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(PHP 4, PHP 5)
list — Weist Variablen zu, als wären sie ein Array
Wie array() ist auch dies keine wirkliche Funktion, sondern ein Sprachkonstrukt. list() wird verwendet, um eine Liste von Variablen innerhalb einer Operation zuzuweisen.
Hinweis:
list() funktioniert nur bei numerischen Arrays und basiert auf der Annahme, dass die numerischen Indizes bei 0 beginnen.
Beispiel #1 list()
<?php
$info = array('Kaffee', 'braun', 'Koffein');
// Auflisten aller Variablen
list($drink, $color, $power) = $info;
echo "$drink ist $color und $power macht es zu etwas besonderem.\n";
// Ein paar davon auflisten
list($drink, , $power) = $info;
echo "$drink hat $power.\n";
// Oder nur die dritte ausgeben
list( , , $power) = $info;
echo "Ich brauche $power!\n";
?>
Beispiel #2 list()
<table>
<tr>
<th>Mitarbeitername</th>
<th>Gehalt</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>
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
<?php
$info = array('coffee', 'brown', 'caffeine');
list($a[0], $a[1], $a[2]) = $info;
var_dump($a);
?>
Die Ausgabe (Beachten Sie die Reihenfolge der Elemente, verglichen mit der in list() eingetragenen Reihenfolge):
array(3) {
[2]=>
string(8) "caffeine"
[1]=>
string(5) "brown"
[0]=>
string(6) "coffee"
}
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.
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, $var2;
?>
A simple way to swap variables (correction of a note of mario dot mueller dot work at gmail dot com below):
<?php
list($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"
?>
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;
}
?>
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);
?>
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.
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
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
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
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.
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?)
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
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($y);
?>
Results in:
string(1) "x"
NULL
list, coupled with while, makes for a handy way to populate arrays.
while (list($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].";
}
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);
?>
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( $a );
This is a function simulair to that of 'list' it lists an array with the 'key' as variable name and then those variables contain the value of the key in the array.
This is a bit easier then list in my opinion since you dont have to list up all variable names and it just names them as the key.
<?php
function lista($a) {
foreach ($a as $k => $v) {
$s = "global \$".$k;
eval($s.";");
$s = "\$".$k ." = \"". $v."\"";
eval($s.";");
}
}
?>
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".
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
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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