如何使用过滤器在数组中搜索对象的多个键值？

Posted 2023-02-22

【中文标题】如何使用过滤器在数组中搜索对象的多个键值？

【英文标题】：How to use filter to search in multiple key values of objects in an array?

【发布时间】：2018-02-09 23:45:00

【问题描述】：

我有一个葡萄酒数组，其中包含带有每种葡萄酒数据的对象：

var wines = [
   _id: '59a740b8aa06e549918b1fda',
    wineryName: 'Some Winery',
    wineName: 'Pinot Noir',
    wineColor: 'Red',
    imageLink: '/img/FortBerensPN.png' ,
   _id: '59a7410aaa06e549918b1fdb',
    wineryName: 'Some Winery',
    wineName: 'Pinot Gris',
    wineColor: 'White',
    imageLink: '/img/FortBerensPG.png' ,
   _id: '59a74125aa06e549918b1fdc',
    wineryName: 'Some Winery',
    wineName: 'Rose',
    wineColor: 'Rose',
    imageLink: '/img/FortBerensRose.png' ,
   _id: '59a74159aa06e549918b1fdd',
    wineryName: 'Some other Winery',
    wineName: 'Rose',
    wineColor: 'Rose',
    imageLink: '/img/FortBerensRose.png' ,
   _id: '59a7417aaa06e549918b1fde',
    wineryName: 'Some other Winery',
    wineName: 'Pinot Gris',
    wineColor: 'White',
    imageLink: '/img/FortBerensPG.png' ,
   _id: '59a8721f4fd43b676a1f5f0d',
    wineryName: 'Some other Winery',
    wineName: 'Pinot Gris',
    wineColor: 'White',
    imageLink: '/img/FortBerensPG.png' ,
   _id: '59a872244fd43b676a1f5f0e',
    wineryName: 'Winery 3',
    wineName: 'Pinot Noir',
    wineColor: 'Red',
    imageLink: '/img/FortBerensPN.png'  ]

我可以弄清楚如何搜索（不区分大小写）葡萄酒对象，同时指定要搜索的对象的哪个键，如下所示：

var search = 'Noir'

filteredWines = function () 
  return wines.filter(function(wine)
    return (wine.wineName.toLowerCase().indexOf(search.toLowerCase())>=0;
  );
;

返回：

[  _id: '59a740b8aa06e549918b1fda',
    wineryName: 'Some Winery',
    wineName: 'Pinot Noir',
    wineColor: 'Red',
    imageLink: '/img/FortBerensPN.png' ,
   _id: '59a872244fd43b676a1f5f0e',
    wineryName: 'Winery 3',
    wineName: 'Pinot Noir',
    wineColor: 'Red',
    imageLink: '/img/FortBerensPN.png'  ]

但是，如果var search = 'Winery 3' 或var search = 'red' 显然不会返回任何结果，因为它正在查看数组中每个对象的wineName 的值。

那么有没有办法使用过滤器（或其他方法？）来搜索所有键值，甚至更好的是，多个指定的键值并返回匹配对象的数组？

类似：

filteredWines = function () 
  return wines.filter(function(wine)
    return ((wine.wineName.toLowerCase() && wine.wineName.toLowerCase() 
          && wine.wineName.toLowerCase()).indexOf(search.toLowerCase())>=0;
  );
;

还是我完全找错树了？

PS。我正在使用 Vue.js 2，所以如果在 vue 中有更好的方法，那么我会全力以赴！

【问题讨论】：

是否应该为每个属性搜索一个值？或者您是否为每个属性指定了不同的搜索值？

也许我误解了你的问题，但我认为简单地使用 OR 运算符就可以了？您可以通过创建函数 (wine.wineName.toLowerCase().indexOf(search.toLowerCase())>=0 来清理这部分。或者，您可以使用正则表达式。

顺便说一句，从人们最终发布的任何过滤方法中进行选择，但 computed property 将是在 Vue 中使用它的理想方式。

@Bert 我将其填充到计算属性中并返回结果 :) 感谢您向其他人说明这一点！

【参考方案1】：

你可以有一个更通用的函数来扫描所有字符串的属性。使用Object.values() 循环所有属性值，并在匹配后立即使用some 退出：

filteredWines = function (search) 
    var lowSearch = search.toLowerCase();
    return wines.filter(wine =>
        Object.values(wine).some(val => 
            String(val).toLowerCase().includes(lowSearch) 
        )
    );

如果您希望传递特定的键进行搜索：

filteredWines = function (search, keys) 
    var lowSearch = search.toLowerCase();
    return wines.filter(wine =>
        keys.some(key => 
            String(wine[key]).toLowerCase().includes(lowSearch) 
        )
    );

调用为

filteredWines('Winery 3', ['wineryName', 'wineName']);

【讨论】：

这太棒了，谢谢！再次感谢@Bert 为我精心制作了它 :)

节省了很多时间。谢谢。

【参考方案2】：

使用“trincot”中的解决方案并将其更改为我的 angular5 应用程序：

filter(search, list): Observable<IFilteredList> 
  return list.filter(item => 
    return Object.values(item).some(val =>
      String(val).includes(search)
    );
  )

【参考方案3】：

也可以这样：

    this.wines = this.wines.filter((item) => 
                return (item.wineryName.toString().toLowerCase().indexOf(val.toLowerCase()) > -1 ||
                item.wineName.toLowerCase().indexOf(val.toLowerCase()) > -1 ||
                item.wineColor.toLowerCase().indexOf(val.toLowerCase()) > -1);
            )

【参考方案4】：

我还建议尝试更通用的方法：

function getMatchingWhine(keys, searchTerm, wines) 

  function extractTextFromKeys(keys, object) 
    let text = '';
    keys.forEach(key => 
        text += ' ' + object[key];
    );
    return text.toLowerCase();
  

  return wines.filter(wine => 
    const relevantText = extractTextFromKeys(keys, wine);
    return relevantText.includes(searchTerm.toLowerCase());
  );

【参考方案5】：

过滤器有效。糟糕，我更仔细地阅读了这个问题。过滤器仍然有效，但您也必须过滤值。

let wines = [
    
        _id: '59a740b8aa06e549918b1fda',
        wineryName: 'Some Winery',
        wineName: 'Pinot Noir',
        wineColor: 'Red',
        imageLink: '/img/FortBerensPN.png'
    ,
    
        _id: '59a7410aaa06e549918b1fdb',
        wineryName: 'Some Winery',
        wineName: 'Pinot Gris',
        wineColor: 'White',
        imageLink: '/img/FortBerensPG.png'
    ,
    
        _id: '59a74125aa06e549918b1fdc',
        wineryName: 'Some Winery',
        wineName: 'Rose',
        wineColor: 'Rose',
        imageLink: '/img/FortBerensRose.png'
    ,
    
        _id: '59a74159aa06e549918b1fdd',
        wineryName: 'Some other Winery',
        wineName: 'Rose',
        wineColor: 'Rose',
        imageLink: '/img/FortBerensRose.png'
    ,
    
        _id: '59a7417aaa06e549918b1fde',
        wineryName: 'Some other Winery',
        wineName: 'Pinot Gris',
        wineColor: 'White',
        imageLink: '/img/FortBerensPG.png'
    ,
    
        _id: '59a8721f4fd43b676a1f5f0d',
        wineryName: 'Some other Winery',
        wineName: 'Pinot Gris',
        wineColor: 'White',
        imageLink: '/img/FortBerensPG.png'
    ,
    
        _id: '59a872244fd43b676a1f5f0e',
        wineryName: 'Winery 3',
        wineName: 'Pinot Noir',
        wineColor: 'Red',
        imageLink: '/img/FortBerensPN.png'
    
];

let search = (val) => wines.filter(w => Object.values(w).filter(v => v.toLowerCase().indexOf(val.toLowerCase()) !== -1).length > 0);

console.log(search('some'));