在嵌套對象數組中按屬性值對對象數組進行排序

Question

我試圖弄清楚如何通過嵌套對象數組中的屬性值是否包含值stopped對對象數組進行排序。 當該值存在於任何嵌套的對象數組中時，我需要將父對象排序到頂部，從那里開始，我嘗試按id對該排序列表進行二次排序。

const arr = [
    {
        id: 1,
        things: [
            {
                thing_id: 1, 
                status: 'started',
            },
            {
                thing_id: 1, 
                status: 'started',
            },
        ],
    },
    {
        id: 2,
        things: [
            {
                thing_id: 1, 
                status: 'started',
            },
            {
                thing_id: 1, 
                status: 'started',
            },
            {
                thing_id: 1, 
                status: 'stopped',
            },
        ],
    },
    {
        id: 3,
        things: [
            {
                thing_id: 1, 
                status: 'started',
            },
            {
                thing_id: 1, 
                status: 'started',
            },
            {
                thing_id: 1, 
                status: 'started',
            },
            {
                thing_id: 1, 
                status: 'started',
            },
        ],
    }
]

// desired result
[
    {
        id: 2,
        things: [
            {
                thing_id: 1, 
                status: 'started',
            },
            {
                thing_id: 1, 
                status: 'started',
            },
            {
                thing_id: 1, 
                status: 'stopped',
            },
        ],
    },
    {
        id: 1,
        things: [
            {
                thing_id: 1, 
                status: 'started',
            },
            {
                thing_id: 1, 
                status: 'started',
            },
        ],
    },
    {
        id: 3,
        things: [
            {
                thing_id: 1, 
                status: 'started',
            },
            {
                thing_id: 1, 
                status: 'started',
            },
            {
                thing_id: 1, 
                status: 'started',
            },
            {
                thing_id: 1, 
                status: 'started',
            },
        ],
    }
]

Answer 1

arr.sort((a, b) => {
    if (a.things.some(thing => thing.status === "stopped")) {
        return -1;
    } else {
        return a.id - b.id;
    }
});

您只需進行排序，檢查當前正在檢查的對象是否至少有一個狀態為“已停止”的“事物”，否則為正常的數字順序。

Answer 2

arr.sort((a, b) => { const stoppeds_in_a = a.things.map(obj => obj.status).filter(status => status === 'stopped').length const stoppeds_in_b = b.things.map(obj => obj.status).filter(status => status === 'stopped').length // I want that who has more 'stoppeds' occurrences first return stoppeds_in_b - stoppeds_in_a })

Answer 3

 const checkStopped = (things) => things.some((el) => el.status === 'stopped'); const desired = arr.sort((a, b) => checkStopped(b.things) - checkStopped(a.things));

Answer 4

您可以引入一個輔助函數，該函數根據提供的回調對集合進行分區。 然后將它們連接在一起以創建所需的結果。

 const arr = [{id:1,things:[{thing_id:1,status:'started'},{thing_id:1,status:'started'}]},{id:2,things:[{thing_id:1,status:'started'},{thing_id:1,status:'started'},{thing_id:1,status:'stopped'}]},{id:3,things:[{thing_id:1,status:'started'},{thing_id:1,status:'started'},{thing_id:1,status:'started'},{thing_id:1,status:'started'}]}]; const [withStopped, withoutStopped] = partition(arr, item => item.things.some(item => item.status == "stopped") ); const result = withStopped.concat(withoutStopped); console.log(result); // helper function partition(iterable, fn) { const partitions = { "true": [], "false": [] }; for (const item of iterable) partitions[!!fn(item)].push(item); return [partitions[true], partitions[false]]; }

Answer 5

這實際上是一個很好的遞歸任務。 但是如果結構是固定的，我就采用了兩個循環和一個條件。 如果這是真的，我將數組推送到全局 res 變量中。

 const arr = [ { id: 1, things: [ { thing_id: 1, status: 'started', }, { thing_id: 1, status: 'started', }, ], }, { id: 2, things: [ { thing_id: 1, status: 'started', }, { thing_id: 1, status: 'started', }, { thing_id: 1, status: 'stopped', }, ], }, { id: 3, things: [ { thing_id: 1, status: 'started', }, { thing_id: 1, status: 'started', }, { thing_id: 1, status: 'started', }, { thing_id: 1, status: 'started', }, ], } ] const res = []; arr.forEach(function(e) { let val = Object.values(e.things) val.forEach((t) => { if(t.status == "stopped") { res.push(e) } }) }) console.log('res', res)

Answer 6

你應該做這樣的事情

const Firstresult = [];//with stopped
const SecontPartresult = [];//without stopped

arr.forEach(element => {
    element.things.forEach(thing => {
        if (thing.status == "stopped") {
            if (Firstresult.filter(x => x.id == element.id).length == 0) {
                Firstresult.push(element)
            }
        } else if (element.things.filter(x => x.status == "stopped").length === 0) {
            if (SecontPartresult.filter(x => x.id == element.id && x.things !== "stopped").length == 0) {
                SecontPartresult.push(element)
            }
        }

    });

});

SecontPartresult.forEach(element => {
    Firstresult.push(element)
});
console.table(Firstresult)