2015-04-04 51 views
1

我需要幫助獲取點擊按鈕的值並使用AJAX發送此信息。這兩個腳本單獨工作。但我需要將「buttonvalue」傳遞給AJAX「data」值「action」。我正在使用的是以下。檢測哪個按鈕已被點擊並用ajx提交

$("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; 
    }); 
}); 

回答

0

我會阻止表單提交做你想要的然後通過ajax調用發佈數據。請嘗試以下操作:

$("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; 
    }); 
}); 

未測試。