如何删除嵌套在另一个数组中的数组subBrands中的对象，其中对象的id属性为= 31。我试图在不删除该子标记的情况下返回整个父数组。
数组是：
[
{
"id": 10,
"name": "Parent Brand 1",
"parent": null,
"author": 1,
"deleted_at": null,
"created_at": "2017-02-02 09:55:51",
"updated_at": "2017-02-02 09:55:51",
"subBrands": [
{
"id": 31,
"name": "Sub Brand 6",
"parent": 10,
"author": 1,
"deleted_at": null,
"created_at": "2017-02-02 11:24:49",
"updated_at": "2017-02-02 11:42:02"
},
{
"id": 32,
"name": "Sub Brand 7",
"parent": 10,
"author": 1,
"deleted_at": null,
"created_at": "2017-02-02 11:24:57",
"updated_at": "2017-02-02 11:42:18"
},
{
"id": 33,
"name": "Sub Brand 8",
"parent": 10,
"author": 1,
"deleted_at": null,
"created_at": "2017-02-02 11:25:04",
"updated_at": "2017-02-02 11:42:34"
},
{
"id": 34,
"name": "Sub Brand 9",
"parent": 10,
"author": 1,
"deleted_at": null,
"created_at": "2017-02-02 11:25:39",
"updated_at": "2017-02-02 11:42:43"
},
{
"id": 35,
"name": "Sub Brand 10",
"parent": 10,
"author": 1,
"deleted_at": null,
"created_at": "2017-02-02 11:25:46",
"updated_at": "2017-02-02 11:42:52"
},
{
"id": 36,
"name": "Sub Brand 4",
"parent": 10,
"author": 1,
"deleted_at": null,
"created_at": "2017-02-02 11:43:53",
"updated_at": "2017-02-02 11:43:53"
}
]
},
{
"id": 12,
"name": "Parent Brand 2",
"parent": null,
"author": 1,
"deleted_at": null,
"created_at": "2017-02-02 09:56:16",
"updated_at": "2017-02-02 09:56:16",
"subBrands": []
},
{
"id": 16,
"name": "Brand no children",
"parent": null,
"author": 1,
"deleted_at": null,
"created_at": "2017-02-02 10:37:40",
"updated_at": "2017-02-02 10:37:40",
"subBrands": []
},
{
"id": 37,
"name": "Whoops brand",
"parent": null,
"author": 1,
"deleted_at": null,
"created_at": "2017-02-02 11:44:10",
"updated_at": "2017-02-02 11:44:10",
"subBrands": []
}
]
我想要得到的是：
[
{
"id": 10,
"name": "Parent Brand 1",
"parent": null,
"author": 1,
"deleted_at": null,
"created_at": "2017-02-02 09:55:51",
"updated_at": "2017-02-02 09:55:51",
"subBrands": [
{
"id": 32,
"name": "Sub Brand 7",
"parent": 10,
"author": 1,
"deleted_at": null,
"created_at": "2017-02-02 11:24:57",
"updated_at": "2017-02-02 11:42:18"
},
{
"id": 33,
"name": "Sub Brand 8",
"parent": 10,
"author": 1,
"deleted_at": null,
"created_at": "2017-02-02 11:25:04",
"updated_at": "2017-02-02 11:42:34"
},
{
"id": 34,
"name": "Sub Brand 9",
"parent": 10,
"author": 1,
"deleted_at": null,
"created_at": "2017-02-02 11:25:39",
"updated_at": "2017-02-02 11:42:43"
},
{
"id": 35,
"name": "Sub Brand 10",
"parent": 10,
"author": 1,
"deleted_at": null,
"created_at": "2017-02-02 11:25:46",
"updated_at": "2017-02-02 11:42:52"
},
{
"id": 36,
"name": "Sub Brand 4",
"parent": 10,
"author": 1,
"deleted_at": null,
"created_at": "2017-02-02 11:43:53",
"updated_at": "2017-02-02 11:43:53"
}
]
},
{
"id": 12,
"name": "Parent Brand 2",
"parent": null,
"author": 1,
"deleted_at": null,
"created_at": "2017-02-02 09:56:16",
"updated_at": "2017-02-02 09:56:16",
"subBrands": []
},
{
"id": 16,
"name": "Brand no children",
"parent": null,
"author": 1,
"deleted_at": null,
"created_at": "2017-02-02 10:37:40",
"updated_at": "2017-02-02 10:37:40",
"subBrands": []
},
{
"id": 37,
"name": "Whoops brand",
"parent": null,
"author": 1,
