根据所有属性值过滤对象数组 [英] Filter array of objects on all properties value

问题描述

我真的很惊讶我找不到与我的问题相关的任何内容.我正在寻找一种快速的方法来根据用户文本输入过滤我的对象数组.

I am really surprised I haven't been able to find anything related to my question. I am looking for a fast way to filter my array of objects based on a user text input.

假设我有这个数组:

let data = [{
  "id": 1,
  "first_name": "Jean",
  "last_name": "Owens",
  "email": "[e-mail]",
  "gender": "Female"
}, {
  "id": 2,
  "first_name": "Marie",
  "last_name": "Morris",
  "email": "[e-mail]",
  "gender": "Female"
}, {
  "id": 3,
  "first_name": "Larry",
  "last_name": "Wallace",
  "email": "[e-mail]",
  "gender": "Male"
}];

用户写s"，预期结果是:

User writes "s", the expected result would be:

let result = [{
  "id": 1,
  "first_name": "Jean",
  "last_name": "Owens",
  "email": "[e-mail]",
  "gender": "Female"
}, {
  "id": 2,
  "first_name": "Marie",
  "last_name": "Morris",
  "email": "[e-mail]",
  "gender": "Female"
}]

我可以这样使用过滤器功能:

I could use the filter function in such a way:

let = searchText = "s";
    let result = data.filter(object=>{
      for (var property in object) {
        if (object.hasOwnProperty(property)) {
          return object[property].toString().toLowerCase().indexOf(searchText) !== -1;
        }
      }
    });

所以我想知道这个解决方案是否有更好的替代方案?

So I am wondering if there are better alternatives to this solution?

--感谢 KoolShams，这是一个有效的 JsFiddle

--Here is a working JsFiddle thanks to KoolShams

--Plunker 用于基准测试(使用 2k 数据测试)

--Plunker for benchmark purposes (tested with 2k data)

推荐答案

你可以使用 Object.keys() 和 some() 代替.

You could use Object.keys() and some() instead.

let data = [{
  "id": 1,
  "first_name": "Jean",
  "last_name": "Owens",
  "email": "[e-mail]",
  "gender": "Female"
}, {
  "id": 2,
  "first_name": "Marie",
  "last_name": "Morris",
  "email": "[e-mail]",
  "gender": "Female"
}, {
  "id": 3,
  "first_name": "Larry",
  "last_name": "Wallace",
  "email": "[e-mail]",
  "gender": "Male"
}];

var result = data.filter(function(o) {
  return Object.keys(o).some(function(k) {
    return o[k].toString().toLowerCase().indexOf('s') != -1;
  })
})

console.log(result)

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