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

COOIS - Saving Object Overview WIP Status layouts
COOIS - Would like to be able to save the layout of an Object Overview - WIP Status report. There i

IS Retail & Manufacturing
Hi

Can IS Retail and Manufacturing activities be done in same instance ie. in same client

Need help: how to catch acess of undefined class properties
Hello. I am learning OO with PHP and have hit a problem.
Some code runs as perfectly valid code,

Putting double spaces instead of single spaces
Im looking at trying to replace all single spacing between fields with double spacing

At pres

Echo Tweaking help!
HI. I would like to have the output of the entered variables repeat forever, but it's stopping at th

Hyperlink is adding an extra gap to variable
Hi, here's my problem..

I have a php generated page with a hyperlink which opens in a new win

update the selected existing records of database
Hi frndz,
I am new to php and get stuck....

Edit.php > Update.php > Updatea

Help with PHP and checkboxes
I am posting this for a friend of mine, i wrote a small script for her to process her form data to a

constructor ?
i have written this program and made 3 constructors in the class and i want to call them in the main

session checking in page load
hai all I have a web site is www.Mryas.com in this my login page is Page1.aspx its co