使用JavaScript对JSON数据进行排序

我正在寻找从我的JSON提供程序订购数据的方法：

[{"Username":"Mehmet","UserID":2,"OkeyTablePlayerChairNumber":1},
{"Username":null,"UserID":0,"OkeyTablePlayerChairNumber":2},
{"Username":"Erçin","UserID":1,"OkeyTablePlayerChairNumber":3},
{"Username":null,"UserID":0,"OkeyTablePlayerChairNumber":4},
{"Username":null,"UserID":0,"OkeyTablePlayerChairNumber":5},
{"Username":null,"UserID":0,"OkeyTablePlayerChairNumber":6},
{"Username":null,"UserID":0,"OkeyTablePlayerChairNumber":7},
{"Username":null,"UserID":0,"OkeyTablePlayerChairNumber":8},
{"Username":null,"UserID":0,"OkeyTablePlayerChairNumber":9}]

订购时还有两件更重要的事情：

OkeyTablePlayerHairNumber

首先使用用户ID；让我解释一下：

我想先按OkeyTableChairNumber订购；这已经由服务器端完成。数据由OkeyTablePlayerChairNumber ASC订购

现在公牛， 我想按用户ID订购，但是；例如，如果我取UserID==1，它应该对它们进行排序，比如：UserID==1字段，将出现在第一个字段中。上面的对象将被删除并添加到列表的末尾

查看：

UserID==1

 [{"Username":"Erçin","UserID":1,"OkeyTablePlayerChairNumber":3},
 {"Username":null,"UserID":0,"OkeyTablePlayerChairNumber":4},
 {"Username":null,"UserID":0,"OkeyTablePlayerChairNumber":5},
 {"Username":null,"UserID":0,"OkeyTablePlayerChairNumber":6},
 {"Username":null,"UserID":0,"OkeyTablePlayerChairNumber":7},
 {"Username":null,"UserID":0,"OkeyTablePlayerChairNumber":8},
 {"Username":null,"UserID":0,"OkeyTablePlayerChairNumber":9},
 {"Username":"Mehmet","UserID":2,"OkeyTablePlayerChairNumber":1},
 {"Username":null,"UserID":0,"OkeyTablePlayerChairNumber":2}]

C#我以前做过的解决方案；但我需要JS解决方案：

int yourSitPositionIndex = playersOnTheTableWithEmpytPositions.ToList().FindIndex(x => x.UserID == userID);
var beforePlayers = playersOnTheTableWithEmpytPositions.ToList().GetRange(0, yourSitPositionIndex);
IEnumerable<tbl_Okey_TablePlayer> afterPlayers = playersOnTheTableWithEmpytPositions.Except(beforePlayers);
IEnumerable<tbl_Okey_TablePlayer> newPositions = afterPlayers.Concat(beforePlayers);

int-yourSitPositionIndex=playersonetablewithempypositions.ToList（）.FindIndex（x=>x.UserID==UserID）；
var beforePlayers=playersonetablewithempypositions.ToList（）.GetRange（0，yourSitPositionIndex）；
IEnumerable afterPlayers=具有临时位置的etables上的玩家。除了（在玩家之前）；
IEnumerable newPositions=afterPlayers.Concat（beforePlayers）；

转换您的C#解决方案：

var wantedId = 1;

var index = 0, result;
//int yourSitPositionIndex = playersOnTheTableWithEmpytPositions.ToList().FindIndex(x => x.UserID == userID);
while (index < data.length && data[index].UserID != wantedId) index += 1;
if (index < data.length) {
    //var beforePlayers = playersOnTheTableWithEmpytPositions.ToList().GetRange(0, yourSitPositionIndex);
    var beforePlayers = data.slice(0, index);
    //IEnumerable<tbl_Okey_TablePlayer> afterPlayers = playersOnTheTableWithEmpytPositions.Except(beforePlayers);
    var afterPlayers = data.slice(index);
    //IEnumerable<tbl_Okey_TablePlayer> newPositions = afterPlayers.Concat(beforePlayers);
    result = afterPlayers.concat(beforePlayers);
} else {
    result = data;
}

var wantedId=1；
var指数=0，结果；
//int yourSitPositionIndex=playersonetablewithempypositions.ToList（）.FindIndex（x=>x.UserID==UserID）；
而（index

• javascript的核心中没有

  FindIndex

  ，所以我做了一个while循环来查找索引
• 我用于替换

  GetRange

• 代码的其余部分应该足够清晰。所有代码都可以通过以下库简化

---

如果目标是按用户ID描述订购，则OkeyTable。。。ASC

您可以定义客户排序功能：

在行动中：

结果:

