weixin_33709590 2015-03-13 04:15 采纳率: 0%
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发送的ajax数据将无法正常工作

below is an ajax but it wont work, what seems be the problem? assume that I have requestparser.php and inside it i have "echo "yes ajax work!";".

$(document).ready(function(){

    // start ajax form submission
    $.ajax({
    url: "requestparser.php",
    type:"POST",
    data: ({ "pulldata" : "oy" }),
    success:function(e){
        if(($.trim(e)=="success")){
            alert("yes");
        }else{
            alert("no");     
        }
    },error:function(){
       alert("error");
    }
   }); 
});

as above ajax function ,the process should be, on load it will send first a post request to requestparser.php with a post name of "pulldata" and post content of "oy" and when the requestparser receive that post request from the ajax so then respond with "yes ajax work!" and since the respond from requestparser.php is not equal to success then it should display "no".

any help would be greatly appreciated

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1条回答 默认 最新

  • weixin_33711647 2015-03-13 04:49
    关注

    I tried to debug your code and it worked fine for me. So can you try this code and say what error it shows ?

    $(document).ready(function () {
        $.ajax({
            url: "test.php",
            type: "POST",
            data: ({
                "pulldata": "oy"
            }),
            success: function (e) {
                console.log(e);
            },
            error: function (e) {
                console.error(e);
            }
        });
    });
    

    test.php

    <?php
    var_dump($_POST);
    

    One more thing, just for confirmation, you have included jQuery in your page before using ajax right ?

    评论

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