限制JSON结果

Limit JSON Results

本文关键字:结果 JSON 限制      更新时间:2023-09-26

写一个循环,最好的方法是什么,最好使用jQuery,来限制json对象/数组的结果?

例如,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"}
    ];

我如何获得前5个结果(id 1,2,3,4,&5)?这类似于c#/Linq的。take(5)或SQL的TOP(5)方法。

我希望这里有一个通用的方法,这个示例JSON与我的实际应用程序没有任何关系。

谢谢!

好吧,它们已经排序好了,所以只要…

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

或者如果您确实需要对它们进行排序,您可以这样做…

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

虽然斜眼的回答是完美的。我是http://www.jinqJs.com的创建者。

在jinqJs中你会这样做:

jinqJs().from(myData).top(5).select();
如果我能帮上忙,请告诉我。