PHP using IF to display error

Posted on 16th Feb 2014 07:03 pm by admin

i have a MySQL query and i want to display 1 thing only if the number of affected rows is >=1 and if not then display the error, here is what i have so far and nothing is being displayed ...

Code: if (mysql_num_rows() >= "1")
{
echo "Update OK!<br />";
}
else
{
die("Update failed: " . mysql_error());
}

Your Answer:

Login to answer

Other forums

Table control is disappearing from Screen
Hi Experts,

I have a screen with multiple table control(TC). Each table control has two c

Displaying Multidimensional Arrays...about to put my head in a vice, PLEASE HELP
I am fairly new to PHP and MySQL and I have a problem which I have been trying to resolve for days.

need help in update query
i create a form for update. there are 8 columns in my mysql table. on my main page all the data is r

Help with search.php
Hello All

I need the search engine on a site I am building at the moment to display the searc

Pls help with PHP
Pls can anyone direct me on how to create a user account for a customer. For example i want know if

please help
HTML Code:

Code: <span id="ctl00"><span>

Show specific day of the week
I have been racking my brain trying to get this function to work properly and it is doing so for eve

Problem with DB connection
Hello there! I'm new to this forum and I'm new to PHP coding also. I wrote something that doesn't ma

problems with php variables in mysql query
i can't seem to get the following query to work.

select $q1c from $vote_rate where id = $re

Please help - should be a simple fix.. driving me nuts
Everything seemed to be working fine. I have a table, it alphabetically lists a bunch of cities and