doubi3996
2014-10-16 07:25
浏览 63
已采纳

使用$ .post从表单接收php服务器中的数据

Getting following error in place of the value

Notice: Undefined index: data in /home/ashutosh/public_html/xyz/about/testimonials/serviceTestimonials.php on line 6 dddd

The code segment:

On webpage: . . .

               <form id="formTesti" name="formTesti" action="serviceTestimonial.php" method="post">
                    <input type="hidden" name="page" id="page" value="1">
                    <input type="hidden" name ="row" id="row" value="10">
                    <button type="submit" name="button" id="buttonPrev" value="Prev">Previous</button>
                    <button type="submit" name="button" id="buttonNext" value="Next">Next</button>
                </form>
            </div>
        </div>

    </div>

    <script src="//ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
    <script>
        //get the data when loading the page
        $(document).ready(
                function()
                {
                    $("#formTesti").submit(function(){
                        return false;
                        //event.preventDefault();
                    });
                    $("#buttonPrev").click(function(){
                        $("#page").val(parseInt($("#page").val())-1);
                        Doit();
                    });

                    $("#buttonNext").click(function(){
                        $("#page").val(parseInt($("#page").val())+1);
                        Doit();
                    });

                    function Doit(){
                        var data = $("#formTesti:input").serializeArray();
                        //alert(data);
                        $.post(
                                "serviceTestimonials.php",
                                data,
                                function(json){
                                    if(json.status="fail")
                                        alert(json.message());
                                    else {

                                    }
                                },
                                "json"
                        );
                    }
                }
        );

    </script>

server side code:

print_r($_POST["data"]);

die(" dddd ");

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3条回答 默认 最新

  • duandazhen7306 2014-10-16 08:21
    已采纳

    Thanks all for support. It seems when $.post is used for json data transfer, we need to use the parameter in callback function for client to server data transfer as well as server to client.

    So code changes required was: var json = $("#formTesti:input").serializeArray();

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  • duanba2001 2014-10-16 07:33

    You have to pass key value pair for data in post.. not data variable.. something like this

    $.post( "test.php", { name: "John", time: "2pm" } );
    
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  • douzi1350 2014-10-16 08:27
    var data = $("#formTesti").serialize();
    $.post(
        "serviceTestimonials.php",
        data: data,
        function(json){
          if(json.status="fail")
              alert(json.message());
          else {
    
          }
       },
    "json"
    );
    
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