使用$ .post（）将值发布到PHP并使用$ .getJSON返回

I'm trying to use $.getJSON inside $.post function and display returned value in textbox (id=desc). Here is my code:

我正在尝试在$ .post函数中使用$ .getJSON并在textbox中显示返回值(id = desc)。这是我的代码:

function dataload(){
    var currentId = $(this).attr('id'); 
    var e = document.getElementById(""+currentId);
    var sel = e.options[e.selectedIndex].text; 
    $("#"+currentId).change(function (){
        $.post("test.php", {data:sel}, function(data){
            $.getJSON("test.php", function(data){
                $("#desc").val(data[0].description);
            });
        });
    });
}

test.php:

<?php
   include('config.php');//db connection
   $itemvalue=($_POST["data"]);
   $item = mysql_query("SELECT description FROM invoice WHERE item='$itemvalue'");
   $data = array();
   while($row = mysql_fetch_array($item, true)) {
        $data[] = $row;
   }
   $data = json_encode($data);
   echo $data;
?>

but test.php is not taking posted value. So how to post sel?

但是test.php没有采用发布值。那么如何发布sel?

1 个解决方案

#1

You call test.php twice, once using $.post and once using $.getJSON. On the first call, the POST param should be passed correctly to your PHP - but you don't do anything with the result of the request. On the second call, you don't transmit any POST param and $itemvalue just is empty.

你调用test.php两次,一次使用$ .post,一次使用$ .getJSON。在第一次调用时,POST参数应该正确传递给您的PHP - 但是您不会对请求的结果做任何事情。在第二次调用时,您不传输任何POST参数,$ itemvalue只是为空。

Just use:

$.getJSON("test.php", {data:sel}, function(data){
    // [...]

and in your PHP:

在你的PHP中:

$itemvalue=$_GET["data"]; // $.getJSON performs a GET request

btw: Your code is vulnerable to SQL injections.

顺便说一句:您的代码容易受到SQL注入攻击。

#1