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up vote 0 down vote favorite

I have a code like this.

mysql_connect("localhost", "root", "") or die(mysql_error());
mysql_select_db("joomla") or die(mysql_error());

$result = mysql_query("SELECT id,name FROM jos_samples") or die(mysql_error());  



while($row = mysql_fetch_array( $result )) {
print_r($row);
}

And my print array is like this.

Array ( [0] => 3 [id] => 3 [1] => User1 [name] => User1 )
Array ( [0] => 4 [id] => 4 [1] => User2 [name] => User2 ) 
Array ( [0] => 5 [id] => 5 [1] => User3 [name] => User3 ) 
Array ( [0] => 6 [id] => 6 [1] => User4 [name] => User4 )

I want this array something like this

Array ( [3] => User1 [4] => User2 [5] => User3 [6] => User4 )

I am pretty not good in this .Could any1 help with me with a piece of code.

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5 Answers

up vote 3 down vote accepted

You can reshape your array:

/* ... */

$rows = array();
while($row = mysql_fetch_array($result)) {
    $rows[$row['id']] = $row['name'];
}

$rows is your result.

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Thanks a lot man.. This worked. Thank you again.. Out of words –  hjaffer2001 May 27 '11 at 7:25
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up vote 3 down vote

If I get this right, you want:

while($row = mysql_fetch_array( $result, MYSQL_ASSOC )) {

http://php.net/manual/en/function.mysql-fetch-array.php

(You're currently using the default, which is MYSQL_BOTH.

And then you need to rework your array:

$rows = array();
while($row = mysql_fetch_array( $result, MYSQL_ASSOC )) {
  $rows[$row['id']] = $row['name'];
}
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+1 for explaining the use of the result_type parameter. –  Mike May 27 '11 at 7:46
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up vote 2 down vote

Try this, it builds an array containing the usernames as you want it.

$aUsers = array();

// Use _assoc, it filters numeric key's.
while($aRow = mysql_fetch_assoc( $result )) {
    $aUsers[$aRow['id']] = $aRow['name'];
}

echo '<pre>';
print_r($aUsers);
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I think the asker wants the mysql field id to represent the array's key. –  proximus May 27 '11 at 7:21
    
@proximus, that's what this does –  Dormouse May 27 '11 at 7:23
    
@proximus I edited it the second after I posted it :) you were correct! –  Wesley van Opdorp May 27 '11 at 7:25
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up vote 1 down vote

Then dont ask for the ID to be returned in your query, the reason you get 2 items is because ID is in there, secondly if you dont want the name of the field, change the request for the fetch array to

mysql_fetch_array ($result, MYSQL_NUM)

this only gets numbers not the name of the field

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If you just want to print the id and name, then you can do a foreach rather than print_r, you control exactly how and what you see then. –  BugFinder May 27 '11 at 7:18
    
Thanks a lot BugFinder –  hjaffer2001 May 27 '11 at 8:45
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up vote 1 down vote 
while($row = mysql_fetch_array( $result )) {
$array[$row['id']] = $row['name'];
}
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