array()

[dreamwest]

Im trying to get a name from an array

 

Array
(
    [0] => Array
        (
            [name] => acid jazz
            [id] => 32
        )

    [1] => Array
        (
            [name] => african
            [id] => 78
        )

    [2] => Array
        (
            [name] => alternative country
            [id] => 62
        )

    [3] => Array
        (
            [name] => alternative dance
            [id] => 19
        )

    [4] => Array
        (
            [name] => alternative metal
            [id] => 33
        )

    [5] => Array
        (
            [name] => alternative rap
            [id] => 38
        )

)

 

Basically all ive got is the id, so if array[5] has id of 38 i want to return the name, aka "alternative rap"

 

[marcus]

echo $array[5]['name'];

[dreamwest]

Close but i only have the id "38" not the array number, is there a way i can number the array so i can just call the name?

 

Array
(
    [32] => Array
        (
            [name] => acid jazz
         )

    [78] => Array
        (
            [name] => african
        )

    [62] => Array
        (
            [name] => alternative country
        )

    [19] => Array
        (
            [name] => alternative dance
        )

    [4] => Array
        (
            [name] => alternative metal
            [id] => 33
        )

    [38] => Array
        (
            [name] => alternative rap
        )

)

[PFMaBiSmAd]

In you want to be able to lookup/convert any number of values on a page, you would build a simple key/value array (see the code in this post - http://www.phpfreaks.com/forums/php-coding-help/using-$row-and-changing-values/msg1484306/#msg1484306)

 

If these are in a database and you are only getting one id/value on a page, you would do this in the query WHERE id = 38

[dreamwest]

Yeah i finally figured it out

 

$res = mysql_query("SELECT  * FROM styles_q"); 
 	while ($new = mysql_fetch_assoc( $res )){
	$tmb[$new['st_id']] = array('name' => $new['st_name']);
	}
	echo $tmb['32']['name'];