ksort
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.
list
(PHP 4, PHP 5)
list — Değişkenlere bir dizi gibi atama yapar
Açıklama
array() gibi bu da bir işlev değil bir dil oluşumudur. list() işlevi tek deyimde birden fazla değişkene atama yapmak için kullanılır.
Değiştirgeler
-
değişken -
Bir değişken.
Dönen Değerler
Atanan dizi döner.
Örnekler
Örnek 1 - list() örnekleri
<?php
$veri = array('kahve', 'uyku', 'kafein');
// Değişkenlerin tamamını listeleyelim
list($içecek, $neyi, $güç) = $veri;
echo "$içecek $neyi kaçırır, çünkü içinde $güç vardır.\n";
// Bir kısmını listeleyelim
list($içecek, , $güç) = $veri;
echo "$içecek $güç içerir.\n";
// Sadece üçüncü kalsın
list( , , $güç) = $veri;
echo "Bana $güç lazım!\n";
// list() dizgelerle çalışmaz
list($bar) = "abcde";
var_dump($bar); // NULL
?>
Örnek 2 - list() kullanım örneği
<table>
<tr>
<th>Çalışanın adı</th>
<th>Ücreti</th>
</tr>
<?php
$sonuç = mysql_query("SELECT id, ad, ucret FROM personel", $conn);
while (list($id, $ad, $ücret) = mysql_fetch_row($sonuç)) {
echo " <tr>\n" .
" <td><a href=\"info.php?id=$id\">$ad</a></td>\n" .
" <td>$ücret</td>\n" .
" </tr>\n";
}
?>
</table>
Örnek 3 - İç içe list() kullanımı
<?php
list($a, list($b, $c)) = array(1, array(2, 3));
var_dump($a, $b, $c);
?>
int(1) int(2) int(3)
Örnek 4 - Dizi indisleri ile list() kullanımı
<?php
$veri = array('kahve', 'uyku', 'kafein');
list($a[0], $a[1], $a[2]) = $veri;
var_dump($a);
?>
Aşağıdaki çıktıyı verir (dikkatı ederseniz, eleman sırası list() ile belirtilen sıra ile aynıdır):
array(3) {
[2]=>
string(6) "kafein"
[1]=>
string(4) "uyku"
[0]=>
string(5) "kahve"
}
Notlar
Uyarı
list() değerleri en sağdaki değiştirgeden itibaren atar. Düz değişken kullanıyorsanız bundan dolayı endişelenmeniz gerekmez. Fakat indisli diziler kullanıyorsanız, dizi indislerinin, list()'teki yazılış sırasına göre soldan sağa sıralanacağını düşünürsünüz. Fakat atama sağdan sola doğru yapılır.
Bilginize:
list() sadece sayısal indisli dizilerle çalışır ve indislerin 0'dan başladığı varsayılır.
add a note
User Contributed Notes
list
dan dot lugg at gmail dot com
14-Jan-2012 05:43
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;
...
}