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

captcha error
I have been trying to implement a captcha in php...

here is the code..
Code: (php) [Select

SQl num_rows problem
when i try to count rows from an SQL select i get an warning

Code: [Select]$countviews = mysq

A href problem under php
Hello...

I tried to explain the issue in an earlier post.. but was not clear enough....

Do While statement
hi guys,

This may sound trivial but im new to php and as part of an assignmenti have to const

generating all possible random letters
hi'

how can i randomize the letters a,b,c,d,e all possible ways, and i want to print the res

Issue Parsing XML into table
Hello all,

Im a bit new to php and new to phpfreaks. But thanks in advance for the help!
<

update 2 columns by doing inner 2-column query
Hi,
is something like this possible?

update contract_all set col1,col2 =
(

Sharing PHP Sessions Across Domains
I am in the process of writing a script to share a php session across various domains I have.
The

PHP Display Telephone Number On Referrer
I have used the php below to show a different telephone number in the header of the site depending u

Why do I get this error?
Error:

Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or