natcasesort

ksort

Last updated: Sun, 25 Nov 2007

list

(PHP 4, PHP 5)

list — Wijs variabelen toe alsof zij een array zijn

Beschrijving

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

Net als array() is dit niet echt een functie, maar een language construct. list() wordt gebruikt voor het toewijzen van een lijst van variabelen in een enkele handeling.

Note: list() werkt alleen op numerieke arrays en neemt aan dat numerieke indices beginnen bij 0.

Example#1 list() voorbeelden


<?php

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

// Wijs alle variabelen toe
list($drink, $color, $power) = $info;
echo "$drink is $color and $power makes it special.\n";

// Wijs enkele variabelen toe
list($drink, , $power) = $info;
echo "$drink has $power.\n";

// Wijs alleen de derde variabele toe
list( , , $power) = $info;
echo "I need $power!\n";

?>

Example#2 Een gebruiksvoorbeeld van list()


<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() wijst de waarden toe beginnende bij de meest rechtse parameter. Wanneer je normale variabelen gebruikt, hoef je hier geen rekening mee te houden. Echter, wanneer je arrays gebruikt met indices, dan zou je gewoonlijk verwachten dat de volgorde van de indices in de array gelijk is aan de volgorde waarin je het beschreef in list() van links naar rechts, wat niet het geval is. De toewijzing gebeurt in omgekeerde volgorde.

Example#3 list() gebruiken met array indices


<?php

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

list($a[0], $a[1], $a[2]) = $info;

var_dump($a);

?>

Geeft het volgende resultaat (let op de volgorde van de elementen in vergelijking tot de volgorde waarin zij waren geschreven in de list() syntax):

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

Zie ook each(), array() en extract().

natcasesort

ksort

Last updated: Sun, 25 Nov 2007

add a note User Contributed Notes
list

claude dot pache at gmail dot com
13-May-2009 10:26


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"

?>

sasha
29-Apr-2009 05:08


Note that list assigns values from right to left and it does so as soon as it reached argument.



$var = array ("test" ,"blah") ;

list ($a,$var) = $var;

echo $a ; // prints "b"

echo $var ; // prints "blah"



In this case $var has been assigned new value "blah" before value was assigned to $a. 



$a was assigned value "b" (I assume string "blah" was converted to an array and $a was assigned first element of that array).

mario dot mueller dot work at gmail dot com
12-Feb-2009 02:08


Some simple method to exchange values with two variables:



<?php

$var1 = 'John';

$var2 = 'Doe';



list($var2,$var1) = list($var1,$var2);

?>

tristan in oregon
09-Apr-2008 02:44


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 05:50


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 07:08


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 10:55


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 06:29


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 06:08


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 02:34


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 06:05


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 03:29


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 08:08


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;


  ...


}

add a note

natcasesort

ksort

Last updated: Sun, 25 Nov 2007