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Manual: list
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list

(unknown)

list --  配列と同様の形式で、複数の変数への代入を行います

説明

void list(...);

array() と同様に、この関数は実際には関数 ではなく言語の構成要素です。list()は、 単一の操作で一連の変数に値を割り当てるために使われます。

例 1. list()の例


<table>
 <tr>
  <th>従業員名</th>
  <th>給与</th>
 </tr>

<?php

$result = mysql($conn, "SELECT id, name, salary FROM employees");
while (list($id, $name, $salary) = mysql_fetch_row($result)) {
    print(" <tr>\n".
          "  <td><a href=/old?u=http%3A%2F%2Fwww.php.net%2Fmanual%2Fja%2F%5C&y=1999"info.php3?id=$id\">$name</a></td>\n".
          "  <td>$salary</td>\n".
          " </tr>\n");
}

?>

</table>

each(),array() も参照下さい。


User Contributed Notes: list

arni@linux.is
23-Aug-1999 08:53
I've been trying to put the field names of the rows I want to select from the table in a $fields variable, like this:


<pre>
$fields = 'id, name, salary';
$result = mysql($conn, "SELECT $fields FROM employees");
$fields = '$'.str_replace(", ",', $',"$fields");

while (list($fields) = mysql_fetch_row($result)) {
print("$fields");
}
</pre>



I also tried to eval() the $values, but that does not seem to work either. I spent few hours trying to find out how to do this, but...there is no solution. list() re-creates the $values variable as a one variable, not three variables as shown in earlier code.


siggma@xs4all.be
08-Oct-1999 09:09
At first I thought that list(...) creates an array of references, so that
<pre>
$a='x'; $b='y'; $c='z';
$d=list($a,$b,$c);
$d[1]='OK';
print $b;
</pre>
would give 'OK', the new value of $b put there through the reference in $d[1];
But no, list(...) is a function that requires to be assigned an array to...

evergod@dcnetsolutions.net
20-Jan-2000 09:55
I am trying to edit a certain portion of a file, which has lines individually containing something like this:
<PRE>user;name;email;score</PRE>
I need to change the "score" part of one of those lines when I come to a specific "user". I tried to use a while loop with list($key, $val) and change the $val variable to one incremented when it came to $key == 4, but it seems like it did not change the value of it. I use implode() to put the variables back together... here is my code:

<PRE>$fp = file("Agents/list.dat");
while(list($key,$val)=each($fp))
{
$array = explode(";",$val);
while(list($key2,$val2)=each($array))
if ($key2 == 3)
{
$val2 = 1;
$array2[$x+=1] = implode(";", $array);
}
}

$newfp = implode("
\n", $array2);

print("$newfp");</PRE>

gs@e-nova.com.sg
18-Feb-2000 02:32
Suppose a list of three items is returned and you are only interested in items one and three, you can skip assigning item two by
<pre>
list($one, , $three) = some_function();
</pre>

martin@hkhotmail.com
02-Aug-2000 04:53
Is that had any easy way to convert the value in assocative array to a new var. that the name is its key? Like:

/* $array = $mysql_fetch_array($handler, MYSQL_ASSOC); */

And want every val assigned to its key,

$col1=$array['col1']; /*a new var named $col1 with value $array['col1']*/


I tried use while...each and list to do, but both no work

Thx a lot.

moosh@phpfrance.com
05-Aug-2000 01:47
list($a,$b,$c)=$myArray



$myArray must be a scalar array and not an associative array.

moosh@phpfrance.com
05-Aug-2000 01:48
list($a,$b,$c)=$myArray



$myArray must be a scalar array and not an associative array.

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