这是我正在使用的服务方法。它从项目中找到所有收入（与此问题相关：Angular service calling another service）并创建一个新的对象数组。
我希望这个对象数组按年和三个月对收入进行排序，因此我可以在视图中轻松地循环（使用ng-repeat）。
我当前的数据（由Incomes.buildAndGetIncomes()返回，您可以在下面看到）：
[
{
"projectName":"Deuxième test",
"clientName":"Deuxième client",
"typeIncome":"Accompte",
"amount":1000,
"date":"2014-09-10",
"notes":"Chèque / LDD",
"trim":"third",
"year":"2014"
},
{
"projectName":"efzefze",
"clientName":"zfezefzef",
"typeIncome":"Accompte",
"amount":30,
"date":"2014-09-10",
"notes":"fzefzef",
"trim":"third",
"year":"2014"
},
{
"projectName":"Nouveau test",
"clientName":"Nouveau client",
"typeIncome":"Accompte",
"amount":16,
"date":"2014-09-19",
"notes":"CC",
"trim":"third",
"year":"2014"
},
{
"projectName":"Nouveau projet",
"clientName":"Nouveau client",
"typeIncome":"Accompte",
"amount":1200,
"date":"2014-05-17",
"notes":"Chèque cc",
"trim":"second",
"year":"2014"
},
{
"projectName":"Projet test",
"clientName":"Client test",
"typeIncome":"Accompte",
"amount":1500,
"date":"2014-01-15",
"notes":"Chèque cc",
"trim":"first",
"year":"2014"
},
{
"projectName":"Deuxième test",
"clientName":"Deuxième client",
"typeIncome":"Reliquat",
"amount":4500,
"date":"2014-09-27",
"notes":"Virement",
"trim":"third",
"year":"2014"
},
{
"projectName":"efzefze",
"clientName":"zfezefzef",
"typeIncome":"Reliquat",
"amount":8,
"date":"2014-09-05",
"notes":"zefzefzef",
"trim":"third",
"year":"2014"
},
{
"projectName":"Nouveau test",
"clientName":"Nouveau client",
"typeIncome":"Reliquat",
"amount":7,
"date":"2014-09-27",
"notes":"LDD",
"trim":"third",
"year":"2014"
}
]
我想要的数据结构：
[
{
year: 2014,
trim: [
{
name : 'first',
content : [
// some content
]
},
{
name : 'second',
content : [
// some content
]
}
]
},
{
year: 2013,
trim: [
{
name : 'first',
content : [
// some content
]
}
]
}
]
以下是我现在正在使用的方法：
self.trimestral = function(){
var deferred = $q.defer();
self.buildAndGetIncomes().then(function(result) {
var trimestral = [];
var incomes = result;
angular.forEach(incomes, function(income, i){
var year = income.year,
trim = income.trim,
newTrim = {},
newTrimContent = {};
if( i === 0 ){
newTrim.year = year;
newTrim.trim = [];
newTrimContent ={};
newTrimContent.name = trim;
newTrimContent.content = [];
newTrimContent.content.push(income);
trimestral.push(newTrim);
console.log(trimestral);
trimestral[0].trim.push(newTrimContent);
}
else {
var maxLength = incomes.length;
for (var h = 0; h<maxLength; h++){
if(trimestral[h].year === year){
for (var j = 0; j<trimestral[h].trim.length; j++){
console.log(h,j,trimestral[h].trim[j].name === trim);
if(trimestral[h].trim[j].name === trim){ // trimester already exists
trimestral[h].trim[j].content.push(income);
}
else {// trimester doesn't exist, create it
var createTrim = {};
createTrim.name = trim;
createTrim.content = [];
createTrim.content.push(income);
trimestral[h].trim.push(newTrimContent);
}
}
}
else {
newTrim.year = year;
newTrim.trim = [];
newTrimContent ={};
newTrimContent.name = trim;
newTrimContent.content = [];
newTrimContent.content.push(income);
trimestral.push(newTrim);
console.log(trimestral);
trimestral[0].trim.push(newTrimContent);
}
}
}
});
deferred.resolve(trimestral);
});
return deferred.promise;
};
这段代码按照假设运行，它检查我们是否在循环的第一个索引上，并且它推动了那个三个月的年/三个月/内容。那个结构还可以。
现在我的问题是我需要检查年份是否已经存在，然后检查该年份对象中是否存在三个月，构建像我粘贴在上面的对象数组。 < / p>
我尝试了很多方法，但对我的JS技能来说似乎太难了......有什么帮助吗？
答案 0 :(得分：1)
这肯定可以使用几个循环，但代码将很难遵循。如果您可以引入另一个库，请考虑使用underscore.js。有了它，你可以做这样的事情......
var trimestral = _(incomes).chain()
.groupBy('year')
.map(function (items, year) {
return {
year: year,
trim: _(items).chain()
.groupBy('trim')
.map(function(content, trimester) {
return {
trim: trimester,
content: content
};
})
.sortBy(function(i) {
var trimesters = {
first: 1,
second: 2,
third: 3
};
return trimesters[i.trim];
}).value()
};
})
.sortBy('year')
.value();
答案 1 :(得分：0)
我终于要求一位优秀的网络开发人员帮我解决这个问题。他告诉我在两个步骤中执行此操作：构造对象，然后按照我需要的方式格式化。
所以我这样做了：
self.trimestral = function(){
var deferred = $q.defer();
var global = [];
self.buildAndGetIncomes().then(function(result) {
var trimestral = {};
var incomes = result;
// First loop, construct the object
angular.forEach(incomes, function(income, i){
var year = income.year,
trim = income.trim,
newTrim = {},
newTrimContent = {};
if(trimestral[year] === undefined){
trimestral['year' , year] = {};
}
if(trimestral[year][trim] === undefined) {
trimestral[year][trim] = [];
}
trimestral[year][trim].push(income);
});
deferred.resolve(global);
// Second loop, format the data
for ( prop in trimestral ) {
newYear = {};
newYear.name = prop;
newYear.content = [];
for ( trim in trimestral[prop]){
newTrim = {};
newTrim.name = trim;
newTrim.content = [];
newTrim.content = trimestral[prop][trim];
newYear.content.push(newTrim);
}
global.push(newYear);
}
});
return deferred.promise;
};
我想它可能更干净，但它有效并且比我预期的要简单得多。