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:
$sql = "SHOW TABLE STATUS FROM db_name LIKE 'bookmark_unit4'";
Mike
jtjohnston w
I'm still getting "Supplied argument is not a valid MySQL result resource"
for:
while ($data = mysql_fetch_array($result)) {
mysql_free_result($result);
presumably $result
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