"deleted_at": null,
"created_at": "2017-02-02 11:44:10",
"updated_at": "2017-02-02 11:44:10",
"subBrands": []
}
]
我打开使用下划线。我最接近的是：
var brands = _.filter(brands, function(n) {
return _.some(n.subBrands, function(subBrand){
return subBrand.id != brand.id;
});
});
但是这会删除不包含id为31的子品牌的数组。所以它与我所需要的不太接近。
干杯！
答案 0 :(得分：1)
你可以迭代父部分和子部分，如果找到则拼接对象。
var data = [{ id: 10, name: "Parent Brand 1", parent: null, author: 1, deleted_at: null, created_at: "2017-02-02 09:55:51", updated_at: "2017-02-02 09:55:51", subBrands: [{ id: 31, name: "Sub Brand 6", parent: 10, author: 1, deleted_at: null, created_at: "2017-02-02 11:24:49", updated_at: "2017-02-02 11:42:02" }, { id: 32, name: "Sub Brand 7", parent: 10, author: 1, deleted_at: null, created_at: "2017-02-02 11:24:57", updated_at: "2017-02-02 11:42:18" }, { id: 33, name: "Sub Brand 8", parent: 10, author: 1, deleted_at: null, created_at: "2017-02-02 11:25:04", updated_at: "2017-02-02 11:42:34" }, { id: 34, name: "Sub Brand 9", parent: 10, author: 1, deleted_at: null, created_at: "2017-02-02 11:25:39", updated_at: "2017-02-02 11:42:43" }, { id: 35, name: "Sub Brand 10", parent: 10, author: 1, deleted_at: null, created_at: "2017-02-02 11:25:46", updated_at: "2017-02-02 11:42:52" }, { id: 36, name: "Sub Brand 4", parent: 10, author: 1, deleted_at: null, created_at: "2017-02-02 11:43:53", updated_at: "2017-02-02 11:43:53" }] }, { id: 12, name: "Parent Brand 2", parent: null, author: 1, deleted_at: null, created_at: "2017-02-02 09:56:16", updated_at: "2017-02-02 09:56:16", subBrands: [] }, { id: 16, name: "Brand no children", parent: null, author: 1, deleted_at: null, created_at: "2017-02-02 10:37:40", updated_at: "2017-02-02 10:37:40", subBrands: [] }, { id: 37, name: "Whoops brand", parent: null, author: 1, deleted_at: null, created_at: "2017-02-02 11:44:10", updated_at: "2017-02-02 11:44:10", subBrands: [] }];
data.some(function (a) {
return a.subBrands.some(function (b, i, bb) {
if (b.id === 31) {
bb.splice(i, 1);
return true;
}
});
});
console.log(data);.as-console-wrapper { max-height: 100% !important; top: 0; }
答案 1 :(得分：1)
var arr = [{"id":10,"name":"Parent Brand 1","parent":null,"author":1,"deleted_at":null,"created_at":"2017-02-02 09:55:51","updated_at":"2017-02-02 09:55:51","subBrands":[{"id":31,"name":"Sub Brand 6","parent":10,"author":1,"deleted_at":null,"created_at":"2017-02-02 11:24:49","updated_at":"2017-02-02 11:42:02"},{"id":32,"name":"Sub Brand 7","parent":10,"author":1,"deleted_at":null,"created_at":"2017-02-02 11:24:57","updated_at":"2017-02-02 11:42:18"},{"id":33,"name":"Sub Brand 8","parent":10,"author":1,"deleted_at":null,"created_at":"2017-02-02 11:25:04","updated_at":"2017-02-02 11:42:34"},{"id":34,"name":"Sub Brand 9","parent":10,"author":1,"deleted_at":null,"created_at":"2017-02-02 11:25:39","updated_at":"2017-02-02 11:42:43"},{"id":35,"name":"Sub Brand 10","parent":10,"author":1,"deleted_at":null,"created_at":"2017-02-02 11:25:46","updated_at":"2017-02-02 11:42:52"},{"id":36,"name":"Sub Brand 4","parent":10,"author":1,"deleted_at":null,"created_at":"2017-02-02 11:43:53","updated_at":"2017-02-02 11:43:53"}]},{"id":12,"name":"Parent Brand 2","parent":null,"author":1,"deleted_at":null,"created_at":"2017-02-02 09:56:16","updated_at":"2017-02-02 09:56:16","subBrands":[]},{"id":16,"name":"Brand no children","parent":null,"author":1,"deleted_at":null,"created_at":"2017-02-02 10:37:40","updated_at":"2017-02-02 10:37:40","subBrands":[]},{"id":37,"name":"Whoops brand","parent":null,"author":1,"deleted_at":null,"created_at":"2017-02-02 11:44:10","updated_at":"2017-02-02 11:44:10","subBrands":[]}];
