Solved

Querying mysql with multiple checkbox values

Posted on 2004-09-07
2
1,446 Views
Last Modified: 2013-12-12
I'm trying to query a table named property with two fields names city and rent.
The user will select multiple checkboxes with a maximum rent they are willing to pay to see if there are any matching properties. On the form page, the name of the checkbox options are city[] while the name of the rent field is rent. I've written the code to pass the checkbox values to an array, but when I query the db, I get a negative result if I select more than one check box. If I select just one box, I get a successful result.

Here is my code:
<?php
$city = $_POST['city'];
$rent = $_POST['rent'];
foreach ($city as $value){
$search=$search.",".$value;
}
/* Remove Leading comma*/
$searchit=substr($search,1);
/*Show the selected fields*/
$db="petfri_petrent";
$link = mysql_connect("localhost","user","pass");
if (! $link)
die("Couldn't connect to MySQL");
mysql_select_db($db , $link)
or die("Couldn't open $db: ".mysql_error());


$result=mysql_query("SELECT * FROM property WHERE city LIKE '%$searchit%' AND rent <= '$rent'")
or die("SELECT Error: ".mysql_error());
$numrows=mysql_num_rows($result);
if ($numrows > 0)
{
echo "we've found one or more matches, please continue to sign up <a href=\"join_form.html\"> Here </a>";
}
else
{
echo "I am sorry we do not have any listings in the area(s) or rent range you chose.";
echo  "You will need to widen your search area and raise your rent maximum.<br \> Click <a href=\"join_form1.php\">HERE</a> to try again.";
}
?>

Thank you for your time....
0
Comment
Question by:ksecor
[X]
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
  • 2
2 Comments
 
LVL 17

Expert Comment

by:akshah123
ID: 12000941
the problem is with your query. Try this.

$result=mysql_query("SELECT * FROM property WHERE city IN ('%$searchit%') AND rent <= '$rent'")
or die("SELECT Error: ".mysql_error());
0
 
LVL 17

Accepted Solution

by:
akshah123 earned 250 total points
ID: 12000967
sorry above is wrong. You need two things

first:

foreach ($city as $value){
$search=$search.",\"".$value . "\"";
}

second your query:


$result=mysql_query("SELECT * FROM property WHERE city IN ($searchit) AND rent <= '$rent' ")
or die("SELECT Error: ".mysql_error());

0

Featured Post

Industry Leaders: We Want Your Opinion!

We value your feedback.

Take our survey and automatically be enter to win anyone of the following:
Yeti Cooler, Amazon eGift Card, and Movie eGift Card!

Question has a verified solution.

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

Generating table dynamically is the most common issue faced by php developers.... So it seems there is a need of an article that explains the basic concept of generating tables dynamically. It just requires a basic knowledge of html and little maths…
Introduction This article is intended for those who are new to PHP error handling (https://www.experts-exchange.com/articles/11769/And-by-the-way-I-am-New-to-PHP.html).  It addresses one of the most common problems that plague beginning PHP develop…
The viewer will learn how to count occurrences of each item in an array.
This tutorial will teach you the core code needed to finalize the addition of a watermark to your image. The viewer will use a small PHP class to learn and create a watermark.

742 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