我有一个像这样的对象数组:
对象数组总是看起来像这样未排序。首先是运输编号,然后是时间。这是由于api我正在使用。
我的问题是:我如何只按时间排序这个数组?
我尝试在我的变量中使用sort函数,如下所示,其中存储了对象数组但没有成功:
allBuses.sort(function(a,b){
var c = a.time;
var d = b.time;
if(c > d){
return d
}
else return c
答案 0 :(得分:3)
您可以将time
视为字符串,并使用String#localeCompare
进行排序。
var data = [{ transportnumber: '45', time: '10:28:00', date:"2017-01-16"}, { transportnumber: '45', time: '10:38:00', date:"2017-01-16" },{ transportnumber: '45', time: '10:48:00', date:"2017-01-16" }, { transportnumber: '14', time: '10:12:00', date:"2017-01-16" }, { transportnumber: '14', time: '10:24:00', date:"2017-01-16" }, { transportnumber: '14', time: '10:52:00', date:"2017-01-16"}];
data.sort(function (a, b) {
return a.time.localeCompare(b.time);
});
console.log(data);
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答案 1 :(得分:1)
var data = [{ transportnumber: '45', time: '10:28:00', date:"2017-01-16"}, { transportnumber: '45', time: '10:38:00', date:"2017-01-16" },{ transportnumber: '45', time: '10:48:00', date:"2017-01-16" }, { transportnumber: '14', time: '10:12:00', date:"2017-01-16" }, { transportnumber: '14', time: '10:24:00', date:"2017-01-16" }, { transportnumber: '14', time: '10:52:00', date:"2017-01-16"}];
data.sort(function(a,b){
return b.time>a.time;
})
console.log(data)
答案 2 :(得分:0)
您将返回 c 或 d ,这是时间值。你的职能需要回归......
答案 3 :(得分:0)
var array = [ { transportnumber: '45', time: '10:28:00', date:"2017-01-16"}
,{ transportnumber: '45', time: '10:38:00', date:"2017-01-16"}
,{ transportnumber: '45', time: '10:48:00', date:"2017-01-16"}
,{ transportnumber: '14', time: '10:12:00', date:"2017-01-16"}
,{ transportnumber: '14', time: '10:24:00', date:"2017-01-16"}
,{ transportnumber: '14', time: '10:52:00', date:"2017-01-16"}];
array.sort(function(a,b){
return new Date(b.date + " "+b.time) - new Date(a.date + " "+a.time);
});
console.log(array);
您可以将日期和时间结合起来并按此排序
答案 4 :(得分:0)
您可以使用Array.prototype.sort():
var allBuses = [{ transportnumber: '45', time: '10:28:00', date:"2017-01-16"}, { transportnumber: '45', time: '10:38:00', date:"2017-01-16" },{ transportnumber: '45', time: '10:48:00', date:"2017-01-16" }, { transportnumber: '14', time: '10:12:00', date:"2017-01-16" }, { transportnumber: '14', time: '10:24:00', date:"2017-01-16" }, { transportnumber: '14', time: '10:52:00', date:"2017-01-16"}];
allBuses.sort((a, b) => (a.time < b.time) ? -1 : ((a.time > b.time) ? 1 : 0));
console.log(allBuses);
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答案 5 :(得分:0)
如果您还想考虑添加日期
data = []
data[0] = { transportnumber: '45', time: '10:28:00', date:"2017-01-16"}
data[1] = { transportnumber: '45', time: '10:38:00', date:"2017-01-16"}
data[2] = { transportnumber: '45', time: '10:48:00', date:"2017-01-16"}
data[3] = { transportnumber: '14', time: '10:12:00', date:"2017-01-16"}
data[4] = { transportnumber: '14', time: '10:24:00', date:"2017-01-16"}
data[5] = { transportnumber: '14', time: '10:52:00', date:"2017-01-16"}
data.sort((a, b) => {
if (a.date < b.date)
return -1
if (a.date > b.date)
return 1
if (a.date == b.date) {
if (a.time < b.time)
return -1
if (a.time > b.time)
return 1
return 0
}
});
答案 6 :(得分:0)
试试这个
allBuses.sort(function(a,b){
var c = new Date(a.date + ":"+ a.time).getTime();
var d = new Date(b.date + ":"+b.time).getTime();
console.log(c,d);
return c -d;
});
答案 7 :(得分:0)
如果您的应用程序有lodash,则可以非常轻松地完成。
var sorted = _.orderBy(data,function(item){
return new Date(item.date + " " + data.time);
});