Re: how to handle exception in php

On 02/05/10 15:26, Bavithra R wrote:

hi friends I am doing a simple student mark details project. for calculating rank I need to compare the total marks one by one. To do so i use *for loop.* so atlast while reaching the end of the table it shows the following warning. * Warning*: mysql_result() [function.mysql-result<http://127.0.0.1/pro/function.mysql-result>]: Unable to jump to row 18 on MySQL result index on line *50* It is because it has no next value to compare.How to use try..catch exception handling in this.?Or any other suggestion. Can anybody help me

This isn't an exception, so a try catch won't work. (Exceptions are fatal and would say something like "uncaught exception ..."). This is a warning about trying to read something that doesn't exist from mysql. I'd do something like $query = "select blah"; $result = mysql_query($query); while ($row = mysql_fetch_assoc($result)) { echo "do stuff here"; } -- Postgresql & php tutorials http://www.designmagick.com/