|  | Posted by Rik Wasmus on 02/01/08 16:07 
On Wed, 30 Jan 2008 05:02:31 +0100, Sudhakar <sudhakararaog@gmail.com>  =
 wrote:
 > my question is about "SHOW TABLE STATUS LIKE 'tablename'";
 
 Which is MySQL, crossposted, f'upped to there.
 
 > 1.
 > echo "The number of allocated but unused bytes: ".$array['Data_free'];=
 
 > =3D This does not display any number at all.
 
 http://bugs.mysql.com/bug.php?id=3D32440
 
 > 2.
 > echo "The maximum length of the data file. This is the total number of=
 
 > bytes of data that can be stored in the table :  " .
 > $array['Max_data_length'];
 >
 > displays 4294967295 which comes to 3.99 GB
 >
 > Does this mean that the Entire capacity for the MySql Database that is=
 
 > installed on the unix server including all the
 > a) Tables and from all the Databases is 3.99 GB
 >
 > Because i have used this SHOW TABLE STATUS LIKE 'tablename' for all
 > the tables i have have in 3 databases that i have, in each case the
 > value displayed for
 > echo $array['Max_data_length']; is 4294967295 which is 3.99 GB so i am=
 
 > assuming the total capacity for "MySql Database VERSION 3.23.58" that
 > is installed on the unix server can hold a maximum of 3.99 GB of data
 > each time data is inserted into the various tables which are in
 > different databases
 
 I'll leave this for more MySQL savvy users at comp.databases.mysql..
 
 > 3.
 > echo $array['Data_length'] =3D The value displayed is ex=3D 36872 for
 > this, does it mean that 36872 number of bytes have been consumed by a
 > particular table at a particular time, everytime an insert statement
 > had been executed
 
 No idea what this means.
 
 > 4.
 > Also i have 2 databases and each have couple of tables so i guess i
 > will have to use the following statement 2 times
 >
 > mysql_select_db("database1", $conn);
 > mysql_select_db("database2", $conn);
 
 Or just mysql_query('USE database2'); will do the trick.
 -- =
 
 Rik Wasmus
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