[ { Username: 'Mehmet', UserID: 2, OkeyTablePlayerChairNumber: 1 },
  { Username: 'Erçin', UserID: 1, OkeyTablePlayerChairNumber: 3 },
  { Username: null, UserID: 0, OkeyTablePlayerChairNumber: 2 },
  { Username: null, UserID: 0, OkeyTablePlayerChairNumber: 4 },
  { Username: null, UserID: 0, OkeyTablePlayerChairNumber: 5 },
  { Username: null, UserID: 0, OkeyTablePlayerChairNumber: 6 },
  { Username: null, UserID: 0, OkeyTablePlayerChairNumber: 7 },
  { Username: null, UserID: 0, OkeyTablePlayerChairNumber: 8 },
  { Username: null, UserID: 0, OkeyTablePlayerChairNumber: 9 } ]

[ { Username: 'Erçin', UserID: 1, OkeyTablePlayerChairNumber: 3 },
  { Username: 'Mehmet', UserID: 2, OkeyTablePlayerChairNumber: 1 },
  { Username: null, UserID: 0, OkeyTablePlayerChairNumber: 2 },
  { Username: null, UserID: 0, OkeyTablePlayerChairNumber: 4 },
  { Username: null, UserID: 0, OkeyTablePlayerChairNumber: 5 },
  { Username: null, UserID: 0, OkeyTablePlayerChairNumber: 6 },
  { Username: null, UserID: 0, OkeyTablePlayerChairNumber: 7 },
  { Username: null, UserID: 0, OkeyTablePlayerChairNumber: 8 },
  { Username: null, UserID: 0, OkeyTablePlayerChairNumber: 9 } ]

这是用户ID ASC，OkeyTable…ASC:

function idSort(a,b){
    var notZero = (a["UserID"] > 0 && b["UserID"] > 0);
    var notEqual = (a["UserID"] !== b["UserID"]);
    if(notZero && notEqual){
        return a["UserID"] - b["UserID"];
    } else {
        return a["OkeyTablePlayerChairNumber"] - b["OkeyTablePlayerChairNumber"];
    }
}
var sorted = json.sort(idSort);

结果:

[ { Username: 'Mehmet', UserID: 2, OkeyTablePlayerChairNumber: 1 },
  { Username: 'Erçin', UserID: 1, OkeyTablePlayerChairNumber: 3 },
  { Username: null, UserID: 0, OkeyTablePlayerChairNumber: 2 },
  { Username: null, UserID: 0, OkeyTablePlayerChairNumber: 4 },
  { Username: null, UserID: 0, OkeyTablePlayerChairNumber: 5 },
  { Username: null, UserID: 0, OkeyTablePlayerChairNumber: 6 },
  { Username: null, UserID: 0, OkeyTablePlayerChairNumber: 7 },
  { Username: null, UserID: 0, OkeyTablePlayerChairNumber: 8 },
  { Username: null, UserID: 0, OkeyTablePlayerChairNumber: 9 } ]

[ { Username: 'Erçin', UserID: 1, OkeyTablePlayerChairNumber: 3 },
  { Username: 'Mehmet', UserID: 2, OkeyTablePlayerChairNumber: 1 },
  { Username: null, UserID: 0, OkeyTablePlayerChairNumber: 2 },
  { Username: null, UserID: 0, OkeyTablePlayerChairNumber: 4 },
  { Username: null, UserID: 0, OkeyTablePlayerChairNumber: 5 },
  { Username: null, UserID: 0, OkeyTablePlayerChairNumber: 6 },
  { Username: null, UserID: 0, OkeyTablePlayerChairNumber: 7 },
  { Username: null, UserID: 0, OkeyTablePlayerChairNumber: 8 },
  { Username: null, UserID: 0, OkeyTablePlayerChairNumber: 9 } ]

---

行动：

为什么要投反对票，请解释一下。（不知道投反对票的情况）你试过什么？您是将数据转换成字符串还是javascript对象？谢谢；我的数据来自jQuery.parseJSON（obj）（编辑：顺便说一句，我尝试了JSON.search，但我对JSON是新手）在你的例子中，你真的需要把

OkeyTablePlayerChairNumber=1

和

OkeyTablePlayerChairNumber=2

的元素放在最后，还是把它们放在

UserID=1

元素之后的任何地方都可以？（再说一遍，你试过什么？）不是重复的，你不能用简单的排序来解决这个问题，因为必须考虑数组的初始状态。非常好。谢谢！