Posted by Sudhakar on 01/29/08 22:11

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.

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