = $t;

    [1] => bar
    [value] => bar
    [0] => foo
    [key] => foo
string(3) "foo"
string(3) "bar"
tenz699 at hotmail dot com
18.09.2007 19:50
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:

= array ("1" => "one", "2" => "two","3"=>"three");
$keys,$values) = each($tenzin))
$keys." ".$values."<br>");

gives O/P
1 one 
2 two
3 three

mick at wireframe dot com
8.08.2007 21:08
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:

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

$Letter_1, $Letter_2) = $Array_Letters;

$Letter_1 . $Letter_2;

Will output: AB

tobylewis at logogriph dot com
8.05.2007 12:55
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'


$arr = array('red'=>'stop','green'=>'go');
list($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
4.05.2006 20:29
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
6.04.2006 20:08
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) **

= 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");

$data as $record)
$var1, $var2, $var3, $record) = $record;
"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
3.01.2006 5:49
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,

= 0;
$x, $y) = array("x");

Results in:

string(1) "x"
25.07.2005 16:34
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.


for ($i=0;$i<$rowcount;$i++)
echo "The title number $repcnt[$i] was written on $date[$i].";
webmaster at miningstocks dot com
1.06.2005 8:05
One way to use the list function with non-numerical keys is to use the array_values() function

= array ("value1" => "one", "value2" => "two");
list (
$value1, $value2) = array_values($array);
mortoray at ecircle-ag dot com
16.02.2005 10:29
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
21.10.2004 17:29
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.

function lista($a) {
  foreach (
$a as $k => $v) {
$s = "global \$".$k;
$s = "\$".$k ." = \"". $v."\"";
14.08.2004 22:08
The list() construct can be used within other list() constructs (so that it can be used to extract the elements of multidimensional arrays):
= array(array(1,2),

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

"$tl $tr $bl $br";
Outputs "1 2 3 4".
jeronimo at DELETE_THIS dot transartmedia dot com
29.01.2004 4:28
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.
rubein at earthlink dot net
29.12.2000 2:15
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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