如何在jQuery DataTable的AJAX调用上传递额外的参数

Question

我使用DataTable文档中指示的以下代码传递参数。

视图：

$('#example').dataTable( {
      "ajax": {
           "url": "/Student/GetStudents",
           "data": function ( d ) {
               d.test= "some data";
           }
      }
});

控制器：

public ActionResult GetStudents(JQueryDataTableParamModel param, string test)
{
    //code omitted for brevity

    return Json(new
    {
        sEcho = param.sEcho,
        iTotalRecords = allRecords.Count(),
        iTotalDisplayRecords = filteredRecords.Count(),
        aaData = result
    },
    JsonRequestBehavior.AllowGet);
}

尽管将“ test”参数传递给控制器​​，但“ param”参数中的值为null或0，并且导致数据表的返回空数据。 另一方面，如果我在数据表参数中使用以下行而不是AJAX调用，则所有param值都将正确传递给Controller（但使用AJAX调用，此行也会导致错误）。 我需要将额外的参数传递给Controller，并且必须使用AJAX调用。 传递参数值时如何传递呢？

"ajaxSource": "/Student/GetStudents",

Answer 1

您的JavaScript代码：

$('#example').dataTable( {
      "ajax": {
           "url": "/Student/GetStudents",
            type: 'GET',
            data: {
                    test1: "This test1 data ",
                    test2: "This test2 data"
                }
        }
});


public ActionResult GetStudents(JQueryDataTableParamModel param, string test)
{
    //code omitted for brevity

     //printing in params in controller with asp.net code. 
     print_r("Data from" ,param.test1 ,param.test2);

    return Json(new
    {
        sEcho = param.sEcho,
        iTotalRecords = allRecords.Count(),
        iTotalDisplayRecords = filteredRecords.Count(),
        aaData = result
    },
    JsonRequestBehavior.AllowGet);
}

Answer 2

最后，我通过使用fnServerData方法解决了该问题，如下所示。

"ajaxSource": "/Student/GetStudents",

//fnServerData used to inject the parameters into the AJAX call sent to the server-side
"fnServerData": function (sSource, aoData, fnCallback) {
    aoData.push({ "name": "test", "value": "some data" });
    $.getJSON(sSource, aoData, function (json) {
        fnCallback(json)
    });
},

...

无论如何，非常感谢您提供的有用答案。 投票+有用的...

Answer 3

var finalArray = [];
var data = {'test':"some data","test1":"some data1"};
finalArray.push(data);
var rec = JSON.stringify(finalArray);

$('#example').dataTable( {
"ajax": {
   "url": "/Student/GetStudents",
   "data": rec
}
});

public ActionResult GetStudents(JQueryDataTableParamModel param,string test)
{
//code omitted for brevity

 //printing in params in controller with asp.net code. 
 print_r(json_decode(param));

return Json(new
{
    sEcho = param.sEcho,
    iTotalRecords = allRecords.Count(),
    iTotalDisplayRecords = filteredRecords.Count(),
    aaData = result
},
JsonRequestBehavior.AllowGet);
}

Answer 4

您可以创建一个json数据字符串，您可以在其中传递其他参数。

var data = {'test':"some data","test1":"some data1"};
    $('#example').dataTable( {
    "ajax": {
       "url": "/Student/GetStudents",
       "data": data
    }
});