populating form option list with an array of an array

Question

I've been at it a few hours now and just can't seem to get what I'm looking for. Here is an example of what I'm up to:

$query1=mysql_query("Select * From table1 Where name='blah'");
while($queryvalue1=mysql_fetch_array($query1)) {
    $array1 = $queryvalue1['Column1']
    $Options="";
    $query2=mysql_query("Select * From table2 Where id In ($array1)");
    while($queryvalue2=mysql_fetch_array($query2)) {
        $var1 = $queryvalue2['Column2'];
        $var2 = $queryvalue2['Column3'];
        $Options.="<OPTION VALUE=\"$var1\">".$var2;
    }
}

If I echo $var1 and $var2 in the loop I get all the appropriate values, but when I wrap them in the html tag and echo options it only gives me the first value. I can use the $Options code fine if it takes place in the first array loop, it's just when I use it in the loop for the array of the array.

I'm at my wits end. Does anyone know why this problem is? Does anyone know of a way to fix it? Any help would be much appreciated!

Phil · Accepted Answer · 2011-03-17 02:41:46Z

1

Apart from the missing semi-colon on line #3, the problem is you're clearing the $Options string inside the outer loop.

Try something like this

$options = array();
$query1=mysql_query("Select * From table1 Where name='blah'");
while($queryvalue1=mysql_fetch_array($query1)) {
    $array1 = $queryvalue1['Column1'];
    $query2=mysql_query("Select * From table2 Where id In ($array1)");
    while($queryvalue2=mysql_fetch_array($query2)) {
        $var1 = $queryvalue2['Column2'];
        $var2 = $queryvalue2['Column3'];
        $options[] = sprintf('<option value="%s">%s</option>',
                             htmlspecialchars($var1),
                             htmlspecialchars($var2));
    }
}
$options = implode(PHP_EOL, $options);

answered Mar 17, 2011 at 2:41

Phil

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1 Comment

Oldtimer Over a year ago

Took me a while to discover anyone replied, but boy am I glad! Thank you so much; the script works perfectly now! Just, wow, can't thank you enough -- I was COMPLETELY lost.

mario · Accepted Answer · 2011-03-17 02:41:46Z

1

The problem is probably that you have this in the while loop:

$Options="";

This will reset the collected <option> fields and is why you only see the last entry from the second loop in the page output.

answered Mar 17, 2011 at 2:41

mario

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