mysql trouble

[Anon]

I have this code

 $tables = mysql_list_tables("mydatabase");
 $i = 0;

 // Check table
 while ($i < mysql_num_rows($tables)) {
     if (mysql_tablename($tables, $i) == "$mysql_tb") {
         print("passed");
     } else {
         print("failed");
     }

     $i++;
 }

I want to do a check to see if the database i'm looking for is in there or not. Pass if it found it or fail if it didn't.

The problem is, i get failed passed failed when the second table is a match and the first and last aren't. How can I fix this so it would print passed or failed only once.

[Anon]

instead of a print statements in a while loop, you can try :

$failed = false;
while (...)
{

if (mysql_tablename($tables, $i) != "$mysql_tb") {
$failed = true;
}

//end while
}

-- while loop ends

if ($failed == false){
echo "passed";
}

-m.

[dgrinberg]

Its a good way of solving the problem Mustafa. Except that your code has a minor problem.

Assuming that you are searching for one table in many, there will be many false matches and only one true match.

Thus your code will always return a fail, unless there is only one table which is the correct table.

It would be better to do it the other way around, using $passed instead of $failed and setting $passed to true when you find the correct table. Of course initiate $passed as false.

Or better still stop the loop when the table is found, unless your searching for more than one table.

David

[Anon]

true, depending on what Mark really wants to do.

your method does make more sense.

Regards,
Mustafa.