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使用jquery ajax在按钮单击上发布表单

[英]POST form on button click using jquery ajax

i'm trying to load form via ajax and submit form then write response to "#pagediv". 我正在尝试通过Ajax加载表单并提交表单,然后将响应写入“ #pagediv”。

Here is my working code 这是我的工作代码

    $(function() {              
    $.ajax({
        url: "<?php echo $_SERVER['PHP_SELF']; ?>",
        type: "GET",
        data: { <?php echo $ajxparam; ?> },
        dataType: "html",
        cache: false
    }).done(function( msg ) {
        $("#pagediv").html(msg);
        $("#savebtn").click(function(event){ event.preventDefault(); $("#testform").submit(); });       
    }).fail(function() {
        $("#pagediv").html( "Request Failed. Please Try Again Later." );
    });
});

modified code in form submit part(refer below); 表单提交部分中的修改代码(请参阅下文); its not working, did i missed anything? 它不起作用,我有什么错过吗?

    $(function() {              
    $.ajax({
        url: "<?php echo $_SERVER['PHP_SELF']; ?>",
        type: "GET",
        data: { <?php echo $ajxparam; ?> },
        dataType: "html",
        cache: false
    }).done(function( msg ) {
        $("#pagediv").html(msg);
        $("#savebtn").click(function(event){
            event.preventDefault();
            $("#testform").submit(function(evt) {
              evt.preventDefault();
              var $form = $( this ),
                  url = $form.attr( 'action' );
              var posting = $.post( url, $(this).serialize() );
              posting.done(function( dte ) {
                $("#successdiv").html(dte);
              });
            });
        });
    }).fail(function() {
        $("#pagediv").html( "Request Failed. Please Try Again Later." );
    });
});

Note:- Form POST URL is different from ajax load main page. 注意:-表单POST URL与ajax加载主页不同。

Inside your click event handler you're adding a submit event handler and not actually submitting the form. 在click事件处理程序中,您要添加一个Submit事件处理程序,而不是实际提交表单。 Since you're uploading the data via ajax on a button click just call the ajax in the button click callback. 由于您是通过按钮上的ajax上传数据的,因此只需在按钮click回调中调用ajax。

    $("#savebtn").click(function(event){
          event.preventDefault();
          var $form = $("#testform"),
              url = $form.attr( 'action' );
          var posting = $.post( url, $form.serialize() );
          posting.done(function( dte ) {
            $("#successdiv").html(dte);
          });
    });

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