如何獲得JSON字符串變量

Question

我得到了我的JSON字符串阿賈克斯內的功能像這樣

$.ajax({ 
     type: "POST", 
     url: "http://localhost/./Service/GetPageInfo", 
     dataType: "json", 
     contentType: 'application/json', 
     data: JSON.stringify({ 

      filename: filename 
     }), 
     success: function (data) { 
      alert('Success'); 

     }, 
     error: function() { 
      alert('Error'); 
     } 
    }); 

在這裏我得到的數據像

[{"main":{"sub":[],"tittle":"manu","startvalue":"","stopvalue":"","status":"","accumalated":"","comment":""}}] 

我希望它在一個變量像

var myjsonobject =[{"main":{"sub":[],"tittle":"manu","startvalue":"","stopvalue":"","status":"","accumalated":"","comment":""}}] 

Answer 1

你去那裏：

$.ajax({ 
    type: "POST", 
    url: "http://localhost/./Service/GetPageInfo", 
    dataType: "json", 
    contentType: 'application/json', 
    data: JSON.stringify({ 

     filename: filename 
    }), 
    success: function (data) { 
     alert('Success'); 
     var jsonobject = data; 

    }, 
    error: function() { 
     alert('Error'); 
    } 
}); 

此外，我強烈建議你使用的承諾，使API調用：https://developer.mozilla.org/fr/docs/Web/JavaScript/Reference/Objets_globaux/Promise

Answer 2

var jsonobject= null; 

$.ajax({ 
     type: "POST", 
     url: "http://localhost/./Service/GetPageInfo", 
     dataType: "json", 
     contentType: 'application/json', 
     data: JSON.stringify({ 

      filename: filename 
     }), 
     success: function (data) { 
      jsonobject=data; 
      alert('Success'); 

     }, 
     error: function() { 
      alert('Error'); 
     } 
    }); 

如果你想等待Ajax響應，並填寫了變量然後傳遞async: false在AJAX請求選項。

Answer 3

基於您的評論，您需要解析JSON在成功處理程序，

success: function (data) { 
    alert('Success'); 
    var myjsonobject = JSON.parse(data); 
},