Question

我想一一过滤具有多个搜索条件的对象数组。

假设第一个过滤条件是“ x”，然后应用“ x + y”等。只要x或“ x + y”与对象数组匹配，就会显示匹配的结果。

如果用户要删除“ x”或“ y”，则数组应做出相应的响应。

例如：这里 x = mumbai 然后是 y = 25

   var users = [{
   name: 'nitin',
   email: 'nitin@mail.com',
   age: 25,
   address: 'mumbai'
   },
   {
    name: 'pawan',
    email: 'pan@mail.com',
    age: 35,
    address: 'kolkata'
   },
   {
    name: 'sudip',
    email: 'sudip@mail.com',
    age: 28,
    address: 'mumbai'
   }];

  this.searchGridText = "";
  this.users = this.users.filter((data) => {
  const stringified = JSON.stringify(data);
  const filtered = stringified.replace(/"\w+":/gm, '');

  return (filtered.toLowerCase()
    .indexOf(this.searchGridText.toLowerCase().trim()) !== -1);

  });

当搜索条件为“孟买”时，应返回两条记录，然后搜索条件为25，则应返回一条记录。如果删除了孟买搜索条件，则仅显示与25匹配的记录。搜索条件将在文本框中输入按下Enter键后，过滤器将应用于对象数组。

Answer 1

这是通过Array.filter，Array.every，Array.some和Object.values进行操作的一种方式：

var users = [{ name: 'nitin', email: 'nitin@mail.com', age: 25, address: 'mumbai' }, { name: 'pawan', email: 'pan@mail.com', age: 35, address: 'kolkata' }, { name: 'sudip', email: 'sudip@mail.com', age: 28, address: 'mumbai' } ]; 

let findUsers = (arr, filters) => {
  let fs = filters.split('+').map(x => x.trim().toLowerCase())
  return arr.filter(x => {
    let vs = Object.values(x)
    return fs.every(f => vs.some(v => v.toString().trim().toLowerCase().includes(f)))
  })
}

console.log(findUsers(users, 'mumbai'))
console.log(findUsers(users, 'mumbai + 25'))
console.log(findUsers(users, 'sudip'))

想法是过滤并在内部获取每个对象的值，并对其进行迭代，以尝试将传递给函数过滤器的内容进行匹配。

Answer 2

您可以使用相同的想法，但可以将其转变为功能，这将通过您所寻找的特征逐渐加以区分

var users = [{
    name: 'nitin',
    email: 'nitin@mail.com',
    age: 25,
    address: 'mumbai'
  },
  {
    name: 'pawan',
    email: 'pan@mail.com',
    age: 35,
    address: 'kolkata'
  },
  {
    name: 'sudip',
    email: 'sudip@mail.com',
    age: 28,
    address: 'mumbai'
  }
];

this.searchGridText = "mumbai 25";

let filters = searchGridText.split(" ");
let res = [];
for (let i = 0; i < filters.length; i++) {
  res = filtereds(filters[i]);

  users = res;
}


function filtereds(c) {

  this.users = this.users.filter((data) => {
    const stringified = JSON.stringify(data);
    const filtered = stringified.replace(/"\w+":/gm, '');
    //console.log(c);
    return (filtered.toLowerCase()
      .indexOf(c.toLowerCase().trim()) !== -1);

  });
  return this.users;
}



console.log(users);

Answer 3

创建一个将过滤键作为数组并根据这些值进行过滤的函数。您可以使用.split()将输入的值拆分为数组，并将其传递给函数。

function searchJson(array, keys) {

  return array.filter(item => {
    const string = item && typeof item === 'object' && JSON.stringify(item).toLowerCase();

    return keys.filter(key => string && string.includes(key.toString().toLowerCase())).length === keys.length;
  })
}

var users = [{
   name: 'nitin',
   email: 'nitin@mail.com',
   age: 25,
   address: 'mumbai'
   },
   {
    name: 'pawan',
    email: 'pan@mail.com',
    age: 35,
    address: 'kolkata'
   },
   {
    name: 'sudip',
    email: 'sudip@mail.com',
    age: 28,
    address: 'mumbai'
   }];
   
   function searchJson(array, keys) {

  return array.filter(item => {
    const string = item && typeof item === 'object' && JSON.stringify(item).toLowerCase();

    return keys.filter(key => string && string.includes(key.toString().toLowerCase())).length === keys.length;
  })
}

console.log(searchJson(users, ['mumbai', 25]))

Answer 4

没有字符串化，没有正则表达式，也没有问题：

function search(array, query) {
    let filters = query.split("+").map(x => x.trim().toLowerCase())
    return users.filter(user => {
        for (let filter of filters) {
            let match = false
            for (let key in user) {
                let value = user[key]
                if (value && value.toString().toLowerCase().includes(filter)) {
                    match = true
                    break
                }
            }
            if (!match) { return false }
        }
        return true
    })
}

let users = [{
    name: undefined,
    email: 'nitin@mail.com',
    age: 25,
    address: 'mumbai'
}, {
    name: 'pawan',
    email: 'pan@mail.com',
    age: 35,
    address: 'kolkata'
}, {
    name: 'sudip',
    email: 'sudip@mail.com',
    age: null,
    address: 'mumbai'
}]

let result = search(users, "Mumbai + 25");

console.log(result)

一对一过滤JavaScript数组添加/删除条件

4 个答案: