ajax发送数据不起作用

I tried using ajax to send data from onclick event on image This is the ajax code

function get_img(name) {
            $.ajax({
                type: "POST",
                url: "img_main.php",
                data: "img_name="+name,
                success: function(response)
                {
                alert("main image selected");
                }
            });
        }

This is the img tag where I put the onclick event

while($row = mysql_fetch_array($query))
        {
        $test = $row['img_name'];
        $result .= "<td><img src='".$uploaddir.$row['img_name']."' class='imgList' onclick='get_img(\"".$test."\")' /></td>";

This is the img_main.php

<?php
    include("connect.inc.php");
    echo "<script type='text/javascript'> alert('test'); </script>";


        $img_name = $_POST['img_name'];

        $query = "UPDATE upload set status = 1 where img_name = $img_name";
        $result = mysql_query($query);
        $query2 = "UPDATE upload set status = 0 where img_name != $img_name";
        $result2 = mysql_query($query2);

?>

What I want is when I clicked on the image, it update the field status on the database.

The success function in the ajax code give me the alert function but the status field is not updated.

My console returns no error.

Can somebody please help?

try change

data: "img_name="+name,

to

data: {img_name:name},

or

$.ajax({
                type: "POST",
                url: "img_main.php", or use full url like "http://domain/img_main.php"
                data: {img_name:name},
                success: function(response)
                {
                alert("main image selected");
                }
            });

and img_main.php (add quotes to your query variable)

$query = "UPDATE upload set status = 1 where img_name = '$img_name'";
        $result = mysql_query($query);
        $query2 = "UPDATE upload set status = 0 where img_name != '$img_name'";
        $result2 = mysql_query($query2);

Add quotes to $img_name in the update queries like:

$query = "UPDATE upload set status = 1 where img_name='$img_name'";
$result = mysql_query($query);
$query2 = "UPDATE upload set status = 0 where img_name!='$img_name'";

As you are getting alert as response, I don't think that there is error in ajax call.