解析JSON數組對象

Question

我需要將從C＃代碼獲得的一組值存儲到一個javascript數組中。 我得到一個錯誤..有人可以告訴錯誤是什麼？ 即時通訊使用此jcode。

$.get('Dataextract.aspx', function (data, textStatus) { 
     alert('Status is ' + textStatus);//success 
     alert('JSON data string is: ' + data);//string as below 

    var JSONdata = data; 
    eval(JSONdata);//error here-> expected ; 
    alert(JSONdata.rowval[0].CustomerID); 

}, 'text'); 

我正在使用ajax查詢來檢索JSON對象的數組。我得到的數據對象值就像字符串一樣。

{"rowval" :[{"CustomerID":12"Title":"Mr.""FirstName":"Johnny""MiddleName":"A.""LastName":"Caprio""CompanyName":"Bikes and Motorbikes""RowNumber":10},{"CustomerID":16"Title":"Mr.""FirstName":"Christopher""MiddleName":"R.""LastName":"Beck""CompanyName":"Bulk Discount Store""RowNumber":11},{"CustomerID":18"Title":"Mr.""FirstName":"David""MiddleName":"J.""LastName":"Liu""CompanyName":"Catalog Store""RowNumber":12},{"CustomerID":19"Title":"Mr.""FirstName":"John""MiddleName":"A.""LastName":"Beaver""CompanyName":"Center Cycle Shop""RowNumber":13},{"CustomerID":20"Title":"Ms.""FirstName":"Jean""MiddleName":"P.""LastName":"Handley""CompanyName":"Central Discount Store""RowNumber":14},{"CustomerID":21"Title":"FirstName":"Jinghao""MiddleName":"LastName":"Liu""CompanyName":"Chic Department Stores""RowNumber":15},{"CustomerID":22"Title":"Ms.""FirstName":"Linda""MiddleName":"E.""LastName":"Burnett""CompanyName":"Travel Systems""RowNumber":16},{"CustomerID":23"Title":"Mr.""FirstName":"Kerim""MiddleName":"LastName":"Hanif""CompanyName":"Bike World""RowNumber":17},{"CustomerID":24"Title":"Mr.""FirstName":"Kevin""MiddleName":"LastName":"Liu""CompanyName":"Eastside Department Store""RowNumber":18},{"CustomerID":25"Title":"Mr.""FirstName":"Donald""MiddleName":"L.""LastName":"Blanton""CompanyName":"Coalition Bike Company""RowNumber":19},{"CustomerID":28"Title":"Ms.""FirstName":"Jackie""MiddleName":"E.""LastName":"Blackwell""CompanyName":"Commuter Bicycle Store""RowNumber":20}]} 

這裏是生成JSON我的C＃代碼

sb.Append("{\"rowval\" :"); 
      sb.Append("["); 
      if (table != null) 
      { 
       foreach (DataRow row in table.Rows) 
       { 
        sb.Append("{"); 
        if (row.Table != null && row.Table.Columns != null && row.Table.Columns.Count > 0) 
        { 
         foreach (DataColumn column in row.Table.Columns) 
         { 
          parseMember(row[column], column.ColumnName, sb); 
         } 
        } 
        sb.Append("},"); 
       } 
      } 
      sb.Append("]"); 
      sb.Append("}"); 

      sqlcon.Close(); 
      Response.Write(sb); 



     } 

     private static void parseMember(object val, string memberName, StringBuilder sb) 
     { 
      Type t = val.GetType(); 

      if (memberName != null && memberName.Trim().Length > 0) 
       sb.AppendFormat("\"{0}\":", memberName); 
      if (typeof(string) == t || typeof(char) == t) 
       sb.AppendFormat("\"{0}\"", val.ToString()); 
      else 
       sb.AppendFormat("{0}", val.ToString()); 
     } 

Answer 1

如果你得到json然後指定dataType等於json as 4th argument in $不用彷徨will parse the json which you can iterate using the each`的jQuery的方法，像

$.get('Dataextract.aspx', function (data, textStatus) { 
     alert('Status is ' + textStatus);//success 
     alert('JSON data string is: ' + data);//string as below 
    // no need for eval 
    // var JSONdata = data; 
    // eval(JSONdata);//error here-> expected ; 
    alert(JSONdata.rowval[0].CustomerID); 

}, "json"); // <-- 

，或者您可以解析JSON明確就像

$.get('Dataextract.aspx', function (data, textStatus) { 
     alert('Status is ' + textStatus);//success 
     alert('JSON data string is: ' + data);//string as below 

    var JSONdata = $.parseJSON(data); 
    // eval(JSONdata);//error here-> expected ; again no need for the eval 
    alert(JSONdata.rowval[0].CustomerID); 

}, 'text'); 

---

更新

the json您正在形成不正確，爲驗證您的JSON，您可以訪問www.jsonlint.com，這是有效的JSON

{ 
    "rowval": [ 
     { 
      "CustomerID": 12, // <-- you are missing the commas 
      "Title": "Mr.", 
      "FirstName": "Johnny", 
      "MiddleName": "A.", 
      "LastName": "Caprio", 
      "CompanyName": "Bikes and Motorbikes", 
      "RowNumber": 10 
     } 
    ] 
} 

Answer 2

你需要兩個類-call他們ROVAL和客戶，爲示例 - 並將其定義爲這樣：

class RowVal 
{ 

    public List<Customer> Customers {get;set;} 
} 

class Customer 
{ 
    public int ID {get;set;} 
    public string Title{get;set;} 
    public string FirstName {get;set;} 
    //and so on 

} 

現在您可以反序列化該JSON字符串了

JavascriptSerializer serializer = new JavascriptSerializer(); 
RowVal rowVal = serializer.Deserialize<RowVal>(yourJSonstring);