Insert query not working in WAMP Server - php

Question

We have one android application which access WAMP Server using PHP. All queries like, Select* etc working fine. But insert query not working properly in specific condition,
When WAMP already have some data in database we failed to insert new data in database. But when WAMP server don't have any existing data then insert query work successfully. following is PHP code we are using to insert a data,
<?php
/*
* Following code will create a new product row
* All product details are read from HTTP Post Request
*/
// array for JSON response
$response = array();
// check for required fields
if (isset($_POST['UserName']) && isset($_POST['UserID']) && isset($_POST['Password']) && isset($_POST['Age'])&& isset($_POST['ContactNumber']) && isset($_POST['expiryDate'])) {
$UserName = $_POST['UserName'];
$UserID = $_POST['UserID'];
$Password = $_POST['Password'];
$Age = $_POST['Age'];
$ContactNumber = $_POST['ContactNumber'];
$expiryDate = $_POST['expiryDate'];
// include db connect class
require_once __DIR__ . '/db_connect.php';
// connecting to db
$db = new DB_CONNECT();
// mysql inserting a new row
$result = mysql_query("INSERT INTO normaluser(UserName, UserID, Password,Age,ContactNumber,expiryDate) VALUES('$UserName', '$UserID', '$Password','$Age','$ContactNumber','$expiryDate')");
// check if row inserted or not
if ($result) {
// successfully inserted into database
$response["success"] = 1;
$response["message"] = "New user successfully created.";
// echoing JSON response
echo json_encode($response);
} else {
// failed to insert row
$response["success"] = 0;
$response["message"] = "Oops! error is there.";
// echoing JSON response
echo json_encode($response);
}
} else {
// required field is missing
$response["success"] = 0;
$response["message"] = "Required fields is missing";
// echoing JSON response
echo json_encode($response);
}
?>
Can you guys please help use to solved this problem?

Answer 1

Ignoring the using of mysql_* which is deprecated and lack of sanitization of data.
You do not check if this User is already on the database. Assuming at least one of your columns is a unique or Primary key then this query is likely to fail if you try and create the same user twice.
But regardless of that possible issue if you add some useful error processing to your code you will be told exactly what the problem is. This is a little simplistic as you probably dont want these errors sent back to a user but it will tell you the exact error being returned from MySQL:
$result = mysql_query("INSERT INTO normaluser(UserName, UserID, Password,Age,ContactNumber,expiryDate)
VALUES('$UserName', '$UserID', '$Password','$Age','$ContactNumber','$expiryDate')");
// check if row inserted or not
if ($result) {
// successfully inserted into database
$response["success"] = 1;
$response["message"] = "New user successfully created.";
// echoing JSON response
echo json_encode($response);
} else {
// failed to insert row
$response["success"] = 0;
$response["message"] = 'Database error ' . mysql_errno() . ' ' . mysql_error();
// echoing JSON response
echo json_encode($response);
}