我有一個這種型別的 json 物件。
{
"id": "001",
"type": "A",
"value": "aaaaa",
"data:": {},
"path": ["001"],
"children": [
{
"id": "002",
"type": "A",
"value": "aaaaa",
"data:": {},
"path": ["001", "002"],
"children": []
},
{
"id": "003",
"type": "A",
"value": "aaaaa",
"data:": {},
"path": ["001", "003"],
"children": [
{
"id": "00001",
"type": "B",
"children": []
}
]
},
{
"id": "004",
"type": "A",
"value": "aaaaa",
"data:": {},
"path": ["001", "004"],
"children": [
{
"id": "005",
"type": "A",
"value": "aaaaa",
"data:": {},
"path": ["001", "004", "005"],
"children": []
},{
"id": "005",
"type": "A",
"value": "aaaaa",
"data:": {},
"path": ["001", "004", "005"],
"children": [
{
"id": "00002",
"type": "B",
"children": []
}
]
}
]
},
{
"id": "00003",
"type": "B",
"children": []
}
]
}
我需要type: "B"用這個（下面提到的）型別的物件替換所有的物件，我可以從一個具有 ids 作為型別 B 的鍵的物件中獲取該型別的物件。這個型別 B 物件可以嵌套在任何地方作為第一個孩子或嵌套子陣列的第五個子元素
{
"id": "002",
"type": "A",
"value": "aaaaa",
"data:": {},
"children": []
},
我怎樣才能做到這一點？這可以是深度嵌套的，并且沒有我們應該事先替換物件的特定位置。所以，我需要遍歷整個物件并做到這一點。我應該如何完成它？
編輯
我稍微更新了問題中的代碼。除了型別化的B物件外，每個物件中都有一個嵌套的路徑屬性。因此，當用另一個物件替換鍵入的 B 屬性時，我還需要在其中添加路徑。
例如：id 的路徑：“00001”，鍵入的 B 物件應為：[“001”、“003”、“00001”]
編輯：預期結果
{
"id": "001",
"type": "A",
"value": "aaaaa",
"data:": {},
"path": ["001"],
"children": [
{
"id": "002",
"type": "A",
"value": "aaaaa",
"data:": {},
"path": ["001", "002"],
"children": []
},
{
"id": "003",
"type": "A",
"value": "aaaaa",
"data:": {},
"path": ["001", "003"],
"children": [
{
"id": "002",
"type": "A",
"value": "aaaaa",
"data:": {},
"path": ["001", "003", "002"],
"children": []
},
]
},
{
"id": "004",
"type": "A",
"value": "aaaaa",
"data:": {},
"path": ["001", "004"],
"children": [
{
"id": "005",
"type": "A",
"value": "aaaaa",
"data:": {},
"path": ["001", "004", "005"],
"children": []
},{
"id": "005",
"type": "A",
"value": "aaaaa",
"data:": {},
"path": ["001", "004", "005"],
"children": [
{
"id": "002",
"type": "A",
"value": "aaaaa",
"data:": {},
"path": ["001", "004", "005", "002"],
"children": []
}
]
}
]
},
{
"id": "002",
"type": "A",
"value": "aaaaa",
"data:": {},
"path": ["001", "002"],
"children": []
}
]
}
uj5u.com熱心網友回復：
如果你不介意的話。
使用cloneDeepWith克隆整個樹并替換特定值。
const data = {"id":"001","type":"A","value":"aaaaa","data:":{},"children":[{"id":"002","type":"A","value":"aaaaa","data:":{},"children":[]},{"id":"003","type":"A","value":"aaaaa","data:":{},"children":[{"id":"00001","type":"B","children":[]}]},{"id":"004","type":"A","value":"aaaaa","data:":{},"children":[{"id":"005","type":"A","value":"aaaaa","data:":{},"children":[]},{"id":"005","type":"A","value":"aaaaa","data:":{},"children":[{"id":"00002","type":"B","children":[]}]}]},{"id":"00003","type":"B","children":[]}]};
const result = _.cloneDeepWith(data, (value) => {
const newObj = {"id": "002", "type": "A", "value": "---NEW VALUE FOR 'B' TYPE---", "data:": {} };
return (value.type === 'B') ? { ...value, ...newObj} : _.noop();
});
console.dir(result, { depth: null } );.as-console-wrapper{min-height: 100%!important; top: 0}<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.21/lodash.js" integrity="sha512-2iwCHjuj PmdCyvb88rMOch0UcKQxVHi/gsAml1fN3eg82IDaO/cdzzeXX4iF2VzIIes7pODE1/G0ts3QBwslA==" crossorigin="anonymous" referrerpolicy="no-referrer"></script>--- 更新 2---（沒有 lodash）
