Question

下面的代码中有一个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，具体取决于返回的过滤器数据。

最后的输出存储在userColl变量中。

当前我的输出为null，但我期望第一和第三 userList对象；同样的条件是我不想更改usersKey和usersValue。

谢谢。

Answer 1

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

然后使用从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期望布尔谓词进行测试，但是在您的代码forEach中将返回undefined。

第二个键也应该是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 2

您可以使用filter来过滤数组。使用every检查所有元素是否都包含某个字符串。

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);

如果确实需要使用 usersKey 和 usersValue ，则可以使用reduce来汇总过滤器，例如：

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

forEach不返回任何内容-稍微更改代码以使用every和some。我还用三元数对其进行了一些修改：

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);

Answer 4

您可以规范化约束并获取一个数组，因为使用另一个键值数组和其他数组中的所需值很容易迭代。

然后迭代constraints并检查对象中的值是否为数组。根据此检查，迭代值或进行一次检查。

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 5

您可以考虑将用作过滤器参数的变量更改为object或Map()。

请注意，使用变量的方法在某些情况下可能会出现问题，例如，如果某些用户名为IND，会发生什么情况。您如何知道数组usersValue的某些元素与过滤属性user或属性langSpeak有关？

如果使用对象，则可以这样进行：

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;}

使用if else条件

5 个答案: