我有一个包含任务员工任务的数组,它看起来像这样:
$scope.assignments = [
{
employee: {
id:"1", firstname:"John", lastname:"Rambo"
},
task: {
name:"Kill everyone", project:"Destruction"
},
date: {
day:"01/01", year:"1985"
}
},
{
employee: {
id:"2", firstname:"Luke", lastname:"Skywalker"
},
task: {
name:"Find daddy", project:"Star Wars"
},
date: {
day:"65/45", year:"1000000"
}
},
{
employee: {
id:"1", firstname:"John", lastname:"Rambo"
},
task: {
name:"Save the world", project:"Destruction"
},
date: {
day:"02/01", year:"1985"
}
}
];
我想按员工分组,因为有这样的事情:
$scope.assignmentsByEmployee = [
{ //First item
id:"1",
firstname:"John",
lastname:"Rambo",
missions: [
{
name:"Kill everyone",
date:"01/01",
year:"1985"
},
{
name:"Save the world",
date:"02/01",
year:"1985"
}
]
},
{ //Second item
id="2",
firstname:"Luke",
lastname:"Skywalker",
missions: [
name:"Find daddy",
date:"65/45",
year:"1000000"
]
}
];
他们是一个简单的方法吗?我用双forEach
尝试了一些东西,但它让我无处可去。
希望我能理解:)
谢谢!
答案 0 :(得分:2)
您应该能够遍历赋值数组并在员工ID上创建“键控数组”(这意味着在JavaScript中使用对象)。然后你只需要填写任务阵列。
像
这样的东西// initialise a holding object
var assignmentsByEmployee = {};
// loop through all assignemnts
for(var i = 0; i < $scope.assignments.length; i++) {
// grab current assignment
var currentAssignment = $scope.assignments[i];
// grab current id
var currentId = currentAssignment.employee.id;
// check if we have seen this employee before
if(assignmentsByEmployee[currentId] === undefined) {
// we haven't, so add a new object to the array
assignmentsByEmployee[currentId] = {
id: currentId,
firstname: currentAssignment.employee.firstname,
lastname: currentAssignment.employee.lastname,
missions: []
};
}
// we know the employee exists at this point, so simply add the mission details
assignmentsByEmployee[currentId].missions.push({
name: currentAssignment.task.name,
date: currentAssignment.date.day,
year: currentAssignment.date.year
});
}
这些留下assignmentsByEmployee
作为对象,但您可以简单地通过它,并在需要时将其转换回数组。 E.g:
$scope.assignmentsByEmployee = [];
for(var employeeId in assignmentsByEmployee) {
$scope.assignmentsByEmployee.push(assignmentsByEmployee[employeeId]);
}