使用 .filter() 函数过滤嵌套数组

Question

我有一个看起来像这样的数组：

const data = [
 {"total": 24,"items":[{"id":1,"name":"foo1", "items": [{"label": "TEST", "total": 50}, {"label": "TEST2", "total": 50}]}]},
{"total": 25,"items":[{"id":2,"name":"foo2", "items": [{"label": "FOO", "total": 60}, {"label": "ANOTHER2", "total": 50}]}]},
{"total": 26,"items":[{"id":3,"name":"foo3", "items": [{"label": "BAR", "total": 70}, {"label": "LAST2", "total": 50}]}]},
];

我想在 angular 中使用 .filter() 函数，所以当我传入字符串 '2' 时，它返回只包含包含字符串 '2' 的嵌套数组的对象，换句话说，应该返回这个：

[
 {"total": 24,"items":[{"id":1,"name":"foo1", "items": [{"label": "TEST2", "total": 50}]}]},
{"total": 25,"items":[{"id":2,"name":"foo2", "items": [{"label": "ANOTHER2", "total": 50}]}]},
{"total": 26,"items":[{"id":3,"name":"foo3", "items": [{"label": "LAST2", "total": 50}]}]},
];

我试过了

let values = data.items.filter(d => d.items.forEach(c => c.label.toLowerCase().includes(searchText.toLowerCase())
}))

它只返回一个空数组，

我也试过

 let values = data.items.forEach(d => d.items.filter(c => c.label.toLowerCase().includes(searchText.toLowerCase())
})) 

和

let values = data.items.filter(d => d.items.every(c => c.label.toLowerCase().includes(searchText.toLowerCase())
}))

并且都返回 undefined 和一个空数组。

这样做的正确方法是什么？

Answer 1

 const data = [ {"total": 24,"items":[{"id":1,"name":"foo1", "items": [{"label": "TEST", "total": 50}, {"label": "TEST2", "total": 50}]}]}, {"total": 25,"items":[{"id":2,"name":"foo2", "items": [{"label": "FOO", "total": 60}, {"label": "ANOTHER2", "total": 50}]}]}, {"total": 26,"items":[{"id":3,"name":"foo3", "items": [{"label": "BAR", "total": 70}, {"label": "LAST2", "total": 50}]}]}, ]; let searchText='TEST' let values = JSON.parse(JSON.stringify(data)).map(d => { d.items.map(i=> { i.items = i.items.filter(c =>c.label.toLowerCase().includes(searchText.toLowerCase())); return i; }) return d; }) console.log(values)

let values =  JSON.parse(JSON.stringify(data)).map(d => {
d.items.map(i=> {
i.items = i.items.filter(c =>c.label.toLowerCase().includes(searchText.toLowerCase()));
return i;
})
return d;
})

forEach不返回任何内容， map返回数组，其中每个元素都被您传递给它的函数的返回值替换。此外，您访问的深度不够，因为您的目标数组位于 data[0]。 items[0].items，例如。 我可能在这里错过了括号或括号