Solved

Will someone help with this MySQL Query?

Posted on 2008-10-21
2
206 Views
Last Modified: 2013-11-19
I can't figure out why this query still shows results where PaidID=''.

The url calling this would be: ...?page=listing/results&State=FL

OR

$SEARCH="FL"; << Session

if (isset($_SESSION['search'])){
$SEARCH=$_SESSION['search'];
 
if (!ereg('[^0-9]', $SEARCH)){$searchResult = mysql_query("SELECT * FROM members WHERE ZipCodes LIKE '%$SEARCH%' AND PaidID!='' ");}
 
elseif (!ereg('[^A-Z]', $SEARCH) && (strlen($SEARCH) == 2)){
$searchResult = mysql_query("SELECT * FROM members WHERE PaidID!='' AND LicState LIKE '%$SEARCH%' OR LicState2 LIKE '%$SEARCH%' OR LicState3 LIKE '%$SEARCH%' OR LicState4 LIKE '%$SEARCH%' ");
}
 
elseif (isset($_GET['State'])){
 
$searchResult = mysql_query("SELECT * FROM members WHERE
 
(LicState='$State' AND Counties LIKE '%$SEARCH%') OR 
(LicState2='$State' AND Counties2 LIKE '%$SEARCH%') OR 
(LicState3='$State' AND Counties3 LIKE '%$SEARCH%') OR 
(LicState4='$State' AND Counties4 LIKE '%$SEARCH%')
 
AND PaidID!=''
 ");
 
}
 
elseif ($SEARCH=="EXAMPLE"){
$searchResult = mysql_query("SELECT * FROM members WHERE PaidID!='' ORDER BY RAND() LIMIT 5");
}
 
else{
$searchResult = mysql_query("SELECT * FROM members WHERE Counties LIKE '%$SEARCH%' OR Counties2 LIKE '%$SEARCH%' OR Counties3 LIKE '%$SEARCH%' OR Counties4 LIKE '%$SEARCH%'  PaidID!='' ");
}
 
if (!$searchResult){die(mysql_error());}
	
$num_rows= mysql_num_rows($searchResult);
 
}
 
if ($num_rows==0){include ("pages/listing/results_none.php");}
else{include ("pages/listing/results_found.php");}

Open in new window

0
Comment
Question by:phpretard
2 Comments
 
LVL 1

Accepted Solution

by:
yeruhn earned 500 total points
ID: 22767102
Use brackets in your query. Like:

SELECT * FROM members WHERE PaidID!='' AND (LicState LIKE '%$SEARCH%' OR LicState2 LIKE '%$SEARCH%' OR LicState3 LIKE '%$SEARCH%' OR LicState4 LIKE '%$SEARCH%' )

Dependent on what you want to select...
0
 

Author Closing Comment

by:phpretard
ID: 31508229
Thank you!
0

Featured Post

Master Your Team's Linux and Cloud Stack!

The average business loses $13.5M per year to ineffective training (per 1,000 employees). Keep ahead of the competition and combine in-person quality with online cost and flexibility by training with Linux Academy.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Suggested Solutions

Boost your ability to deliver ambitious and competitive web apps by choosing the right JavaScript framework to best suit your project’s needs.
There’s a good reason for why it’s called a homepage – it closely resembles that of a physical house and the only real difference is that it’s online. Your website’s homepage is where people come to visit you. It’s the family room of your website wh…
The viewer will learn how to dynamically set the form action using jQuery.
The viewer will learn how to count occurrences of each item in an array.

770 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question