dropdown list and search field php - php

Question

:)
I'm new here and I'm very new with php.
I am trying to make a search form with:
a dropdown list with two items: category and location;
a text field;
a search button.
It should work like this:
When "category" is selected, you enter a text and it will be searched only into categories.
When "location" is selected, your term will be searched among countries, states, zip codes.
I have a table with columns: id, name, category, country, zipcode, state.
Could somebody help me to understand why it doesn't display any results?
Here is my code:
<form action='search4.php' method='POST' name='form_filter'>
<b>Search</b><br>
<select name="selectVal">
<option value="category">category</option>
<option value="location">Country, state or zipcode</option>
</select>
<input type='text' name='search' placeholder='Enter text here...' size='50'><br>
<input type='submit' value='Send'>
</form>
<?php
// database connection
$db_host = "myhost";
$db_user = "myuser";
$db_password = "mypsw";
$db_name = "myname";
//connecting to database
$db = mysql_connect($db_host, $db_user, $db_password) or die ('Error - connection failed');
mysql_select_db($db_name, $db) or die ('Database selection error');
// retrieving search value we sent using get
$research = $_GET['research'];
// check if it has been sent, then it is ok
if ( $research == 'ok' ) {
// retrieving search value we sent using post
$search = $_POST['search'];
// check if the field has been filled
if ( $search == TRUE && $search != "" ) {
// character lenght more than 3
if ( strlen($search) >= 3 ) {
$search = mysql_escape_string(stripslashes($search));
}
if(isset($_POST['value'])) {
if($_POST['value'] == 'category') {
// query to get all categories
$query = "SELECT * FROM table_name WHERE category='$search'";
}
elseif($_POST['value'] == 'location') {
// query to get all country/state/zipcode records
$query = "SELECT * FROM table_name WHERE country='$search' OR zip_code='$search' OR state='$search'";
} else {
// query to get all records
$query = "SELECT * FROM table_name";
}
$sql = mysql_query($query);
while ($row = mysql_fetch_array($query)){
$Id = $row["Id"];
$country = $row["country"];
$category = $row["category"];
$name = $row['name'];
$zip_code = $row['zip_code'];
$state = $row['state'];
echo "Name: $name<br>";
echo "Zip_code : $zip_code<br>";
echo "State : $state<br>";
echo "Country: $country<br>";
echo "Category: $category<hr>";
}
}
}
}
?>
Thank you very much for your help.

Answer 1

You need to understand how to use <select> with php.
if you have this form:
<form method='post'>
<select name='example'>
<option value='e1'>example1</option>
<option value='e2'>example2</option>
</select>
</form>
You need to print it like that:
echo $_POST['example'];
In case the user selcted example1, the value will be e1.
In case the user selcted example2, the value will be e2.
You are using in your script $_POST['value']. It's just dosen't exist.

Answer 2

Try this, instead:
HTML FORM:
<form action='search4.php' method='POST' name='form_filter'>
<b>Search</b><br>
<select name="selectVal">
<option value="category">category</option>
<option value="location">Country, state or zipcode</option>
</select>
<input type='text' name='search' placeholder='Enter text here...' size='50'><br>
<input type='submit' value='Send'>
</form>
FORM PROCESSING:
<?php
// database connection
$db_host = "myhost";
$db_user = "myuser";
$db_password = "mypsw";
$db_name = "myname";
//connecting to database
$db = mysql_connect($db_host, $db_user, $db_password) or die ('Error - connection failed');
mysql_select_db($db_name, $db) or die ('Database selection error');
/*********************************************/
/***WHY DO YOU NEED THIS RESEARCH VARIABLE?***/
/*****WHAT IS ITS PURPOSE IN THIS SCRIPT?*****/
/*********************************************/
//GET CLEAN VERSIONS OF ALL NECESSARY VARIABLES:
$search = isset($_POST['search']) ? htmlspecialchars(trim($_POST['search'])) : null;
$catLocation = isset($_POST['selectVal']) ? htmlspecialchars(trim($_POST['selectVal'])) : null;
$query = "SELECT * FROM table_name WHERE ";
//YOU INDICATED YOU'D NEED TO RUN THE SEARCH-QUERY IF THE SEARCH-TERM AND SEARCH-SCOPE ARE DEFINED IE: NOT NULL; HOWEVER IF THE SEARCH TERM IS NOT GIVEN, YOU SELECT EVERYTHING IN THAT TABLE... (BAD PRACTICE, THOUGH)
if($catLocation){
if($search){
if($catLocation == "category"){
$query .= " category LIKE '%" . $search . "%'";
}else if($catLocation == "location"){
$query .= " country LIKE '%" . $search . "%' OR zip_code LIKE '%" . $search . "%' OR state LIKE '%" . $search . "%'";
}
}else{
$query .= "1";
}
$sql = mysql_query($query);
//HERE AGAIN WAS AN ERROR... YOU PASSED mysql_fetch_array A STRING $query INSTEAD OF A RESOURCE: $sql
while ($row = mysql_fetch_array($sql)){
$Id = $row["Id"];
$country = $row["country"];
$category = $row["category"];
$name = $row['name'];
$zip_code = $row['zip_code'];
$state = $row['state'];
echo "Name: $name<br>";
echo "Zip_code : $zip_code<br>";
echo "State : $state<br>";
echo "Country: $country<br>";
echo "Category: $category<hr>";
}
}