Data not getting Inserted in MYSQL using Php - php

Question

I am making a web application which allows user to get funds for the project,I have got most part working but I am stuck at one place where the php script returns a positive status code but no data gets inserted into Mysql database. Following is the php script i am using:
Code:
<?php
session_start();
if(!isset($_SESSION['username']))
{
echo "Unauthorised Page Usage Please Relogin to Access All the Page features;";
header('location:login.html');
}
$sponsor=mysql_real_escape_string($_POST['sponsorid']);
$projectid=mysql_real_escape_string($_POST['projectid']);
$pledge=mysql_real_escape_string($_POST['pledgevalue']);
$servername = "localhost";
$usernam = "root";
$password = "";
$dbname = "project";
$httpStatusCode = 400;
$httpStatusMsg = 'Incorrect Username or Password';
$protocol=isset($_SERVER['SERVER_PROTOCOL']) ? $_SERVER['SERVER_PROTOCOL'] : 'HTTP/1.0';
$connection=new mysqli($servername,$usernam,$password,$dbname);
if (!$connection) {
die("Connection failed: " . $connection->connect_error);
}
$sql1="INSERT INTO `sponsor`(`spon_id`,`project_id`,`spon_amt`,`spon_date_time`) VALUES ('$sponsor','$projectid','$pledge',NOW())";
$result=$connection->query($sql1);
if ($result) {
$Success=200;
$httpStatusMsg=mysqli_error($connection);
header($protocol.' '.$Success.' '.$result);
}
else {
$Success=400;
$httpStatusMsg=mysqli_error($connection);
header($protocol.' '.$Success.' '.$httpStatusMsg);
}
?>
Below is the ajax used to post data to a page:
var xhttp = new XMLHttpRequest();
xhttp.onreadystatechange = function() {
if (this.readyState === 4) {
if(this.status===404){
alert(this.responseText);
}
if(this.status===200)
{
alert("Project backed successfully");
window.location.reload(true);
}
}
};
xhttp.open("POST", "sponsor.php", true);
xhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
var param= "sponsorid"+"="+ <?php echo json_encode($_SESSION['username']); ?>+"&"+"pledgevalue"+"="+document.getElementById("pledge").value+"&"+"projectid"+"="+<?php echo json_encode($projectid);?>;
console.log(param);
xhttp.send(param);
}
I have cross referenced my sponsor table to make sure that every field is same.The code works fine on my friend`s computer. Please help me
Update: the $sql1 query is giving me Error code:1644 Problem when I ran it in Database (using XAMMP).
Please help.

Answer 1

Change
$result=$connection->query($sql);
To
$result=$connection->query($sql1);

Answer 2

You have used $sql1 as a query and when you fire query to database you are using $sql which is not available actually.
If make error reporting ON you will see an error about undefined variable $sql
If still it is not solve then print generated sql query using echo $sql1; and try it executing in database.you will get exact error.

Answer 3

Try this:
<?php
session_start();
if(!isset($_SESSION['username']))
{
echo "Unauthorised Page Usage Please Relogin to Access All the Page features;";
header('location:login.html');
}
$sponsor=mysql_real_escape_string($_POST['sponsorid']);
$projectid=mysql_real_escape_string($_POST['projectid']);
$pledge=mysql_real_escape_string($_POST['pledgevalue']);
$servername = "localhost";
$usernam = "root";
$password = "";
$dbname = "project";
$httpStatusCode = 400;
$httpStatusMsg = 'Incorrect Username or Password';
$protocol=isset($_SERVER['SERVER_PROTOCOL']) ? $_SERVER['SERVER_PROTOCOL'] : 'HTTP/1.0';
$connection=new mysqli($servername,$usernam,$password,$dbname);
if (!$connection) {
die("Connection failed: " . $connection->connect_error);
}
$sql1="INSERT INTO `sponsor`(`spon_id`,`project_id`,`spon_amt`,`spon_date_time`) VALUES ('$sponsor','$projectid','$pledge',NOW())";
$result=$connection->query($sql1);
if ($result) {
$Success=200;
$httpStatusMsg=mysqli_error($connection);
header($protocol.' '.$Success.' '.$result);
}
else {
$Success=400;
$httpStatusMsg=mysqli_error($connection);
header($protocol.' '.$Success.' '.$httpStatusMsg);
}
?>