PHP Search Script for Website - php

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 !

Answer 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']) : '';
}