Because you are wiping out your $sponsor variable in that commented-out
code.
On Sat, 28 Dec 2002, James Brennan wrote:
> I am getting a "Warning: mysql_fetch_row(): supplied argument is not a valid
> MySQL result resource" error only when the commented-out code bellow
> (starting from $id = bb_sponsor...) is added. If I omit the code the
> resource is recognized as valid. Can anyone explain?
>
> Thanks,
> loop
>
> P.S. - I know my conditions for comparing values are far from accurate, but
> they work for this particular purpose.
>
> /* variables for sponsor table insert */
> $insert = '';
> $insert_pre = "INSERT INTO `bb_sponsor` ( `id` , `name` , `url` ) VALUES
> (";
> $insert_post = ");\n";
>
> /* if inputed sponsor name exists, define it's index and url value.*/
> while ($row = mysql_fetch_row($sponsor)) {
> foreach ($bb_sponsor as $key=>$value) {
> if (levenshtein($bb_sponsor[$key]['name'], $row[1]) <=3 or
> stristr(($bb_sponsor[$key]['name']), $row[1])) {
> $bb_sponsor[$key]['id'] = $row[0];
> $bb_sponsor[$key]['url'] = $row[2];
> }
> /*
> $id = $bb_sponsor[$key]['id'];
> $name = $bb_sponsor[$key]['name'];
> $sponsor = $bb_sponsor[$key]['url'];
> $insert = $insert . $insert_pre . "'', '$name', '$sponsor'" .
> $insert_post; */
> }
> }
>
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