Re: running query through while loop returns same variables
| From: | PHPGalaxy.com | Date: | Wed, 10 Oct 2001 20:10:24 +0000 |
| Subject: | Re: running query through while loop returns same variables | ||
| References: | 1 | Groups: | php.general |
| Request: | Send a blank email to php-general+get-70730@lists.php.net to get a copy of this message | ||
The mysql_fetch_row() runction has to go inside the loop. Each time it's
accessed (once during each loop) it grabs the next result row =) Since it's
outside the loop (just before it starts), it only grabbed the first result row,
and keeps those values for the whole loop =)
Move the mysql_fetch_row() line of code down 2 line (just under the "while"
function), and it should work =)
Tom Beidler wrote:
> I'm trying to run a query through a while loop and keep getting the same
> variables. I've tried flushing out using mysql_free_result without much
> success. Here's the code to give you an idea of what I'm trying to do.
>
> while (list($key, $value) = each($year_array)) {
> $js_firstgroup .= "group[$counter1][$counter2]=new Option(\"Now Select a
> Make\",\" \");\n";
>
> $make_query = "SELECT make FROM application WHERE year_in >= $value AND
> year_out <= $value ORDER BY make";
> $make_result = mysql_query ($make_query)
> or die ("Cannot get make.");
> $make_row = mysql_fetch_array ($make_result);
> while ($make_row = mysql_fetch_array ($make_result)) {
> $make = $make_row["make"];
> $counter2++;
> $counter3++;
> $js_firstgroup .= "group[$counter1][$counter2]=new
> Option(\"$make\",\"$counter3\");\n";
> }
> $js_firstgroup .= "\n";
> $counter1++;
> $counter2 = 0;
> mysql_free_result ($make_result);
> }
>
> It apears that I keep looping through the first query array everytime.
>
> Any help appreciated.
>
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- Tim
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