Repairing Data Search Code from MySQL Database using Dropdown List
Introduction
In this article, we will discuss how to repair the data search code in a MySQL database using a dropdown list. The code is written in PHP and uses jQuery for dynamic functionality. We will go through the problem, analyze the existing code, and provide a solution to fix the issues.
Problem Description
The problem is that when two or more countries have the same state name (in this case, “xyz”), the program displays all cities in the “xyz” state from both countries. This happens because the code does not differentiate between countries when selecting states.
For example, if we select USA and the state “xyz”, the program will also display the cities for Russia with the same state “xyz”. To fix this issue, we need to add a condition to check the country before displaying the cities.
Analyzing the Existing Code
Let’s analyze the existing code:
index.php
<?php
//index.php
include('database_connection.php');
$country = '';
$statement = $connect->prepare("
SELECT country
FROM country_state_city
GROUP BY country
ORDER BY country ASC
");
$statement->execute();
while($row=$statement->fetch()){
$country .= '<option value="'.$row["country"].'">'.$row["country"].'</option>';
}
fetch.php
<?php
//fetch.php
if(isset($_POST['action'])){
include('database_connection.php');
$output = '';
if($_POST["action"] == 'country'){
$statement = $connect->prepare("
SELECT state FROM country_state_city
WHERE country = :country
GROUP BY state
");
$statement->execute([
':country' => $_POST["query"]
]);
$output .= '<option value="">Select State</option>';
while($row=$statement->fetch()){
$output .= '<option value="'.$row["state"].'"'.$row["state"].'</option>';
}
}else if($_POST["action"] == 'state'){
$statement = $connect->prepare("
SELECT city FROM country_state_city
WHERE state = :state
AND country=:country
");
$statement->execute([
':state' => $_POST["query"],
':country' => $_POST["country"]
]);
while($row=$statement->fetch()){
$output .= '<option value="'.$row["city"].'"'.$row["city"].'</option>';
}
}
echo $output;
}
Solution
To fix the issue, we need to add a condition to check the country before displaying the cities. We will modify the fetch.php file to include this condition.
Modified fetch.php
<?php
//fetch.php
if(isset($_POST['action'])){
include('database_connection.php');
$output = '';
if($_POST["action"] == 'country'){
$statement = $connect->prepare("
SELECT state FROM country_state_city
WHERE country = :country
GROUP BY state
");
$statement->execute([
':country' => $_POST["query"]
]);
$output .= '<option value="">Select State</option>';
while($row=$statement->fetch()){
$output .= '<option value="'.$row["state"].'"'.$row["state"].'</option>';
}
}else if($_POST["action"] == 'state'){
$country = $_POST['country'];
$statement = $connect->prepare("
SELECT city FROM country_state_city
WHERE state = :state
AND country=:country
");
$statement->execute([
':state' => $_POST["query"],
':country' => $country
]);
while($row=$statement->fetch()){
$output .= '<option value="'.$row["city"].'"'.$row["city"].'</option>';
}
}
echo $output;
}
Explanation
In the modified fetch.php file, we added a variable $country to store the selected country. We then use this variable in the SQL query to filter the cities by country.
With this modification, when we select a state, the program will only display the cities for the selected country.
Conclusion
In this article, we discussed how to repair the data search code in a MySQL database using a dropdown list. We analyzed the existing code and provided a solution to fix the issues. The modified fetch.php file includes a condition to check the country before displaying the cities, ensuring that the program displays accurate results.
Example Use Case
Suppose we have two countries: USA and Russia, both with state “xyz”. When we select USA as the country and then “xyz” as the state, the modified fetch.php file will display only the cities for USA with state “xyz”, not the cities for Russia.
Last modified on 2023-08-22