我正在开发一个Angular应用程序，该应用程序在Angular Bootstrap日历中的cell modifier内使用custom cell template。在每个单元格内部，而不是标准事件，我放置了一组表格，用于在当天的站点注册轮班。表格分为两组; am和pm，在每个am / pm组内，每个站都有一个表，每个表中有三行。
AM
| 1 |职位|姓名|
| | Pos 1 |名称1 |
| | Pos 2 |姓名2 |
| | Pos 3 |姓名3 |
| 2 |职位|姓名|
| | Pos 1 |名称1 |
| | Pos 2 |姓名2 |
| | Pos 3 |姓名3 |
PM
| 1 |职位|姓名|
| | Pos 1 |名称1 |
| | Pos 2 |姓名2 |
| | Pos 3 |姓名3 |
| 2 |职位|姓名|
| | Pos 1 |名称1 |
| | Pos 2 |姓名2 |
| | Pos 3 |姓名3 |
在我的单元格修改器函数中，我得到当天的一组移位对象，每个移位对象包含ampm值和工作站值：
{
"_id": "57776537ac0a88010063b9b9",
"modified": "2016-07-02T06:54:47.518Z",
"data": {
"station": "1",
"date": "2016-07-01T07:00:00.000Z",
"ampm": "pm",
"slots": [
{
"position": "AO",
"name": ""
},
{
"position": "FF",
"name": {
"_id": "57776507ac0a88010063b9b8",
"modified": "2016-07-02T06:53:59.661Z",
"data": {
"group": "suppression",
"driving": {
"n": false,
"d": true,
"ao": false,
"wt": false
},
"emtLevel": "b",
"secondaryPhoneNumber": "",
"primaryPhoneNumber": "5556781234",
"emailAddress": "person.one@mysite.com",
"fullName": "Person One",
"userName": "person.one",
"assignedStation": "18",
"probationary": false
},
"form": "57427ba554ec330100dad645",
"created": "2016-07-02T06:53:59.644Z",
"externalIds": [],
"access": [],
"roles": [
"573511a8ffaa7a0100a5718a"
],
"owner": "57776507ac0a88010063b9b8"
}
},
{
"position": "FF",
"name": {
"_id": "57439d856e67b40100d4c420",
"modified": "2016-05-24T00:17:09.493Z",
"data": {
"userName": "person.two",
"fullName": "Person Two",
"emailAddress": "person.two@mysite.com",
"primaryPhoneNumber": "5555556666",
"secondaryPhoneNumber": "",
"assignedStation": "",
"emtLevel": "b",
"driving": {
"d": true
},
"group": "suppression"
},
"form": "57427ba554ec330100dad645",
"created": "2016-05-24T00:17:09.474Z",
"externalIds": [],
"access": [],
"roles": [
"573511a8ffaa7a0100a5718a"
],
"owner": "5734bba2ffaa7a0100a57029"
}
}
]
},
所以问题是如何获取这些对象并将它们组织成上面提到的两个分组，以便我可以在我的模板中使用ngRepeat循环遍历它们。到目前为止我所拥有的是：
vm.cellModifier = function(cell) {
cell.text = 'Test Text';
var shifts = vm.events;
// Get the date for the cell.
this.cellDate = moment(cell.date).format('YYYY-MM-DD');
// Iterate over shifts to get ones for this day.
this.cell = cell;
this.todayShifts = {};
shifts.forEach(function(shift, index, allShifts) {
var shiftDate = moment(shift.data.date).format('YYYY-MM-DD');
// Now we need to see if this shift belongs to this day.
if (moment(vm.cellDate).isSame(moment(shiftDate))) {
// Shift is today, so let's put it into the appropriate array.
if (typeof vm.todayShifts[shift.data.ampm] == 'undefined') {
vm.todayShifts[shift.data.ampm] = shift;
} else {
vm.todayShifts[shift.data.ampm].push(shift);
}
}
});
// Add arrays to cell object.
cell.todayShifts = vm.todayShifts;
};
这会给vm.todayShifts[am]和vm.todayShifts[pm]，但我也希望获得第二级，以便vm.todayShifts[am][1]，vm.todayShifts[am][2]等。有更简单的方法吗？做我正在尝试做的事情（我相当确定），而不是添加另一部分语句？我想知道自定义指令或组件是否更干净，因为那时我可以将我的数据传递到该控制器，但即便如此，我仍然需要正确安排我的数据，以便它能以正确的顺序显示。
希望这一切都有道理。
感谢。
答案 0 :(得分：1)
我认为你实际上非常接近......试试这个将始终以您要求的格式存储数据。
shifts.forEach(function(shift, index, allShifts) {
var shiftDate = moment(shift.data.date).format('YYYY-MM-DD');
// Now we need to see if this shift belongs to this day.
if (moment(vm.cellDate).isSame(moment(shiftDate))) {
// Shift is today, so let's put it into the appropriate array.
if (typeof vm.todayShifts[shift.data.ampm] == 'undefined') {
// Initialize the shifts as an array.
vm.todayShifts[shift.data.ampm] = [];
}
// Push the shift onto the array.
vm.todayShifts[shift.data.ampm].push(shift);
}
});