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;
>?>
>
>
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