NOPE: [PHP] HELP: Re: Table comments
| From: | jtjohnston | Date: | Sat, 27 Oct 2001 20:19:44 +0000 |
| Subject: | NOPE: [PHP] HELP: Re: Table comments | ||
| References: | 1 2 3 | Groups: | php.general |
| Request: | Send a blank email to php-general+get-72586@lists.php.net to get a copy of this message | ||
I'm still getting "Supplied argument is not a valid MySQL result resource"
for:
while ($data = mysql_fetch_array($result)) {
mysql_free_result($result);
presumably $result
<?php
$myconnection = mysql_pconnect("localhost","","");
mysql_select_db("",$myconnection);
$sql = 'SHOW TABLE STATUS LIKE bookmark_unit4';
$result = mysql_query($sql);
// Only returning 1 row, but I put it in a while() loop in case the
result is empty
// Then I don't get any errors.
while ($data = mysql_fetch_array($result)) {
$table_comment = $data['Comment'];
}
mysql_free_result($result);
mysql_close($myconnection);
echo $table_comment;
?>