Javascript 多條件數組過濾器

Question

我需要幫助來組合基於多個條件的數組搜索。 此外，所有條件都是有條件的，這意味着我可能需要也可能不需要根據這些條件進行過濾。 我擁有的：

要過濾的對象數組：

var data = [{
    "_id" : ObjectId("583f6e6d14c8042dd7c979e6"),
    "transid" : 1,
    "acct" : "acct1",
    "transdate" : ISODate("2012-01-31T05:00:00.000Z"),
    "category" : "category1",
    "amount" : 103
},
{
    "_id" : ObjectId("583f6e6d14c8042dd7c2132t6"),
    "transid" : 2,
    "acct" : "acct2",
    "transdate" : ISODate("2012-01-31T05:00:00.000Z"),
    "category" : "category2",
    "amount" : 103
},
{
    "_id" : ObjectId("583f6e6d14c8042dd7c2132t6"),
    "transid" : 3,
    "acct" : "acct2",
    "transdate" : ISODate("2016-07-31T05:00:00.000Z"),
    "category" : "category1",
    "amount" : 103
},
{
    "_id" : ObjectId("583f6e6d14c8042dd7c2132t6"),
    "transid" : 4,
    "acct" : "acct2",
    "transdate" : ISODate("2012-01-31T05:00:00.000Z"),
    "category" : "category2",
    "amount" : 103
},
{
    "_id" : ObjectId("583f6e6d14c8042dd7c2132t6"),
    "transid" : 5,
    "acct" : "acct2",
    "transdate" : ISODate("2012-01-31T05:00:00.000Z"),
    "category" : "category3",
    "amount" : 103
},
{
    "_id" : ObjectId("583f6e6d14c8042dd7c152g2"),
    "transid" : 6,
    "acct" : "acct3",
    "transdate" : ISODate("2016-10-31T05:00:00.000Z"),
    "category" : "category3",
    "amount" : 103
}]

我正在根據另一個混合元素數組過濾上述對象數組。 這些元素代表以下搜索字段：

• “searchstring”：在數據數組的所有字段上搜索任何匹配的文本序列

• object，鍵值代表帳戶類型，值的真或假指示是否應該用於過濾

• 過濾轉換日期的開始日期

• 過濾轉換的結束日期

• 過濾類別的類別名稱

具有搜索條件的數組如下所示（但如果某些字段不是必需的，它們將被設置為未定義或只是一個空字符串或數組）：

var filtercondition = {
    "p",
    {acct1:true,acct2:false,acct3:true...}
    "2016-06-01",
    "2016-11-30",
    "category3"
}

完成此任務的最佳方法是什么？ 我設計的是單獨搜索過濾器數組中的每個元素，但這似乎不是最佳的並且非常乏味。 我願意重新設計我的設置......

Answer 1

// You wrote that it's an array, so changed the braces 
var filtercondition = ["p",
{acct1:true,acct2:false,acct3:true...}
"2016-06-01",
"2016-11-30",
"category3"
];

var filtered = data.filter(o => {
    if(filtercondition[0] && !o.category.includes(filtercondition[o])) { // checking just the category, but you can check if any of more fields contains the conditions 
        return false;
    }
    if(filtercondition[1]) {
        for(var key in filtercondition[1]) {
        if(filtercondition[1][key] === true && o.acct != key) {
            return false;
        }
        }
    }
    if(filtercondition[2] && o.transdate < filtercondition[2]) {
        return false;
    }
    if(filtercondition[3] && o.transdate > filtercondition[3]) {
        return false;
    }
    if(filtercondition[4] && o.category !== filtercondition[4]) {
        return false;
    }

    return true;
});

兩個注意事項： - 更改了filtercondition的大括號，使其成為一個數組，但是我建議改用一個對象。 - 這個{acct1:true,acct2:false,acct3:true...}樣本對我來說沒有意義，因為它表明acct字段應該同時是acct1和acct3 。

Answer 2

創建一個函數數組，每個函數代表一個條件。

這是一些演示該方法的示例代碼......

 var conditions = [];

 // Dynamically build the list of conditions
 if(startDateFilter) {
    conditions.push(function(item) { 
       return item.transdate >= startDateFilter.startDate;
    });
 };

 if(categoryFilter) {
     conditions.push(function(item) {
         return item.cateogry === categoryFilter.category;
     });
 };
 // etc etc

一旦有了條件數組，就可以使用Array.prototype.every對項目運行每個條件。

 var itemsMatchingCondition = data.filter(function(d) {
     return conditions.every(function(c) {
         return c(d);
     });
 });

Answer 3

首先，您需要為數組使用方括號而不是花括號：

var filtercondition = [
    "p",
    {acct1:true,acct2:false,acct3:true...},
    "2016-06-01",
    "2016-11-30",
    "category3"
];

再說一次，我不認為數組是最好的數據類型。 試試這樣的對象：

var filtercondition = {
    query: "p",
    accounts: {acct1:true,acct2:false,acct3:true...},
    date1: "2016-06-01",
    date2: "2016-11-30",
    category: "category3"
};

