Re: how to handle exception in php

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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";
}

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