使用區域變數來存盤和組合當前路徑。
const data = { "id": "001", "type": "A", "value": "aaaaa", "data:": {}, "path": ["001"], "children": [{ "id": "002", "type": "A", "value": "aaaaa", "data:": {}, "path": ["001", "002"], "children": [] }, { "id": "003", "type": "A", "value": "aaaaa", "data:": {}, "path": ["001", "003"], "children": [{ "id": "00001", "type": "B", "children": [] }] }, { "id": "004", "type": "A", "value": "aaaaa", "data:": {}, "path": ["001", "004"], "children": [{ "id": "005", "type": "A", "value": "aaaaa", "data:": {}, "path": ["001", "004", "005"], "children": [] }, { "id": "005", "type": "A", "value": "aaaaa", "data:": {}, "path": ["001", "004", "005"], "children": [{ "id": "00002", "type": "B", "children": [] }] }] }, { "id": "00003", "type": "B", "children": [] }] }
const deepReplace = (obj, prevPath = []) => {
if (obj.type === 'A') {
if (obj.children.length) {
obj.children = obj.children.map((childObj) => deepReplace(childObj, obj.path))
}
return obj;
};
if (obj.type === 'B') {
const id = '002';
return { id, type: "A", value: "aaaaa", path: [...prevPath, id], data: {}, children: []};
};
};
console.dir(deepReplace(data), { depth: null });.as-console-wrapper{min-height: 100%!important; top: 0}uj5u.com熱心網友回復：
我在控制臺中玩了這個，它做了你想要的（基于提供的 json 陣列，將所有“B”設定為“A”型別）。它是一個遞回函式，因此對于它在“children”陣列中遇到的任何嵌套子元素，它都會在陣列中的每個專案上再次呼叫該函式。
function fixObjects (obj) {
if (obj["type"] === "B") {
obj["type"] = "A";
obj["id"] = "002";
obj["value"] = "aaaaa";
obj["data"] = {};
}
if (obj["children"].length > 0) {
obj["children"].forEach (child => fixObjects (child));
}
}
fixObjects (_yourArray)
uj5u.com熱心網友回復：
這看起來像一個樹遍歷問題。這是一種無需遞回即可通過深度優先搜索處理該問題的方法。
如本答案所述，應盡可能避免遞回，因為它需要更多記憶體并且比迭代實作更難除錯。
根據相關調整更新
const data = {"id":"001","type":"A","value":"aaaaa","data:":{},"path":["001"],"children":[{"id":"002","type":"A","value":"aaaaa","data:":{},"path":["001","002"],"children":[]},{"id":"003","type":"A","value":"aaaaa","data:":{},"path":["001","003"],"children":[{"id":"00001","type":"B","children":[]}]},{"id":"004","type":"A","value":"aaaaa","data:":{},"path":["001","004"],"children":[{"id":"005","type":"A","value":"aaaaa","data:":{},"path":["001","004","005"],"children":[]},{"id":"005","type":"A","value":"aaaaa","data:":{},"path":["001","004","005"],"children":[{"id":"00002","type":"B","children":[]}]}]},{"id":"00003","type":"B","children":[]}]};
const dfs = () => {
const stack = [[data, null]];
while(stack.length) {
const [curr, parent] = stack.pop();
// check for match on type
if (curr.type === "B") {
curr.type = "A";
curr.id = "002";
curr.value = "aaaaa";
curr.data = {};
curr.path = [...parent?.path.slice() ?? [], "002"];
}
curr.children.forEach(child => stack.push([child, curr]));
}
};
dfs();
const output = document.getElementById("output");
output.innerText = JSON.stringify(data, null, 2);<pre id="output" />轉載請註明出處，本文鏈接：https://www.uj5u.com/qukuanlian/411931.html
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