如何使用数据库中的无限父子类别制作 json
How to make json with infinite parent child category from database
我有一个 SQLite 数据库table
+---------------------------------------------------+
| id | Cat_Name | Parent_ID |
|---------------------------------------------------+
| 1 | Asset | NULL |
+---------------------------------------------------+
| 2 | Bank | 1 |
+---------------------------------------------------+
| 3 | Cash | 1 |
+---------------------------------------------------+
| 4 | Petty Cash | 3 |
+---------------------------------------------------+
| 5 | ABC Bank | 2 |
+---------------------------------------------------+
| 6 | Dollar Account | 2 |
+---------------------------------------------------+
我可以获取如下数据
[{ id: 1, Category_Name: "Asset", Parent_ID: 0},
{ id: 2, Category_Name: "Bank", Parent_ID: 1},
{ id: 3, Category_Name: "Cash", Parent_ID: 1},
{ id: 4, Category_Name: "Petty_Cash", Parent_ID: 3},
{ id: 5, Category_Name: "ABC_Bank", Parent_ID: 2},
{ id: 6, Category_Name: "Dollar_Account", Parent_ID: 2}]
在这个table,用户创建的类别和子类别中,我们无法假设在table
中会有多少个父类别和子类别
现在我想将数据作为嵌套 javascript 对象传递给前端
示例
{Asset: {Bank: {ABC Bank: 5}, {Dollar Account: 6}
},
{Cash:{PettyCash: 4}, if any...}
}
任何人都可以帮助以最佳方式获得此结果...
提前致谢
下面介绍的是实现所需 objective 的一种可能方法。不可否认,它不是很优雅(可能不是 most-efficient)。
代码段
// helper method to recursively-add to object
const recurAdd = (arr, idx, res) => {
// when "idx" exceeds length of array "arr",
// simply return existing result "res" object
if (idx >= arr.length) return res;
// de-structure to access parent-id & id for current elt
const { Parent_ID, id } = arr[idx];
if (Parent_ID in res) {
// parent-id exists at current object,
// so, add "id" to same object (mutate)
res[Parent_ID][id] = {};
// make recursive call for "next" elt in "arr"
return recurAdd(arr, idx+1, res);
} else {
// find next-level object where current elt will fit
const foundIt = Object.values(res).map(obj => recurAdd(arr, idx, obj));
// NOTE: "obj" is part of "res" and it gets mutated
// if found, make recursive call
if (foundIt.some(x => x !== false)) return recurAdd(arr, idx+1, res);
};
// in case parent-id is not found, simply return false
return false;
};
// helper method to substitute "id" with "category names"
const recurNamify = (obj, myMap) => (
// reconstruct object from key-value pairs of intermediate result
Object.fromEntries(
// generate intermediate result of key-value pairs
Object.entries(obj)
.map(([k, v]) => (
// substitute key (ie, "id") with category-name
Object.keys(v).length === 0
? [myMap[k], k]
: [myMap[k], recurNamify(v, myMap)]
))
// when "v" is not an empty object, make recursive call
)
);
// transform the array into nested object
const myTransform = arr => {
// first transform "Number" to "string" for id and parent-id
// because JS-object keys are string type
const myArr = arr.map(ob => ({
...ob,
id: ob.id.toString(),
Parent_ID: ob.Parent_ID.toString()
}));
// generate a dictionary/map for "id" to category-name
const myMap = myArr.reduce(
(acc, itm) => {
acc[itm.id] = itm.Category_Name
return acc;
},
{}
);
// find the index of root (ie, parent id is zero)
const rIdx = myArr.findIndex(({ Parent_ID }) => Parent_ID === '0');
// obtain the root & mutate "arr" by removing the root
const [root] = myArr.splice(rIdx, 1);
// use the helper methods to transform
return recurNamify(recurAdd(myArr, 0, {[root.id]: {}}), myMap);
};
const rawData = [
{ id: 1, Category_Name: "Asset", Parent_ID: 0},
{ id: 2, Category_Name: "Bank", Parent_ID: 1},
{ id: 3, Category_Name: "Cash", Parent_ID: 1},
{ id: 4, Category_Name: "Petty_Cash", Parent_ID: 3},
{ id: 5, Category_Name: "ABC_Bank", Parent_ID: 2},
{ id: 6, Category_Name: "Dollar_Account", Parent_ID: 2}
];
console.log('transformed: ', myTransform(rawData));
.as-console-wrapper { max-height: 100% !important; top: 0 }
说明
在上面的代码段中添加了内联评论。
PS:如果您想为 Whosebug 社区增加价值,
请考虑阅读:What to do when my question is answered
谢谢!
