我一直在尝试删除嵌套数组中带有ID的元素。
我不确定如何将filter()
与嵌套数组一起使用。
我只想删除{id: 111,name: "A"}
对象。
这是我的代码:
var array = [{
id: 1,
list: [{
id: 123,
name: "Dartanan"
}, {
id: 456,
name: "Athos"
}, {
id: 789,
name: "Porthos"
}]
}, {
id: 2,
list: [{
id: 111,
name: "A"
}, {
id: 222,
name: "B"
}]
}]
var temp = array
for (let i = 0; i < array.length; i++) {
for (let j = 0; j < array[i].list.length; j++) {
temp = temp.filter(function(item) {
return item.list[j].id !== 123
})
}
}
array = temp
答案 0 :(得分:2)
您可以使用函数forEach
并为每个数组filter
执行函数list
。
var array = [{ id: 1, list: [{ id: 123, name: "Dartanan" }, { id: 456, name: "Athos" }, { id: 789, name: "Porthos" }] }, { id: 2, list: [{ id: 111, name: "A" }, { id: 222, name: "B" }] }];
array.forEach(o => (o.list = o.list.filter(l => l.id != 111)));
console.log(array);
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要保持数据不可变,请使用函数map
:
var array = [{ id: 1, list: [{ id: 123, name: "Dartanan" }, { id: 456, name: "Athos" }, { id: 789, name: "Porthos" }] }, { id: 2, list: [{ id: 111, name: "A" }, { id: 222, name: "B" }] }],
result = array.map(o => ({...o, list: o.list.filter(l => l.id != 111)}));
console.log(result);
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答案 1 :(得分:0)
Array.filter
对元素采取行动:
var myArray = [{something: 1, list: [1,2,3]}, {something: 2, list: [3,4,5]}]
var filtered = myArray.filter(function(element) {
return element.something === 1;
// true = keep element, false = discard it
})
console.log(filtered); // logs [{something: 1, list: [1,2,3]}]
你可以像这样使用它:
var array = [{
id: 1,
list: [{
id: 123,
name: "Dartanan"
}, {
id: 456,
name: "Athos"
}, {
id: 789,
name: "Porthos"
}]
}, {
id: 2,
list: [{
id: 111,
name: "A"
}, {
id: 222,
name: "B"
}]
}]
for (var i = 0; i < array.length; ++i) {
var element = array[i]
// Filter the list
element.list = element.list.filter(function(listItem) {
return listItem.id !== 111 && listItem.name !== 'A';
})
}
console.log(array)
答案 2 :(得分:0)
您可以创建一个新数组,其中包含具有已过滤list
属性的元素。
const result = array.map(element => (
{
...element,
list: element.list.filter(l => l.id !== 111)
}
));
如果运行此代码的运行时不支持扩展运算符,则可以使用Object.assign。