如何在php中通过Ajax发布值？

<?php
session_start();
?>
    <?php
        $con=mysqli_connect("localhost","root","") or die(mysqli_error());
        mysqli_select_db($con, "sada_r_t_m");
        $result=mysqli_query($con, "SELECT * FROM products ");

echo "<h2 fontsize = '26' color = 'yellow' align = 'center'>Flipkart</h2>";

echo "<table border = '5' cellpadding = '12' align = 'center' background-color = '#84ed86' color =  '#761a9b' >";
echo "<tr>";
echo "<th>SNO</th>";
echo "<th>Product Name</th>";
echo "<th>Price</th>";
echo "<th>Cart</th>";

echo "</tr>";
while($data = mysqli_fetch_assoc($result))
{
echo "<tr>";
echo "<th>".$data['id']."</th>";
echo "<th>".$data['name']."</th>";
echo "<th>".$data['price']."</th>";
echo "<th><a href='#' Onclick='add(".$data['id'].")'>Add to Cart</a></th>";

echo "</tr>";

}
echo "</table>";
?>
<script type="text/javascript">
function add(id){
    alert(id);

    $.ajax({
        url: '../ajax.php',
        type: 'POST',
        data: { action: "add",id: id, val:1} ,
        //contentType: 'application/json; charset=utf-8',
        success: function (response) {
            alert(response.status);
        },
        error: function () {
            alert("error");
        }
    }); 
}
</script>

I passed the values through ajax in php. I wrote only in single php page.

How to pass values through ajax in php. While passing values,i am getting below error

Uncaught ReferenceError: $ is not defined

How to achieved this error i am new to php.

Please add jquery library file for use $. Please put this code in your file at top.

<script src="http://code.jquery.com/jquery-latest.min.js" type="text/javascript"></script>

OR

<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.7.1/jquery.min.js" type="text/javascript"></script>

Your JavaScript Change this type. Error alert show what error Occurred

Add Jquery library file at head tag

<script type="text/javascript">
function add(id){
alert(id);

$.ajax({
 url: '../ajax.php',
 type: 'POST',
data: { action: "add",id: id, val:1} ,
success: function (response) {
try{
alert(response.status);
}
catch(e){
alert("Error"+e);
}
},
error: function (e) {
 alert("error"+e); or alert("error"+JSON.stringify(e));
}
});
}
</script>

You use try and catch . this show where error show

this code useful for you

I cannot comment yet, so I posted here. There's a good example on another answer on how to properly do ajax post requests:

http://stackoverflow.com/questions/5004233/jquery-ajax-post-example-with-php

Also, this is a good tutorial that will set you on the right track:

https://www.codeofaninja.com/2013/09/jquery-ajax-post-example.html

http://hayageek.com/jquery-ajax-post/

Also, try to change ../ajax.php to the full url.