Javascript对象数组重组

Question

我正在从连接了 3 个表的关系数据库中获取数据:

var btnObj2 = [
{sect_id: 1, sect_title: 'Navigation', subsect_id: 1, subsect_title:'Übersicht', btn_id: 1, btn_title: 'Inhaltsverzeichnis'},
{sect_id: 1, sect_title: 'Navigation', subsect_id: 1, subsect_title: 'Übersicht', btn_id: 2, btn_title: 'Stichwortverzeichnis'},
{sect_id: 1, sect_title: 'Navigation', subsect_id: 2, subsect_title: 'Praxisphasen', btn_id: 3, btn_title: 'Trainingserfolg'},
{sect_id: 1, sect_title: 'Navigation', subsect_id: 2, subsect_title: 'Praxisphasen', btn_id: 4, btn_title: 'Trainingsablauf'},
{sect_id: 2, sect_title: 'Modul 1', subsect_id: 3, subsect_title: 'Mentor-Gespräche', btn_id: 5, btn_title: 'Lebenszeit'},
{sect_id: 2, sect_title: 'Modul 1', subsect_id: 3, subsect_title: 'Mentor-Gespräche', btn_id: 6, btn_title: 'Lebensplanung'},
{sect_id: 2, sect_title: 'Modul 1', subsect_id: 4, subsect_title: 'Just do it', btn_id: 7, btn_title: 'Vertrauen'},
{sect_id: 2, sect_title: 'Modul 1', subsect_id: 4, subsect_title: 'Just do it', btn_id: 8, btn_title: 'Verantwortung'}
];

必须重组此数据，以便每个部分包含其子部分，每个子部分包含其子子部分(子子子部分是上面代码片段中的按钮)。数据库中的数据已经按照部分、子部分、子部分的正确顺序返回。

输出必须如下所示:

var result = [
{
    sect_id: 1,
    sect_title: 'Navigation',
    subsect: [
        {
            subsect_id: 1,
            subsect_title: 'Übersicht',
            buttons: [
                {
                    btn_id: 1,
                    btn_title: 'Verantwortung'
                },
                {
                    btn_id: 2,
                    btn_title: 'Vertrauen'
                }
            ]
        },
        {
            subsect_id: 2,
            subsect_title: 'Praxisphasen',
            buttons: [
                {
                    btn_id: 3,
                    btn_title: 'Trainingserfolg'
                },
                {
                    btn_id: 4,
                    btn_title: 'Trainingsablauf'
                }
            ]
        }
    ]
},
sect_id: 2,
sect_title: 'Module 1',
//.....
];

以下是我将如何以 PHP 方式完成此操作:

function prepareBtns(arr) {
    var sectId, subsectId = ''
    var newArr = []
    arr.forEach((item) => {
        if(sectId !== item.sect_id){
            sectId = item.sect_id
        }
        if(subsectId !== item.subsect_id){
            subsectId = item.subsect_id
        }
        //In PHP the following line does the Job
        newArr[sectId][subsectId][] = item
    })
    return newArr
}

这是我的 JS 试验之一。这导致空槽:

function prepareBtnsXY(arr) {
var newArr = [];
arr.forEach((item) => {
    if (!newArr[item.sect_id]) {

        var objToPush = {
            sectId: item.sect_id,
            sectName: item.sect_title,
            subsect: []
        };
        newArr[item.sect_id] = objToPush;
    }

    if (!newArr[item.sect_id].subsect[item.subsect_id]) {
        var subObjToPush = {
            subsectId: item.subsect_id,
            subsectName: item.subsect_title,
            buttons: []
        };
        newArr[item.sect_id].subsect[item.subsect_id] = subObjToPush;
    }

    if (!newArr[item.sect_id].subsect[item.subsect_id].buttons[item.btn_id]) {
        var btnObjToPush = {
            btnId: item.btn_id,
            btnName: item.btn_title
        };
        newArr[item.sect_id].subsect[item.subsect_id].buttons[item.btn_id] = btnObjToPush;
    }
});

return newArr
}

Answer 1

您可以将想要的键存储在另一个数组中，其中每行的第一个键是分组的键值，后面的键是该级别的数据键，最后一个键用于子级/子级别。最终数组不包含子值，而是包含 undefined 的占位符。

keys = [
  //  level key          level  properties        children
  // ------------  ----------------------------- ----------
    ['sect_id',    'sect_title' /* more keys */, 'subsect'],
    ['subsect_id', 'subsect_title',              'buttons'],
    ['btn_id',     'btn_title',                  undefined]
],

该算法通过获取实际数组并搜索相同的分组键和值来迭代数据并减少键。如果找不到，它会使用给定的键生成一个新级别并将其推送到数组并返回下一个级别的子数组。

这种方法的优点是，只需更改原始数据和相应的 keys 数组即可轻松维护，该数组充当分组规则集。

var data = [{ sect_id: 1, sect_title: 'Navigation', subsect_id: 1, subsect_title: 'Übersicht', btn_id: 1, btn_title: 'Inhaltsverzeichnis' }, { sect_id: 1, sect_title: 'Navigation', subsect_id: 1, subsect_title: 'Übersicht', btn_id: 2, btn_title: 'Stichwortverzeichnis' }, { sect_id: 1, sect_title: 'Navigation', subsect_id: 2, subsect_title: 'Praxisphasen', btn_id: 3, btn_title: 'Trainingserfolg' }, { sect_id: 1, sect_title: 'Navigation', subsect_id: 2, subsect_title: 'Praxisphasen', btn_id: 4, btn_title: 'Trainingsablauf' }, { sect_id: 2, sect_title: 'Modul 1', subsect_id: 3, subsect_title: 'Mentor-Gespräche', btn_id: 5, btn_title: 'Lebenszeit' }, { sect_id: 2, sect_title: 'Modul 1', subsect_id: 3, subsect_title: 'Mentor-Gespräche', btn_id: 6, btn_title: 'Lebensplanung' }, { sect_id: 2, sect_title: 'Modul 1', subsect_id: 4, subsect_title: 'Just do it', btn_id: 7, btn_title: 'Vertrauen' }, { sect_id: 2, sect_title: 'Modul 1', subsect_id: 4, subsect_title: 'Just do it', btn_id: 8, btn_title: 'Verantwortung' }],
    keys = [['sect_id', 'sect_title', 'subsect'], ['subsect_id', 'subsect_title', 'buttons'], ['btn_id', 'btn_title', undefined]],
    result = data.reduce((r, o) => {
        keys.reduce((a, k) => {
            var temp = a.find(p => o[k[0]] === p[k[0]]);
            if (!temp) {
                a.push(temp = Object.assign(
                    ...k.map((l, i, { length }) =>
                        l && { [l]: i + 1 === length ? [] : o[l] })
                ));
            }
            return temp[k[k.length - 1]];
        }, r);
        return r;
    }, []);

console.log(result);

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