SHOW TABLE STATUS

hi

I am using MySQL - 4.1.22 when i use the following sql query
$result = mysql_query("SHOW tablename STATUS FROM databasename;");

i have also tried = $result = mysql_query("SHOW tablename STATUS FROM
databasename");

i get the following error message

================================================== ==
1064 Error Message : You have an error in your SQL syntax; check the
manual that corresponds to your MySQL server version for the right
syntax to use near 'tablename STATUS FROM databasename' at line 1
================================================== ==

also with the following code
================================================== =
while($array = mysql_fetch_array($result))
{
$total = $array[Data_length]+$array[Index_length];

echo "
Table: ".$array[Name]."<br />
Data Size: ".$array[Data_length]."<br />
Index Size: ".$array[Index_length]."<br />
Total Size: ".$total."<br />
Total Rows: ".$array[Rows]."<br />
Average Size Per Row: ".$array[Avg_row_length]."<br /><br />";
}
=================================================

i get the following error = " mysql_fetch_array(): supplied argument
is not a valid MySQL result resource "

please advice how to fix
1) SHOW TABLE query for version 4.1.22 and also is there a difference
in the SHOW TABLE query for MySQL - 3.23.58
2) while($array = mysql_fetch_array($result))

thanks a lot.

Sudhakar wrote:
> hi
>
> I am using MySQL - 4.1.22 when i use the following sql query
> $result = mysql_query("SHOW tablename STATUS FROM databasename;");
>
> i have also tried = $result = mysql_query("SHOW tablename STATUS FROM
> databasename");
>
> i get the following error message
>
> ================================================== ==
> 1064 Error Message : You have an error in your SQL syntax; check the
> manual that corresponds to your MySQL server version for the right
> syntax to use near 'tablename STATUS FROM databasename' at line 1
> ================================================== ==
>
> also with the following code
> ================================================== =
> while($array = mysql_fetch_array($result))
> {
> $total = $array[Data_length]+$array[Index_length];
>
> echo "
> Table: ".$array[Name]."<br />
> Data Size: ".$array[Data_length]."<br />
> Index Size: ".$array[Index_length]."<br />
> Total Size: ".$total."<br />
> Total Rows: ".$array[Rows]."<br />
> Average Size Per Row: ".$array[Avg_row_length]."<br /><br />";
> }
> =================================================
>
> i get the following error = " mysql_fetch_array(): supplied argument
> is not a valid MySQL result resource "
>
> please advice how to fix
> 1) SHOW TABLE query for version 4.1.22 and also is there a difference
> in the SHOW TABLE query for MySQL - 3.23.58
> 2) while($array = mysql_fetch_array($result))
>
> thanks a lot.
>

This is a MySQL error. Try comp.database.mysql.

--
==================
Remove the "x" from my email address
Jerry Stuckle
JDS Computer Training Corp.
jstucklex@attglobal.net
==================

Jerry Stuckle wrote:
> Sudhakar wrote:
>> hi
>>
>> I am using MySQL - 4.1.22 when i use the following sql query
>> $result = mysql_query("SHOW tablename STATUS FROM databasename;");
>>
>> i have also tried = $result = mysql_query("SHOW tablename STATUS FROM
>> databasename");
>>
>> i get the following error message
>>
>> ================================================== ==
>> 1064 Error Message : You have an error in your SQL syntax; check the
>> manual that corresponds to your MySQL server version for the right
>> syntax to use near 'tablename STATUS FROM databasename' at line 1
>> ================================================== ==
>>
>> also with the following code
>> ================================================== =
>> while($array = mysql_fetch_array($result))
>> {
>> $total = $array[Data_length]+$array[Index_length];
>>
>> echo "
>> Table: ".$array[Name]."<br />
>> Data Size: ".$array[Data_length]."<br />
>> Index Size: ".$array[Index_length]."<br />
>> Total Size: ".$total."<br />
>> Total Rows: ".$array[Rows]."<br />
>> Average Size Per Row: ".$array[Avg_row_length]."<br /><br />";
>> }
>> =================================================
>>
>> i get the following error = " mysql_fetch_array(): supplied argument
>> is not a valid MySQL result resource "
>>
>> please advice how to fix
>> 1) SHOW TABLE query for version 4.1.22 and also is there a difference
>> in the SHOW TABLE query for MySQL - 3.23.58
>> 2) while($array = mysql_fetch_array($result))
>>
>> thanks a lot.
>>
>
> This is a MySQL error. Try comp.database.mysql.
>

Oh sod off jerry.

The syntax is SHOW TABLE STATUS FROM 'database' (LIKE..)

That shows all the tables in the databse with useless info on them

If you want more on a specific table, then SHOW FIELDS FROM table (FROM
database)

etc works. As do SHOW INDEX, SHOW COLUMNS etc etc.

Get yourself an MySQL syntax book. When in doubt use the command line
tool to check syntax.