douguwo2275 2013-07-23 19:32
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为什么我的AJAX get请求不起作用

In my js page I want to get some variables from the php page using ajax(); This is all triggered by the html page load. I've tried to use both GET and POST, but nothing alerts or logs to my console like it supposed to, not even the error.

HTML:

<!DOCTYPE html>
<html>
    <head>

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.9.0/jquery.min.js"></script>
    <script src="http://code.jquery.com/ui/1.10.2/jquery-ui.js"></script>
   <script src="http://XXXXXX/bn/sample.js"></script>
    </head>
    <body>
        <div>Welcome! </div>
    </body>
</html>

JS: (sample.js)

$(function(){

 $.ajax({
        url : "http://XXXXXX/bn/sample.php",
        type : 'GET',
        data : data,
        dataType : 'json',
        success : function (result) {
           alert(result.ajax); // "Hello world!" should be alerted
           console.log(result.advert); 
        },
        error : function () {
           alert("error");
        }
    });

    });

PHP:

<?php

    $advert = array(
        'ajax' => 'Hello world!',
        'advert' => 'Working',
     );

    echo json_encode($advert);
?>
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1条回答 默认 最新

  • doubo6658 2013-07-23 19:52
    关注

    the data you set at "data: data" ist the data u send to the php script. but u dont send any. the data you receive is available in you success function. so print it. plus i removed the alert. alerts are lame. console rocks!

    the " $(document).ready(" part starts your function as soon as your document is completlty loaded. i guess this is what you need and wanted there.

    $(document).ready(function() {    
     $.ajax({
            url : "http://XXXXXX/bn/sample.php",
            type : 'GET',
            data : (),
            dataType : 'json',
            success : function (data) {
               console.log(data.advert); 
            },
            error : function () {
               console.log("error");
            }
        });
    
        });
    

    edit: remove the last comma, as your array ends there. you probably want to set a header before output.

    <?php
    
    header('Cache-Control: no-cache, must-revalidate');
    header('Content-type: application/json');
    
    
        $advert = array(
            'ajax' => 'Hello world!',
            'advert' => 'Working'
         );
    
        echo json_encode($advert);
    ?>
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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