如果对象的值匹配数组，如何过滤对象数组[重复]

Question

我有以下对象表：

 let data = [
   {key:"20-09-2019", skill: [{id: 500, message: "monday"}, {id: 501, message: "tuesday"}]},
   {key:"21-09-2019", skill: [{id: 502, message: "thursday"}, {id: 503, message: "sunday"}]},
   {key:"22-09-2019", skill: [{id: 504, message: "sunday"}]},
]

让搜索 = "太阳"

如果“message”的值与“search”的值匹配，我想过滤表

如果 search = "sun" 过滤器应该返回以下结果：

过滤后的结果：

 data = [
   {key:"21-09-2019", skill: [ {id: 503, message: "sunday"}]},
   {key:"22-09-2019", skill: [{id: 504, message: "sunday"}]},
]

这里的数组只返回消息值与“sun”匹配的对象

我知道过滤方法，但我认为我们不能在过滤器中进行过滤。

我也知道匹配消息的方法：

 message.toLowerCase().includes(search);

但我不知道如何过滤对象数组，如果有人有想法吗？

Answer 1

您可以先通过搜索值对每个元素的skill字段进行过滤，然后过滤仍然有技能的元素。

let search = "sun";
let data = [
   {key:"20-09-2019", skill: [{id: 500, message: "monday"}, {id: 501, message: "tuesday"}]},
   {key:"21-09-2019", skill: [{id: 502, message: "thursday"}, {id: 503, message: "sunday"}]},
   {key:"22-09-2019", skill: [{id: 504, message: "sunday"}]},
];
data.forEach(el => {
  el.skill = el.skill.filter(s => s.message.toLowerCase().includes(search))
});
let result = data.filter(el => el.skill.length);
console.log(result)

编辑不改变原始数组：

let search = "sun";
let data = [
  {key:"20-09-2019", skill: [{id: 500, message: "monday"}, {id: 501, message: "tuesday"}]},
  {key:"21-09-2019", skill: [{id: 502, message: "thursday"}, {id: 503, message: "sunday"}]},
  {key:"22-09-2019", skill: [{id: 504, message: "sunday"}]},
    ];
let result = data.reduce((acc, {skill, ...rest}) => {
   skill = skill.filter(s => s.message.toLowerCase().includes(search));
   if(skill.length) acc.push({skill, ...rest});
   return acc;
}, []);
console.log(result);

Answer 2

您可以在不改变原始数据的情况下创建新对象，并且只返回包含所需子字符串的部分。

function filter(array, value) {
    function subFind(array, [key, ...keys]) {
        return keys.length
            ? array
                  .map(o => {
                      var temp = subFind(o[key], keys);
                      return temp.length && Object.assign({}, o, { [key]: temp });
                  })
                  .filter(Boolean)
            : array.filter(o => o[key].includes(value));
    }
    return subFind(array, ['skill', 'message']);
}

let data = [{ key: "20-09-2019", skill: [{ id: 500, message: "monday" }, { id: 501, message: "tuesday" } ]}, { key: "21-09-2019", skill: [{ id: 502, message: "thursday" }, { id: 503, message: "sunday" } ]}, { key: "22-09-2019", skill: [{ id: 504, message: "sunday" }] }];

console.log(filter(data, 'sun'));

.as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 3

我不认为我们可以在过滤器中做一个过滤器。

是的，你可以，但我会使用forEach()。下面的例子。

let data = [{
    key: "20-09-2019",
    skill: [{ id: 500, message: "monday" }, { id: 501, message: "tuesday" }]
  },
  {
    key: "21-09-2019",
    skill: [ { id: 502, message: "thursday" }, { id: 503, message: "sunday" }]
  },
  { 
    key: "22-09-2019", 
    skill: [{ id: 504, message: "sunday" }] 
  }
];

let search = "sunday";
let result = [];

data.forEach(row => {
  let found = row.skill.filter(skill => skill.message === search);
  if (found.length) result.push({ key: row.key, skill: found });
});

console.log(result);

Answer 4

你可以这样做

let data = [
   {key:"20-09-2019", skill: [{id: 500, message: "monday"}, {id: 501, message: "tuesday"}]},
   {key:"21-09-2019", skill: [{id: 502, message: "thursday"}, {id: 503, message: "sunday"}]},
   {key:"22-09-2019", skill: [{id: 504, message: "sunday"}]},
];

let search = "sun"

const newItems = data.filter(item => item.skill.filter(sk => sk.message.includes(search)).length);
newItems.forEach(item => {
  item.skill = item.skill.filter(x => x.message.includes(search))
})