然后，嘗試使用 Array.prototype.filter：

var filtered = data.filter(function(obj) {
    for (var key in filtercondition) {
        // if condition not met return false
    }
    return true;
});

Answer 4

我會使用一堆小的粒度函數並組合它們。

//only some utilities, from the top of my mind
var identity = v => v;

//string-related
var string = v => v == null? "": String(v);
var startsWith = needle => haystack => string(haystack).startsWith(needle);
var endsWith = needle => haystack => string(haystack).endsWith(needle);
var contains = needle => haystack => string(haystack).contains(needle);

//do sth with an object
var prop = key => obj => obj != null && prop in obj? obj[prop]: undefined;
var someProp = fn => obj => obj != null && Object.keys(obj).some(k => fn(k) );
var someValue = fn => obj => obj != null && Object.keys(obj).some(k => fn(obj[k]) );

//logic
var eq = b => a => a === b;
var not = fn => function(){ return !fn.apply(this, arguments) };
var and = (...funcs) => funcs.reduce((a, b) => function(){
        return a.apply(this, arguments) && b.apply(this, arguments);
    });

var or = (...funcs) => funcs.reduce((a, b) => function(){
        return a.apply(this, arguments) || b.apply(this, arguments);
    });

//composition
var compose = (...funcs) => funcs.reduce((a, b) => v => return a(b(v)));
var chain = (...funcs) => funcs.reduceRight((a, b) => v => return a(b(v)));

//and whatever else you want/need
//but stay granular, don't put too much logic into a single function

和一個示例組合：

var filterFn = and(
    //some value contains "p"
    someValue(contains("p")),

    //and
    chain(
        //property "foo"
        prop("foo"), 
        or(
            //either contains "asdf"
            contains("asdf"),

            //or startsWith "123"
            startsWith("123")
        )
    ),
)

由於我不知道您如何構建過濾條件，因此我無法確切地告訴您如何將它們解析為這樣的組合，但您可以像這樣組合它們：

//start with something basic, so we don't ever have to check wether filterFn is null
var filterFn = identity;

//and extend/compose it depending on some conditions
if(/*hasQuery*/){
    filterFn = and(
        // previous filterFn(obj) && some value on obj contains `query`
        filterFn,
        someValue(contains(query)))
    )
}

if(/*condition*/){
    //extend filterFn
    filterFn = or(
        // (obj.foo === null) || previous filterFn(obj)
        chain(prop("foo"), eq(null)),
        filterFn
    );
}

等等

Answer 5

先說幾點：

• 如果您打算在瀏覽器中使用它，您的data對象將無效。 可能數據來自MongoDB ，對吧？ 您的后端（數據源）應該有一種方法可以對其進行正確編碼並刪除ObjectID和ISODate引用。

• 您的filtercondition條件不是有效的 JavaScript 對象/JSON。 檢查我的例子。

因此，您可以使用Array#filter方法過濾數據數組。

類似的東西：

let data = [{
    "_id" : "583f6e6d14c8042dd7c979e6",
    "transid" : 1,
    "acct" : "acct1",
    "transdate" : "2012-01-31T05:00:00.000Z",
    "category" : "category1",
    "amount" : 103
},
{
    "_id" : "583f6e6d14c8042dd7c2132t6",
    "transid" : 2,
    "acct" : "acct2",
    "transdate" : "2012-01-31T05:00:00.000Z",
    "category" : "category2",
    "amount" : 103
},
{
    "_id" : "583f6e6d14c8042dd7c2132t6",
    "transid" : 5,
    "acct" : "acct2",
    "transdate" : "2012-01-31T05:00:00.000Z",
    "category" : "category3",
    "amount" : 103
}];


let filterToApply = {
    acct: {
        acct1: true,
        acct2: false,
        acct3: true
    },
    initialDate: "2016-06-01",
    finalDate: "2016-11-30",
    category: "category3"
}


let filterData = (array, filter) => {

    return array.filter( (item) => {

        /* here, you iterate each item and compare with your filter,
           if the item pass, you must return true. Otherwise, false */


        /* e.g.: category check (if present only) */
        if (filter.category && filter.category !== item.category) 
            return false;
        }

        /* add other criterias check... */ 

        return true;
});

}

let dataFiltered = filterData(data, filterToApply);
console.log(dataFiltered);

Answer 6

如果你想用多個條件過濾一個數組，並且條件可以是可選的，那么使用下面的方法。

 const data = [ { name: 'John', age: 25, city: 'New York' }, { name: 'John', age: 25, city: 'New' }, { name: 'Jane', age: 32, city: 'Los Angeles' }, { name: 'Bob', age: 45, city: 'New York' }, { name: 'Alice', age: 38, city: 'Los Angeles' } ]; const filteredData = (n, c, a) => data.filter(item => { if (n || c || a) { return (n? item.name === n: true) && (c? item.city === c: true) && (a? item.age === a: true); // keep adding conditons as much as u want } }); console.log(filteredData('John', null, 25)); console.log(filteredData(null, 'Los Angeles', 38)); console.log(filteredData(null, 'Los Angeles', null));

您可以鏈接盡可能多的條件