Javascript 检测已单击的按钮并使用ajx提交

我需要帮助获取单击按钮的值，并使用AJAX发送此信息。这两个脚本分别工作。但是我需要将“buttonvalue”传递给AJAX“data”value“action”。下面是我正在使用的

   $("document").ready(function(){

     $(".js-ajax-php-json").on("click", "button[name=mysqljob]", function (){
    var buttonvalue = $(this).attr("value");
    alert(buttonvalue);
    });

  $(".js-ajax-php-json").submit(function(){ 

    var data = {"action": buttonvalue};

    data = $(this).serialize() + "&" + $.param(data);

    $.ajax({
      type: "POST",
      dataType: "json",
      url: "phpVars/ajaxUpload.php", //Relative or absolute path to response.php file
      data: data,
      success: function(data) {
        $(".the-return").html(
         data["form"]
        );
      //  alert("Form submitted successfully.\nReturned json: " + data["json"]);
      }
    });
    return false;
  });
});

---

我会阻止表单提交，然后通过ajax调用发布数据。请尝试以下操作：

$("document").ready(function(){

    $(".js-ajax-php-json").on("click", "button[name=mysqljob]", function (e){
        e.preventDefault();
        var buttonvalue = $(this).attr("value");
        var data = {"action": buttonvalue, "formData": $(this).serialize() + "&" + $.param(data) };

        $.ajax({
            type: "POST",
            dataType: "json",
            url: "phpVars/ajaxUpload.php", //Relative or absolute path to response.php file
            data: data,
            success: function(data) {
                $(".the-return").html(
                    data["form"]
                );
  //            alert("Form submitted successfully.\nReturned json: " + data["json"]);
            }
        });
    return false;
    });
});

没有测试