In message <fno8c9+ppa6@xxxxxxxxxxx>, Sudhakar <finals27@xxxxxxxxx>
writes
>hi
>
>I am using MySQL - 4.1.22 when i use the following sql query
>$result = mysql_query("SHOW tablename STATUS FROM databasename;");
I am not clear what you are trying to get here, perhaps it should be
SHOW TABLE STATUS FROM $databasename
unless your database is called databasename, in which case it should be
SHOW TABLE STATUS FROM databasename
You can't use a tablename there.
If your table is called tablename, then
SHOW TABLE STATUS FROM databasename LIKE tablename
...but if those are the database and table names that you are using,
then you shouldn't, it's bad practice, and it will cause confusion when
the code needs changing.
>
>also with the following code
>===================================================
>while($array = mysql_fetch_array($result))
>i get the following error = " mysql_fetch_array(): supplied argument
>is not a valid MySQL result resource "
Whatever you have in $result is not the result of a query. You probably
need
$result=mysql_query( $qry );
before your code.
>is there a
>difference in the SHOW TABLE query for MySQL - 3.23.58
Slight differences, in collation and checksum, which probably don't
affect your query.
--
Pete Clark
Sunny Andalucia
http://hotcosta.com/Andalucia.Spain
