假设我有两个帮助函数,将平面的{}数组转换为树形结构。考虑以下平面数据:
const data = [
{
"ID": 1,
"Tier_1": "DataSource1",
"Tier_2": "Area",
"Tier_3": "General",
},
{
"ID": 2,
"Tier_1": "DataSource1",
"Tier_2": "Financial",
"Tier_3": "General",
},
{
"ID": 3,
"Tier_1": "DataSource1",
"Tier_2": "Area",
"Tier_3": "General",
},
{
"ID": 4,
"Tier_1": "DataSource2",
"Tier_2": "Area",
"Tier_3": "General",
},
{
"ID": 5,
"Tier_1": "DataSource2",
"Tier_2": "Area",
"Tier_3": "Management Plan",
}
]
数据包含三行层次信息,我想将其转换为类似树形结构的形式,如下所示(预期输出):
(最后的子级是实际的数据库对象,但分布在树上)
const output = {
"DataSource1: {
"Area": {
{
"ID": 1,
"Tier_1": "DataSource1",
"Tier_2": "Area",
"Tier_3": "General",
},
{
"ID": 3,
"Tier_1": "DataSource1",
"Tier_2": "Area",
"Tier_3": "General",
},
},
"Financial": [
{
"ID": 2,
"Tier_1": "DataSource1",
"Tier_2": "Financial",
"Tier_3": "General",
},
]
},
"DataSource2: {
"Area": [
{
"ID": 4,
"Tier_1": "DataSource2",
"Tier_2": "Area",
"Tier_3": "General",
},
{
"ID": 5,
"Tier_1": "DataSource2",
"Tier_2": "Area",
"Tier_3": "Management Plan",
}
]
}
}
}
我实际上成功创建了函数来实现此目标,但它们不够灵活(深度/维度是固定的,并在每个函数名称中说明)。
返回二维树的函数:
const getDataCategoriesTwoDim = (data, mainCategory) => {
const dataFields = [...data];
let map = {};
for (let i = 0; i < dataFields.length; i += 1) {
const currentField = dataFields[i];
const currentCategory = currentField[mainCategory];
if (!map[currentCategory]) {
map[currentCategory] = [];
}
map[currentCategory].push(currentField);
}
return map;
};
返回三维树的函数:
const getDataCategoriesThreeDim = (data, mainCategory, subCategory) => { // DIFF
const dataFields = [...data];
let map = {};
for (let i = 0; i < dataFields.length; i += 1) {
const currentField = dataFields[i];
const currentCategory = currentField[mainCategory];
const currentSubcategory = currentField[subCategory]; // DIFF
if (!map[currentCategory]) {
map[currentCategory] = {}; /DIFF
}
if (!map[currentCategory][currentSubcategory]) { // DIFF
map[currentCategory][currentSubcategory] = []; // DIFF
} // DIFF
map[currentCategory][currentSubcategory].push(currentField); // DIFF
}
return map;
};
您可以像这样调用两者并获得预期结果:
getDataCategoriesTwoDim(data, 'Tier_2');
getDataCategoriesThreeDim(data, 'Tier_2', 'Tier_3');
如您所见,代码重复和复制粘贴非常常见。我在注释中标记了不同之处。我如何将代码重写为一个函数,以便可以设置2、3或更多维度?