douoyou3348 2012-04-26 02:04
浏览 39
已采纳

jQuery,$ .post()没有访问目标网址

In the code below there is a XMLHTTPRequest with the target url test_location_jquery.php, and this file is never accessed despite the success function being triggered. When run alone for testing test_location_jquery.php has worked successfully returning the JSON object as expected. Any idea as to how the success function could be triggered while the file is not accessed would be appreciated.

JQUERY

<?php

    require_once '../meta/php/dbConn.php';

    $charlie_query = "";
    $charlie_result = "";

    $charlie_query = "
        SELECT  ud.user_id as 'id',
                concat(ud.last_name,', ',ud.first_name) as 'name'
        FROM    `user_detail` AS ud,
                `user_type` AS UT
        WHERE ud.user_type = ut.id
        AND ut.class IN ('charlie')
        AND ut.level IN (4)
        ORDER BY ut.class;
    ";
    $charlie_result = mysql_query($charlie_query);

    if(!(bool)$charlie_result){
        throw new Exception();
    }

?>

<html>
<head>

    <script src="../meta/js/jquery-1.7.2.js"></script>
<!-- <script src="../meta/js/getXML_sir.js"></script>  -->
    <script>
        $(document).ready(function(){
            $('#charlie').change(function(){
                var charlie_id =  $('#charlie option:selected').val();
                var DT = "json";

                $.post(
                        'test_location_jquery.php',
                        {'id':charlie_id,'datatype':DT},
                        function(data){
                            alert('Success');
                            var obj = $.parseJSON(data);
                        }
                );

                if(obj != 'undefined'){
                    var locations = obj.find('location');
                    console.log(locations);
                }else{
                    alert("failure");
                }

            })
        })

    </script>

</head>
<body>

    <form>
        <label>Charlie: </label><select id="charlie" name="charlie">
            <option>&nbsp;</option>
        <?php
            while($row = mysql_fetch_assoc($charlie_result)){
                print("<option value='".$row['id']."'>".$row['name']."</option>");
            } 
        ?>
        </select>
        <label>Location: </label><select id="location" name="location"></select>
    </form>

    <div></div>

</body>
</html>

test_location_jquery.php

<?php

require_once("../meta/php/dbConn.php");

if(!isset($_POST)){
    throw new Exception();
}else{
    $id = $_POST['id'];
    $datatype = $_POST['datatype'];
}

$query = "
        SELECT  l.id as 'id',
            ld.attribute_string as 'name'
        FROM    `location` AS l,
                `location_details` AS ld
        WHERE l.id = ld.location_id
        AND     ld.attribute_label = 'name'
        AND l.id = ".$id.";
";
$result = mysql_query($query);

if(!(bool)$result){
    throw new Exception();
}else{
    switch($datatype){
        case 'xml':
        break;
        case 'json':
            $rows = array();
            while($r = mysql_fetch_assoc($result)) {
                $rows[] = $r;
            }
            return json_encode($rows);
        break;
}


?>
  • 写回答

2条回答 默认 最新

  • doudong2149 2012-04-26 02:07
    关注

    Your .post() function is asynchronous. That means that calling .post() just STARTs the post. The lines following it execute immediately before the post() has completed and before the success handler is called.

    Then, some time later the success handler is called.

    All code that uses the response from the post MUST be in the success handler or called from the success handler.

    So, your problem is that you cannot refer to data set in the success handler where this code is located (right after the .post() call). You have to put this code inside the success handler itself.

                if(obj != 'undefined'){
                    var locations = obj.find('location');
                    console.log(locations);
                }else{
                    alert("failure");
                }
    

    So, after doing that, your code would look like this:

    <script>
        $(document).ready(function(){
            $('#charlie').change(function(){
                var charlie_id =  $('#charlie option:selected').val();
                var DT = "json";
    
                $.post(
                        'test_location_jquery.php',
                        {'id':charlie_id,'datatype':DT},
                        function(data){
                            alert('Success');
                            var obj = $.parseJSON(data);
                            if(obj != 'undefined'){
                                var locations = obj.find('location');
                                console.log(locations);
                            }else{
                               alert("failure");
                            }
                        }
                );
    
    
            })
        })
    
    </script>
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论
查看更多回答(1条)

报告相同问题?

悬赏问题

  • ¥15 #MATLAB仿真#车辆换道路径规划
  • ¥15 java 操作 elasticsearch 8.1 实现 索引的重建
  • ¥15 数据可视化Python
  • ¥15 要给毕业设计添加扫码登录的功能!!有偿
  • ¥15 kafka 分区副本增加会导致消息丢失或者不可用吗?
  • ¥15 微信公众号自制会员卡没有收款渠道啊
  • ¥100 Jenkins自动化部署—悬赏100元
  • ¥15 关于#python#的问题:求帮写python代码
  • ¥20 MATLAB画图图形出现上下震荡的线条
  • ¥15 关于#windows#的问题:怎么用WIN 11系统的电脑 克隆WIN NT3.51-4.0系统的硬盘