var id = prompt("Id of subbrands to remove: ");
arr.forEach(function(o) {
o.subBrands = o.subBrands.filter(s => s.id != id);
});
console.log(arr);
答案 2 :(得分：0)
如果您只是需要查看＆＃34;子品牌＆＃34;您的数组的每个项目的属性，并且您使用下划线，这适用：
var myBrand = _.each(brands, function(brand) {
brand.subBrands = _.filter(brand.subBrands, function(subBrand) {
return subBrand.id != 31;
});
});
答案 3 :(得分：0)
foreach似乎有效：
var brands=[{id:10,name:"Parent Brand 1",parent:null,author:1,deleted_at:null,created_at:"2017-02-02 09:55:51",updated_at:"2017-02-02 09:55:51",subBrands:[{id:31,name:"Sub Brand 6",parent:10,author:1,deleted_at:null,created_at:"2017-02-02 11:24:49",updated_at:"2017-02-02 11:42:02"},{id:32,name:"Sub Brand 7",parent:10,author:1,deleted_at:null,created_at:"2017-02-02 11:24:57",updated_at:"2017-02-02 11:42:18"},{id:33,name:"Sub Brand 8",parent:10,author:1,deleted_at:null,created_at:"2017-02-02 11:25:04",updated_at:"2017-02-02 11:42:34"},{id:34,name:"Sub Brand 9",parent:10,author:1,deleted_at:null,created_at:"2017-02-02 11:25:39",updated_at:"2017-02-02 11:42:43"},{id:35,name:"Sub Brand 10",parent:10,author:1,deleted_at:null,created_at:"2017-02-02 11:25:46",updated_at:"2017-02-02 11:42:52"},{id:36,name:"Sub Brand 4",parent:10,author:1,deleted_at:null,created_at:"2017-02-02 11:43:53",updated_at:"2017-02-02 11:43:53"}]},{id:12,name:"Parent Brand 2",parent:null,author:1,deleted_at:null,created_at:"2017-02-02 09:56:16",updated_at:"2017-02-02 09:56:16",subBrands:[]},{id:16,name:"Brand no children",parent:null,author:1,deleted_at:null,created_at:"2017-02-02 10:37:40",updated_at:"2017-02-02 10:37:40",subBrands:[]},{id:37,name:"Whoops brand",parent:null,author:1,deleted_at:null,created_at:"2017-02-02 11:44:10",updated_at:"2017-02-02 11:44:10",subBrands:[]}];
brands.forEach(function(brand) {
brand.subBrands = brand.subBrands.filter(function(subBrand){
return subBrand.id != 31;
})
});
console.log(brands);
答案 4 :(得分：0)
jQuery方式
function removeById(data, id){
$(data).each(function(i, e){
if(e.subBrands.length > 0){
$(e.subBrands).each(function(_i, _e){
if(_e.id == id){
e.subBrands.splice(_i,1);
return false;
}
});
}
});
return data;
}
console.log(removeById(data,32))
此函数将返回没有指定id对象的整个数据数组
答案 5 :(得分：0)
您可以使用for..of循环来迭代每个subBrands数组，Array.prototype.splice()以从数组中删除具有id 31的对象。
for (let {subBrands} of data) {
let n = 0;
for (let {id} of subBrands) {
if (id === 31) {
subBrands.splice(n, 1);
break;
}
++n;
}
}
答案 6 :(得分：0)
我嵌套了两个forEach循环，并在找到并删除该项目后返回：
let items = [{id: 1, subItems: [{id: 1}, {id: 2}]}];
const subItemToBeRemovedId = 1;
items.forEach((item) => item.subItems.forEach((subItem, index) => {
if (subItem.id === subItemToBeRemovedId) {
return item.subItems.splice(index, 1);
}
}));
console.log(items);