限制JSON結果

Question

編寫循環的短處，最好是使用jQuery來限制json對象/數組的結果，最好的方法是什麼？

舉例來說，如果我有：

var mydata = [ 
     {id:"1",invdate:"2010-05-24",name:"test",note:"note",tax:"10.00",total:"2111.00"} , 
     {id:"2",invdate:"2010-05-25",name:"test2",note:"note2",tax:"20.00",total:"320.00"}, 
     {id:"3",invdate:"2007-09-01",name:"test3",note:"note3",tax:"30.00",total:"430.00"}, 
     {id:"4",invdate:"2007-10-04",name:"test",note:"note",tax:"10.00",total:"210.00"}, 
     {id:"5",invdate:"2007-10-05",name:"test2",note:"note2",tax:"20.00",total:"320.00"}, 
     {id:"6",invdate:"2007-09-06",name:"test3",note:"note3",tax:"30.00",total:"430.00"}, 
     {id:"7",invdate:"2007-10-04",name:"test",note:"note",tax:"10.00",total:"210.00"}, 
     {id:"8",invdate:"2007-10-03",name:"test2",note:"note2",amount:"300.00",tax:"21.00",total:"320.00"}, 
     {id:"9",invdate:"2007-09-01",name:"test3",note:"note3",amount:"400.00",tax:"30.00",total:"430.00"}, 
     {id:"11",invdate:"2007-10-01",name:"test",note:"note",amount:"200.00",tax:"10.00",total:"210.00"}, 
     {id:"12",invdate:"2007-10-02",name:"test2",note:"note2",amount:"300.00",tax:"20.00",total:"320.00"}, 
     {id:"13",invdate:"2007-09-01",name:"test3",note:"note3",amount:"400.00",tax:"30.00",total:"430.00"}, 
     {id:"14",invdate:"2007-10-04",name:"test",note:"note",amount:"200.00",tax:"10.00",total:"210.00"}, 
     {id:"15",invdate:"2007-10-05",name:"test2",note:"note2",amount:"300.00",tax:"20.00",total:"320.00"}, 
     {id:"16",invdate:"2007-09-06",name:"test3",note:"note3",amount:"400.00",tax:"30.00",total:"430.00"}, 
     {id:"17",invdate:"2007-10-04",name:"test",note:"note",amount:"200.00",tax:"10.00",total:"210.00"}, 
     {id:"18",invdate:"2007-10-03",name:"test2",note:"note2",amount:"300.00",tax:"20.00",total:"320.00"}, 
     {id:"19",invdate:"2007-09-01",name:"test3",note:"note3",amount:"400.00",tax:"30.00",total:"430.00"}, 
     {id:"21",invdate:"2007-10-01",name:"test",note:"note",amount:"200.00",tax:"10.00",total:"210.00"}, 
     {id:"22",invdate:"2007-10-02",name:"test2",note:"note2",amount:"300.00",tax:"20.00",total:"320.00"}, 
     {id:"23",invdate:"2007-09-01",name:"test3",note:"note3",amount:"400.00",tax:"30.00",total:"430.00"}, 
     {id:"24",invdate:"2007-10-04",name:"test",note:"note",amount:"200.00",tax:"10.00",total:"210.00"}, 
     {id:"25",invdate:"2007-10-05",name:"test2",note:"note2",amount:"300.00",tax:"20.00",total:"320.00"}, 
     {id:"26",invdate:"2007-09-06",name:"test3",note:"note3",amount:"400.00",tax:"30.00",total:"430.00"}, 
     {id:"27",invdate:"2007-10-04",name:"test",note:"note",amount:"200.00",tax:"10.00",total:"210.00"}, 
     {id:"28",invdate:"2007-10-03",name:"test2",note:"note2",amount:"300.00",tax:"20.00",total:"320.00"}, 
     {id:"29",invdate:"2007-09-01",name:"test3",note:"note3",amount:"400.00",tax:"30.00",total:"430.00"} 
    ]; 

我怎麼會得到公正的前5名成績（IDS 1,2,3,4，& 5）？這與C＃/ Linq的.Take（5）或SQL的TOP（5）方法類似。

我希望在這裏有一個通用的方法，這個示例JSON與我的實際應用無關。

謝謝！

Answer 1

好，他們已經排序，所以只是做...

var t5 = mydata.slice(0,5); 

或者，如果你確實需要對它們進行排序，你可以這樣做......

var t5 = mydata.sort(function(a,b) { 
    return +a.id - +b.id; 
}).slice(0,5); 

Answer 2

儘管斜視的答案非常好。 我是http://www.jinqJs.com的創建者。

在jinqJs你會做：

jinqJs().from(myData).top(5).select(); 

讓我知道如果我可以幫上什麼忙。