PHP Search Script for Website

Question

Alright, So i am trying to code a little PHP Search Script for My website so users can simply do a search from a artist name, song name or a city. My table in my database has 'city', 'artist' and 'city'.

Here is my form:

<div id="search">  
<form name="search" method="post" action="../searchDb.php">  
<input type="text" name="find" placeholder="What are we searching for ?"/> in   
<Select NAME="field">  
<Option VALUE="artist">Artist</option>  
<Option VALUE="song">Song</option>  
<Option VALUE="city">City</option>  
</Select>  
<input type="hidden" name="searching" value="yes" />  
<input type="submit" name="search" value="Search" />  
</form>  
</div>

As you can see, there are three OPTION values (one for each column in my table). Here is my PHP code:

<?php  
$searching = "searching";  
$find = "find";  
$field = "field";  
 //this is to make sure the user entered content  
if ($searching =="yes")   
{   
   echo "<p><h2>Results</h2></p>";   

   //if user did not enter anything in the search box, give error   
   if ($find == "")   
   {   
      echo "<p>You forgot to enter a search term</p>";   
   }   

   include 'connect.php';   

   // strip whitespace, non case sensitive  
   $find = strtoupper($find);   
   $find = strip_tags($find);   
   $find = trim ($find);   

   //perform search in specified field  
   $data = mysql_query("SELECT * FROM artists_table WHERE upper($field) LIKE'%$find%'");   

   //show results   
   while($result = mysql_fetch_array( $data ))   
   {   
      echo $result['artist'];     
      echo " ";   
      echo $result['song'];    
      echo "<br>";    
      echo $result['city'];    
      echo "<br>";    
      echo "<br>";   
   }   

   //counts results. ifnone. error    
  $anymatches=mysql_num_rows($data);    
   if ($anymatches == 0)    
   {    
      echo "Sorry, but we can not find an entry to match your query<br><br>";    
   }    

   //show user what he searched.   
   echo "<b>Searched For:</b> " .$find;     
 }     
 ?>

My connect.php (that is included) works perfectly (i have that same file working on another page, no problems..So its safe to say thats not the problem).

When i do a test and run a search, it loads up my searchDb.php but NOTHING is displayed. Simply a white page...

Any help would be greatly appreciated. I am lost as to why or what is not working... Thanks Guys !

Don't forget to properly sanitize that data in the query, today it's "find" tomorrow it's $_GET... — soulseekah
– soulseekah, Commented Dec 4, 2012 at 0:31
BTW, you don't have to capitalize your string search, as LIKE is case insensitive as long as you aren't dealing with a binary string. — RonaldBarzell
– RonaldBarzell, Commented Dec 4, 2012 at 0:31
I mean don't forget to properly escape $find and $field (given that they're provided by user) to prevent SQL injection. — soulseekah
– soulseekah, Commented Dec 4, 2012 at 0:39
Definitely wont forget this. Thanks. Im only in "testing" for now..Just getting the basics done..Security will be my next step :P — user1853848
– user1853848, Commented Dec 4, 2012 at 0:43

Community · Accepted Answer · 2023-11-17 20:38:18Z

1

If this is your code, then you are hardcoding $searching = "searching", but in your if you are checking if $searching =="yes", so none of the code will show.

<?php  
$searching = "searching";  
...
...  
//this is to make sure the user entered content  
if ($searching =="yes")   
{   
...
}

Edit-

My guess is that you wanted to do something like-

$searching = mysql_real_escape_string($_POST['searching']); // sanitized just to be consistant.  
$find = mysql_real_escape_string($_POST['find']);  
$field = mysql_real_escape_string($_POST['field']);

Note- you should not be writing new code with mysql_* functions. You should learn either mysqli_ or PDO - http://php.net/manual/en/mysqlinfo.api.choosing.php

Here are 2 ways to avoid getting the "Notice: Undefined variable"

Check if submit button was pushed

if (isset($_POST['search'])) {
$searching = mysql_real_escape_string($_POST['searching']); // sanitized just to be consistant.  
$find = mysql_real_escape_string($_POST['find']);  
$field = mysql_real_escape_string($_POST['field']);
}

Or check if each field is set, and set it to the value, and if not set to no/empty

if (isset($_POST['search'])) {  // checks to see if the form submit button was pushed
$searching = isset($_POST['search']) ? mysql_real_escape_string($_POST['searching']) : 'no'; // sanitized just to be consistant.  
$find = isset($_POST['find']) ? mysql_real_escape_string($_POST['find']) : '';  
$field = isset($_POST['field']) ? mysql_real_escape_string($_POST['field']) : '';
}

edited Nov 17, 2023 at 20:38

CommunityBot

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answered Dec 4, 2012 at 0:34

Sean

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5 Comments

user1853848 Over a year ago

I tried removing the first 3 lines ($searching = "searching", $find=......) and when i re-run the code, Notice: Undefined variable: searching in C.......

Sean Over a year ago

See my edit. You have to set the variable $searching or you will get the Notice: Undefined variable: searching..., but you will want to set it to the value of the post from your form input - <input type="hidden" name="searching" value="yes" />

user1853848 Over a year ago

Thank you very Much Sean. I used the code you edited ( $searching = mysql_real_escape_string($_POST['searching']); and I am still getting the Notice: Undefined.....

Sean Over a year ago

Is your search form in the same page as your php script? Are you accessing your php script without submitting your form? I will edit my answer with 2 different ways to check the submission/set the variables.

user1853848 Over a year ago

My search form is on another page then my PHP script. After reading your last comment Sean, i figured out where-what my issue was. I simply added if(isset($_POST['submit'])) and BINGO :) Thank you VERY much for making my brain realize this haha.

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