Php If in MySql query (hiding labels if a field is empty)


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

Okay, I've been trying to do this for a while, and I'm finally going to ask for help so I can get the simple answer and feel like an idoit .

Anyway, I'm building a dynamic driver profile page for my NASCAR website. I have sections for hometown, driver website, spouse, etc.

Did you know?Explore Trending and Topic pages for more stories like this.
This is all displayed down the side of the page. However, if there is a single driver that has no spouse, there is still an area where the Spouse label is displayed when it is on the website.

Each page has labels going down for each field in the MySql table. These show when the MySql database is queried.

Hopefully that makes since?

Anyway, here is what I have tried and all it does is break the page.

<?php
// Connects to your Database
mysql_connect("BLAH BLAH BLAH", BLAH BLAH BLAH", "BLAH BLAH BLAH") or die(mysql_error());
mysql_select_db("BLAH BLAH BLAH") or die(mysql_error());
$data = mysql_query("SELECT * FROM `BLAH BLAH BLAH` WHERE `BLAH BLAH BLAH` = 'BLAH BLAH BLAH'") or die(mysql_error());

while($info = mysql_fetch_array( $data ))
{
if (".$info['spouse']."=="")
{echo "None";}
else {echo "Spouse: ".$info['spouse'] . "";}
}
?>


Any help is really, really, really, really appreciated
No comments posted yet

Your Answer:

Login to answer
126 Like 17 Dislike
Previous forums Next forums
Other forums

A little help needed passing hidden values to next page
I have a page that has hidden values in a form.

example
Code: <input name='signupID

Not loading image
When this function gets loaded it doesn't load the image just trying to figure out why.

Code:

Undefined variable when using $_SERVER['PHP_SELF']
Hi guyz, please suggest me something...
On first.php I have one input field NAME, and on posting

how to easy edit text, with box? Help.
Hi.
I have a little problem. I'm doing a webpage for my aunt and I would like to make it as easy

Using insert variable
need a way to inert variable data to mysql database

$acc = "212121212";
$nok =

Weird problem with SELECT command..Help!
Hi!

It seems I'm having a really weird problem with SQL SELECT command....I have table into a

PHP & Java
Hello,

can PHP code be used inside java code?

Code: <SCRIPT LANGUAGE="Java

Check something, wait, check again, do something!
Hi guys.

I wonder if someone can help me with this.

Basically, what I want to do is (

I am stumped
OK. Here is what I am trying to do. I have the conditions set, and if the conditions are met, I want

Writting a script to arrange images........ need some help
Ok so here is the link

http://hmtotc.com/dev/projects/vrassociates/jeweler_dev/admin/index.ph

Sign up to write
Sign up now if you have flare of writing..
Login   |   Register
Follow Us
Indyaspeak @ Facebook Indyaspeak @ Twitter Indyaspeak @ Pinterest RSS



Play Free Quiz and Win Cash