使用 if else 条件格式化数组内的过滤器函数

Question

在下面的代码中有 usersData 对象，其中包含 usersList 数组。

const usersData = {
  usersList: [
    {
      user: "user1",
      langSpeak: ["IND","SPN"],
    },
    {
      
      user: "user2",
      langSpeak: ["IND"],
    },
    {
     
      user: "user1",
      langSpeak: ["IND", "SPN", "JPN"],
    },
    {
     
      user: "user3",
      langSpeak: ["IND", "SPN"],
    },
    {
      user: "user3"
    }
  ]
};

let usersKey = ["user","langSpeak"];
let usersValue = ["user1","IND","SPN"];
 
 let userColl = usersData.usersList.filter(userObj => {
  
  return usersKey.forEach( usersKey => {
    if(Array.isArray(userObj[usersKey])){
      return userObj[usersKey] && usersValue.forEach(x => {
        console.log("x===" + usersKey)
        return userObj[usersKey].includes(x)
      });
    }
     if(!Array.isArray(userObj[usersKey])){
      console.log(usersKey)
      return usersValue.includes(userObj[usersKey]);
    } 
  
      
  });
});

console.log(userColl); 

有usersKey 和usersValue。

过滤键和值的代码已经存在。

我检查了过滤器函数的值是否包含数组，这取决于返回的过滤器数据。

并且最终输出存储到userColl变量中。

目前我在输出中得到 null，但我期待 first 和 third userList 对象；还有条件是我不想更改usersKey 和usersValue。

提前致谢。

Answer 1

您可以规范化您的约束并获取一个数组，因为它很容易与另一个键值数组和其他数组中的所需值进行迭代。

然后迭代constraints 并检查对象中的值是否为数组。根据此检查迭代值或进行单次检查。

var usersList = [{ user: "user1", langSpeak: ["IND", "SPN"] }, { user: "user2", langSpeak: ["IND"] }, { user: "user1", langSpeak: ["IND", "SPN", "JPN"] }, { user: "user3", langSpeak: ["IND", "SPN"] }, { user: "user3" }],
    constraints = [
        ["user", "user1"],
        ["langSpeak", "IND", "SPN"]
    ],
    result = usersList.filter(o => constraints.every(([key, ...values]) => Array.isArray(o[key])
        ? values.every(v => o[key].includes(v))
        : values.includes(o[key])    
    ));

console.log(result);

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

Answer 2

您可以使用filter 来过滤数组。使用every 检查是否所有元素都包含某个字符串。

const usersData = {"usersList":[{"user":"user1","langSpeak":["IND","SPN"]},{"user":"user2","langSpeak":["IND"]},{"user":"user1","langSpeak":["IND","SPN","JPN"]},{"user":"user3","langSpeak":["IND","SPN"]},{"user":"user3"}]}

let userFilter = "user1";                 //Assign a variable for filtering user
let langFilter = ["IND", "SPN"];          //Assign a variable for filtering language

let result = usersData.usersList.filter(o => o.user == userFilter && o.langSpeak && langFilter.every(e => o.langSpeak.includes(e)));

let finalResult = {                       //Contruct the final object
  usersList: result
}


console.log(finalResult);

---

如果你真的必须使用 usersKey 和 usersValue，你可以使用reduce 来总结过滤器，比如：

const usersData = {"usersList":[{"user":"user1","langSpeak":["IND","SPN"]},{"user":"user2","langSpeak":["IND"]},{"user":"user1","langSpeak":["IND","SPN","JPN"]},{"user":"user3","langSpeak":["IND","SPN"]},{"user":"user3"}]}

let usersKey = ["user", "langSpeak"];
let usersValue = ["user1", "IND", "SPN"];

let filter = usersKey.reduce((c, v, i, a) => {
  c[v] = i + 1 === a.length ? usersValue.slice(i) : [usersValue[i]];
  return c;
}, {});


let result = usersData.usersList.filter(o => Object.entries(filter).every(([k, v]) => v.every(x => o[k].includes(x))));

let finalResult = {
  usersList: result
}

console.log(finalResult);

Answer 3

您可以考虑将用作过滤器参数的变量更改为object，或者Map()。

请注意，在某些情况下，您使用变量的方法可能会出现问题，例如，如果某个用户被命名为 IND，会发生什么情况。您如何知道数组usersValue 的某些元素是否与过滤属性user 或属性langSpeak 相关？

如果你使用一个对象，那么你可以这样进行：

const usersData = {
  usersList: [
    {user: "user1", langSpeak: ["IND","SPN"]},
    {user: "user2", langSpeak: ["IND"]},
    {user: "user1", langSpeak: ["IND", "SPN", "JPN"]},
    {user: "user3", langSpeak: ["IND", "SPN"]},
    {user: "user3"}
  ]
};

let filterObj = {
  "user": ["user1"],
  "langSpeak": ["IND", "SPN"]
};

let userColl = usersData.usersList.filter(userObj =>
{
    return Object.entries(filterObj).every(([k, v]) =>
    {
        let uVal = userObj[k];

        if (Array.isArray(uVal))
            return v.every(x => uVal.includes(x));
        else
            return v.includes(uVal);
    });
});

console.log(userColl);

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

Answer 4

您应该使用Array.every 来检查给定数组中是否存在所需数据，而不是使用Array.forEach。

然后使用every调用返回的布尔值作为过滤条件来过滤数据。

Array.forEach 不返回任何内容，但 Array.filter 需要一个布尔谓词进行测试，但在您的代码中 forEach 将返回 undefined。

另外第二个键应该是langSpeak 而不是code：

const usersData = {usersList:[{user:"user1",langSpeak:["IND","SPN"]},{user:"user2",langSpeak:["IND"]},{user:"user1",langSpeak:["IND","SPN","JPN"]},{user:"user3",langSpeak:["IND","SPN"]},{user:"user3"}]};

let usersKey = ["user","langSpeak"];
let langValues = ["IND","SPN","JPN"];
let userValues = ["user1"];

let userColl = usersData.usersList.filter(userObj => {
return usersKey.every( usersKey => {
    if(Array.isArray(userObj[usersKey])){
      return langValues.every(x => {
          return userObj[usersKey].includes(x)
      });
    }
    else{
      return userValues.every(user => user === userObj[usersKey]);
    } 
 });
});

console.log(userColl);

Answer 5

forEach 不返回任何内容 - 稍微更改您的代码以使用 every 和 some。我还用三元组稍微修改了它：

const usersData = {
  usersList: [{
      user: "user1",
      langSpeak: ["IND", "SPN"],
    },
    {

      user: "user2",
      langSpeak: ["IND"],
    },
    {

      user: "user1",
      langSpeak: ["IND", "SPN", "JPN"],
    },
    {

      user: "user3",
      langSpeak: ["IND", "SPN"],
    },
    {
      user: "user3"
    }
  ]
};

let usersKey = ["user", "langSpeak"];
let usersValue = ["user1", "IND", "SPN"];

let userColl = usersData.usersList.filter(userObj => usersKey.every(k => Array.isArray(userObj[k]) ? usersValue.some(v => userObj[k].includes(v)) : usersValue.includes(userObj[k])));

console.log(userColl);