尝试使用过滤器数组过滤主数组并获得结果答案

【问题标题】：Trying to filter an main array with filter array and get the result尝试使用过滤器数组过滤主数组并获得结果
【发布时间】：2020-07-08 12:45:42
【问题描述】：

我正在尝试根据过滤器值数组过滤结果。下面是主数组

var mainArray = [
    { id: 1, Country: ["India"], Version: "v1" },
    { id: 2, Country: ["US", "UK"], Version: "v2" },
    { id: 3, Country: ["Australia"], Version: "v3" },
    { id: 4, Country: ["UK"], Version: "v4" },
    { id: 5, Country: ["UK"], Version: "v3" },
    { id: 6, Country: ["US"], Version: "v2" },
    { id: 7, Country: ["Australia"], Version: "v4" },
    { id: 8, Country: ["India"], Version: "v2" }
];

这是过滤条件：

var filterValue = [
    {
        name: "Country",
        value: [
            { text: "India", checked: true },
            { text: "US", checked: true },
            { text: "UK", checked: false },
            { text: "Australia", checked: false }
        ]
    },
    {
        name: "Version",
        value: [
            { text: "v1", checked: true },
            { text: "v2", checked: true },
            { text: "v3", checked: false },
            { text: "v4", checked: false }
        ]

    }
]

下面是基于我想要得到的 filterValue 的 resultArray：

var resultArray = [
    { id: 1, Country: ["India"], Version: "v1" },
    { id: 2, Country: ["US", "UK"], Version: "v2" },
    { id: 6, Country: ["US"], Version: "v2" },
    { id: 8, Country: ["India"], Version: "v2" }
];

感谢您的帮助。

【问题讨论】：

您忘记添加遇到问题的实际代码。
另外，你不能有两个resultArray - 实际上预期的结果是什么？

标签： javascript arrays filter

【解决方案1】：

假设您希望在所有过滤器上使用 AND，请使用以下内容：

let results = mainArray.filter(main => {

  // map every filter on each main value
  let filters = filterValue.map(filter => {

    // normalize filter values
    let key = filter.name,
        val = filter.value.filter(r => r.checked).map(r => r.text);
    // console.log(key, val);
    // > "Country", [ 'India', 'US' ]
    // > "Version", [ 'v1', 'v2' ]  
    
    let prop = main[key];

    // filter out empties
    if (!prop)
      return 0;

    // single string to an array
    if (!Array.isArray(prop))
      prop = [prop];

    // intersect prop and val arrays
    return prop.filter(r => val.includes(r)).length
  });
  
  // return wether all conditions match
  return filters.every(r => !!r);
});

【讨论】：

【解决方案2】：

我使用了这些JavaScript Array Methods：filter()、find()和includes

希望这会有所帮助：

        var mainArray = [
            { id: 1, Country: ["India"], Version: "v1" },
            { id: 2, Country: ["US", "UK"], Version: "v2" },
            { id: 3, Country: ["Australia"], Version: "v3" },
            { id: 4, Country: ["UK"], Version: "v4" },
            { id: 5, Country: ["UK"], Version: "v3" },
            { id: 6, Country: ["US"], Version: "v2" },
            { id: 7, Country: ["Australia"], Version: "v4" },
            { id: 8, Country: ["India"], Version: "v2" }];

        var filterValue=[{
            name: "Country",
            value: [
                    { text: "India", checked: true },
                    { text: "US", checked: true },
                    { text: "UK", checked: false },
                    { text: "Australia", checked: false }
                ]}, {
             name: "Version",
             value: [
                    { text: "v1", checked: true },
                    { text: "v2", checked: true },
                    { text: "v3", checked: false },
                    { text: "v4", checked: false }
                ]}];
            
        var result=  mainArray.filter(function(item){
            var country= filterValue[0].value.find(x=>x.checked&&item.Country.includes(x.text));
            var version= filterValue[1].value.find(x=>x.checked&&x.text==item.Version);
            return country&&version;
        });

        for(var i=0;i<result.length;i++)
            console.log(JSON.stringify(result[i]));

【讨论】：