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Posted by Jerry Stuckle on 01/29/08 22:46
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.
--
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Jerry Stuckle
JDS Computer Training Corp.
jstucklex@attglobal.net
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