我建议您更改输出对象的设计。我认为数组方法更适合前端。
const rawData = [
{ id: 1, Category_Name: "Asset", Parent_ID: 0},
{ id: 2, Category_Name: "Bank", Parent_ID: 1},
{ id: 3, Category_Name: "Cash", Parent_ID: 1},
{ id: 4, Category_Name: "Petty Cash", Parent_ID: 3},
{ id: 5, Category_Name: "ABC Bank", Parent_ID: 2},
{ id: 6, Category_Name: "Dollar Account", Parent_ID: 2},
{ id: 7, Category_Name: "Another Wallet", Parent_ID: 4},
];
const getParentDeep = (arr, targetId) => arr.find(({ id }) => id === targetId)
?? arr.flatMap(({ children }) => getParentDeep(children, targetId))
.filter(e => e)
.at(0);
const result = rawData
.sort(({ Parent_ID: a }, { Parent_ID: b }) => a - b)
.reduce((acc, { id, Category_Name, Parent_ID }) => {
const obj = { id, name: Category_Name, children: [] };
const parentObj = getParentDeep(acc, Parent_ID);
if (parentObj) parentObj.children.push(obj)
else acc.push(obj);
return acc;
}, []);
console.log(result);
.as-console-wrapper { max-height: 100% !important; top: 0 }
结果将如下所示:
[{
id: 1,
name: "Asset",
children: [{
id: 2,
name: "Bank",
children: [{
id: 5,
name: "ABC Bank",
children: []
}, {
id: 6,
name: "Dollar Account",
children: []
}]
}, {
id: 3,
name: "Cash",
children: [{
id: 4,
name: "Petty Cash",
children: [{
id: 7,
name: "Another Wallet",
children: []
}]
}]
}]
}]
我有一个 SQLite 数据库table
+---------------------------------------------------+
| id | Cat_Name | Parent_ID |
|---------------------------------------------------+
| 1 | Asset | NULL |
+---------------------------------------------------+
| 2 | Bank | 1 |
+---------------------------------------------------+
| 3 | Cash | 1 |
+---------------------------------------------------+
| 4 | Petty Cash | 3 |
+---------------------------------------------------+
| 5 | ABC Bank | 2 |
+---------------------------------------------------+
| 6 | Dollar Account | 2 |
+---------------------------------------------------+
我可以获取如下数据
[{ id: 1, Category_Name: "Asset", Parent_ID: 0},
{ id: 2, Category_Name: "Bank", Parent_ID: 1},
{ id: 3, Category_Name: "Cash", Parent_ID: 1},
{ id: 4, Category_Name: "Petty_Cash", Parent_ID: 3},
{ id: 5, Category_Name: "ABC_Bank", Parent_ID: 2},
{ id: 6, Category_Name: "Dollar_Account", Parent_ID: 2}]
在这个table,用户创建的类别和子类别中,我们无法假设在table
中会有多少个父类别和子类别现在我想将数据作为嵌套 javascript 对象传递给前端
示例
{Asset: {Bank: {ABC Bank: 5}, {Dollar Account: 6}
},
{Cash:{PettyCash: 4}, if any...}
}
任何人都可以帮助以最佳方式获得此结果...
