處理代碼的正確方法

Question

如何正確處理使用var jquery中的代碼，我怎樣才能正確使用它作爲@url.action中的參數。 我試圖

url: '@Url.Action("CreateDisease", "DiseaseLists", new {'"+diseaseID+"', '"+ assessmentID+"' })', 

，但它沒有工作

$(document).ready(function() { 
    $('#btn-disease').click(function() { 
     var diseaseID = $('#DiseaseID').val(); 
     var assessmentID = $('#AssessmentID').val(); 

     $.ajax({ 
      type: "POST", 
      dataType: "Json", 
      data: {diseaseID:'" + diseaseID + "',assessmentID: '" + assessmentID + "' }, 
      url: '@Url.Action("CreateDisease", "DiseaseLists", new {//parameter here* diseaseID, assessmentID })', 


      success: function (f) { 
       if (f.Result == "success") { 
        alert("success!"); 
       } 
       else { 
        alert("Disease Already Added"); 
       } 
       alert("????????????????????"); 
      } 
     }) 
    }) 
}) 

Answer 1

假設你的控制器DiseaseLists（怪異的名字控制器如果你問我），方法是CreateDisease具有完全相同的參數diseaseID和assessmentID，你可以寫類似

$(document).ready(function() { 
$('#btn-disease').click(function() { 
    var diseaseID = $('#DiseaseID').val(); 
    var assessmentID = $('#AssessmentID').val();  
    $.ajax({ 
     type: "POST", 
     dataType: "Json", 
     data: {'diseaseID': diseaseID , 
       'assessmentID': assessmentID 
     }, 
     url: '@Url.Action("CreateDisease", "DiseaseLists")', 
     success: function (f) { 
      if (f.Result == "success") { 
       alert("success!"); 
      } 
      else { 
       alert("Disease Already Added"); 
      } 
      alert("????????????????????"); 
     } 
    }) 
}) 

}）

Answer 2

你只需要提供在url參數的URL。請求中的數據應作爲對象提供給$.ajax的data屬性。你的語法有點不合適。試試這個：

url: '@Url.Action("CreateDisease", "DiseaseLists")', 
data: { 
    diseaseID: diseaseID, 
    assessmentID: assessmentID 
},