Re: NOPE: [PHP] HELP: Re: Table comments

That's because you're not selecting a database. You need to either put the database name in mysql_select_db, or change the query to: SHOW TABLE STATUS FROM databasename LIKE 'table_name'; so change the line to: $sql = "SHOW TABLE STATUS FROM db_name LIKE 'bookmark_unit4'"; Mike jtjohnston wrote:

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; ?>