Jquery POST传参有关问题

function gomaketo(imgId, url) {
    $.ajax({ //调用jquery的ajax方法   
        type: "POST", //设置ajax方法提交数据的形式   
        url: "a.aspx",
        data: "imgSrc=" + url + "&backImgId=" + imgId,
        dataType: 'text',
        success: function (msg) { //提交成功后的回调
        }
    })
}

public string imgSrc = DNTRequest.GetString("imgSrc");
public string backImgId = DNTRequest.GetString("backImgId");