我有以下函式,它將從陣列中洗掉物件。它還回傳不包含已洗掉專案的陣列樹。我的問題是,當我的 objToFindBy 為空時,它會起作用,洗掉找到 {group: null} 的所有內容,但是如果我設定 objToFindBy {group: 'some string'},它會因承諾拒絕而出錯
此代碼應洗掉 objToFindBy 匹配的所有匹配項,例如 {group: null} 將在組為空的任何地方找到并洗掉所有物件,然后回傳沒有被洗掉物件的完整樹
findAndDeleteAll(tree, 'items', {group: null}) // work and delete all where match. then returns the tree without deleted objects
findAndDeleteAll(tree, 'items', {group: 'd575c91f-4765-4073-a948-5e305116610c'}) // promise rejection
const tree ={
"type": "app",
"info": "Custom Layout",
"items": [
{
"id": "d575c91f-4765-4073-a948-5e305116610c",
"title": "Fc",
"group": null
},
{
"id": "890d5a1e-3f03-42cd-a695-64a17b6b9bea",
"title": null,
"group": null
},
{
"id": "cbe00537-0bb8-4837-8019-de48cb04edd6",
"title": null,
"group": "d575c91f-4765-4073-a948-5e305116610c",
},
{
"id": "b8751c32-2121-4907-a229-95e3e49bcb39",
"title": null,
"group": "d575c91f-4765-4073-a948-5e305116610c"
}
],
"Children": []
}
var findAndDeleteAll = function findAndDeleteAll(tree, childrenKey, objToFindBy) {
var treeModified = false;
var findKeys = Object.keys(objToFindBy);
var findSuccess = false;
findKeys.forEach(function (key) {
(0, _lodash2.default)(tree[key], objToFindBy[key]) ? findSuccess = true : findSuccess = false;
});
if (findSuccess) {
Object.keys(tree).forEach(function (key) {
return delete tree[key];
});
return tree;
}
function innerFunc(tree, childrenKey, objToFindBy) {
if (tree[childrenKey]) {
var _loop = function _loop(index) {
var findKeys = Object.keys(objToFindBy);
var findSuccess = false;
findKeys.forEach(function (key) {
(0, _lodash2.default)(tree[childrenKey][index][key], objToFindBy[key]) ? findSuccess = true : findSuccess = false;
});
if (findSuccess) {
tree[childrenKey].splice(index, 1);
treeModified = true;
}
if (tree[childrenKey][index].hasOwnProperty(childrenKey)) {
innerFunc(tree[childrenKey][index], childrenKey, objToFindBy);
}
};
for (var index = tree[childrenKey].length - 1; index >= 0; index--) {
_loop(index);
}
}
}
innerFunc(tree, childrenKey, objToFindBy);
return treeModified ? tree : false;
};
uj5u.com熱心網友回復:
更短的解決方案怎么樣?
const findAndDeleteAll = (tree, childrenKey, nestedKey, nestedValue) => {
return{...tree, [childrenKey]: tree[childrenKey].filter((row) => {
return row[nestedKey] !== nestedValue;
})}
}
const a = findAndDeleteAll(tree, 'items', 'group', null) // work and delete all where match. then returns the tree without deleted objects
const b = findAndDeleteAll(tree, 'items', 'group', 'd575c91f-4765-4073-a948-5e305116610c') // promise rejection
console.warn(a);
console.warn(b);
uj5u.com熱心網友回復:
如果您發送的不是多余的tree- 而是deleteFrom(tree.items, "group", null);. 想想看。
const deleteFrom = (arr, pr, val) => arr.filter(ob => ob[pr] !== val);
const tree = {
type: "app",
info: "Custom Layout",
items: [
{ id: "10c", title: "Fc", group: null },
{ id: "bea", title: null, group: null },
{ id: "dd6", title: null, group: "10c" },
{ id: "b39", title: null, group: "10c" },
],
Children: []
};
const items = deleteFrom(tree.items, "group", null);
console.log(items); // Only the filtered items array
const newTree = {...tree, items};
console.log(newTree); // Your brand new tree!
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