提前致谢
下面介绍的是实现所需 objective 的一种可能方法。不可否认,它不是很优雅(可能不是 most-efficient)。
代码段
// helper method to recursively-add to object
const recurAdd = (arr, idx, res) => {
// when "idx" exceeds length of array "arr",
// simply return existing result "res" object
if (idx >= arr.length) return res;
// de-structure to access parent-id & id for current elt
const { Parent_ID, id } = arr[idx];
if (Parent_ID in res) {
// parent-id exists at current object,
// so, add "id" to same object (mutate)
res[Parent_ID][id] = {};
// make recursive call for "next" elt in "arr"
return recurAdd(arr, idx+1, res);
} else {
// find next-level object where current elt will fit
const foundIt = Object.values(res).map(obj => recurAdd(arr, idx, obj));
// NOTE: "obj" is part of "res" and it gets mutated
// if found, make recursive call
if (foundIt.some(x => x !== false)) return recurAdd(arr, idx+1, res);
};
// in case parent-id is not found, simply return false
return false;
};
// helper method to substitute "id" with "category names"
const recurNamify = (obj, myMap) => (
// reconstruct object from key-value pairs of intermediate result
Object.fromEntries(
// generate intermediate result of key-value pairs
Object.entries(obj)
.map(([k, v]) => (
// substitute key (ie, "id") with category-name
Object.keys(v).length === 0
? [myMap[k], k]
: [myMap[k], recurNamify(v, myMap)]
))
// when "v" is not an empty object, make recursive call
)
);
// transform the array into nested object
const myTransform = arr => {
// first transform "Number" to "string" for id and parent-id
// because JS-object keys are string type
const myArr = arr.map(ob => ({
...ob,
id: ob.id.toString(),
Parent_ID: ob.Parent_ID.toString()
}));
// generate a dictionary/map for "id" to category-name
const myMap = myArr.reduce(
(acc, itm) => {
acc[itm.id] = itm.Category_Name
return acc;
},
{}
);
// find the index of root (ie, parent id is zero)
const rIdx = myArr.findIndex(({ Parent_ID }) => Parent_ID === '0');
// obtain the root & mutate "arr" by removing the root
const [root] = myArr.splice(rIdx, 1);
// use the helper methods to transform
return recurNamify(recurAdd(myArr, 0, {[root.id]: {}}), myMap);
};
const rawData = [
{ id: 1, Category_Name: "Asset", Parent_ID: 0},
{ id: 2, Category_Name: "Bank", Parent_ID: 1},
{ id: 3, Category_Name: "Cash", Parent_ID: 1},
{ id: 4, Category_Name: "Petty_Cash", Parent_ID: 3},
{ id: 5, Category_Name: "ABC_Bank", Parent_ID: 2},
{ id: 6, Category_Name: "Dollar_Account", Parent_ID: 2}
];
console.log('transformed: ', myTransform(rawData));
.as-console-wrapper { max-height: 100% !important; top: 0 }
说明
在上面的代码段中添加了内联评论。
PS:如果您想为 Whosebug 社区增加价值,
请考虑阅读:What to do when my question is answered 谢谢!
我建议您更改输出对象的设计。我认为数组方法更适合前端。
const rawData = [
{ id: 1, Category_Name: "Asset", Parent_ID: 0},
{ id: 2, Category_Name: "Bank", Parent_ID: 1},
{ id: 3, Category_Name: "Cash", Parent_ID: 1},
{ id: 4, Category_Name: "Petty Cash", Parent_ID: 3},
{ id: 5, Category_Name: "ABC Bank", Parent_ID: 2},
{ id: 6, Category_Name: "Dollar Account", Parent_ID: 2},
{ id: 7, Category_Name: "Another Wallet", Parent_ID: 4},
];
const getParentDeep = (arr, targetId) => arr.find(({ id }) => id === targetId)
?? arr.flatMap(({ children }) => getParentDeep(children, targetId))
.filter(e => e)
.at(0);
const result = rawData
.sort(({ Parent_ID: a }, { Parent_ID: b }) => a - b)
.reduce((acc, { id, Category_Name, Parent_ID }) => {
const obj = { id, name: Category_Name, children: [] };
const parentObj = getParentDeep(acc, Parent_ID);
if (parentObj) parentObj.children.push(obj)
else acc.push(obj);
return acc;
}, []);
console.log(result);
.as-console-wrapper { max-height: 100% !important; top: 0 }
结果将如下所示:
[{
id: 1,
name: "Asset",
children: [{
id: 2,
name: "Bank",
children: [{
id: 5,
name: "ABC Bank",
children: []
}, {
id: 6,
name: "Dollar Account",
children: []
}]
}, {
id: 3,
name: "Cash",
children: [{
id: 4,
name: "Petty Cash",
children: [{
id: 7,
name: "Another Wallet",
children: []
}]
}]
}]
}]