选日期的时候提取相邻的日期天数？_前端问题_站长_知道

checkList:\[  
    {  
        date:'2019-01-01',  
        num:1,  
        roomId:1  
    },  
    {  
        date:'2019-01-02',  
        num:1,  
        roomId:1  
    },  
    {  
        date:'2019-01-03',  
        num:1,  
        roomId:1  
    },  
    {  
        date:'2019-01-05',  
        num:1,  
        roomId:1  
    },  
    {  
        date:'2019-01-03',  
        num:1,  
        roomId:2  
    },  
    {  
        date:'2019-01-04',  
        num:1,  
        roomId:2  
    }  
\],

checkList2:\[  
    {  
        date:'2019-01-01',  
        num:3,  
        roomId:1  
    },  
    {  
        date:'2019-01-05',  
        num:1,  
        roomId:1  
    },  
    {  
        date:'2019-01-03',  
        num:2,  
        roomId:2  
    }  
\]

最终我想把list这个初始数据变成list2这样的数据要怎么弄，num是天数，roomId是房间id QQ截图20191231172213.png

已解决悬赏分：20 - 解决时间 2021-11-27 12:34

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回答1

最佳

@mhwkj88

排序，这样方便统计
判断当前和历史的戳是否连贯。连贯就++。不连贯就push

checkList=[  
    {  
        date:'2019-01-01',  
        num:1,  
        roomId:1  
    },  
    {  
        date:'2019-01-02',  
        num:1,  
        roomId:1  
    },  
    {  
        date:'2019-01-03',  
        num:1,  
        roomId:1  
    },  
    {  
        date:'2019-01-05',  
        num:1,  
        roomId:1  
    },  
    {  
        date:'2019-01-03',  
        num:1,  
        roomId:2  
    },  
    {  
        date:'2019-01-04',  
        num:1,  
        roomId:2  
    }  
].sort((n,m)=>{
    var roomIdsort = n.roomId - m.roomId
    return roomIdsort?roomIdsort:(n.date<m.date?-1:1)
}).reduce((s,v)=>{
    if(s.last && (s.last.roomId == v.roomId) && (new Date(v.date)-new Date(s.last.date)<=1000*60*60*24*s.last.num)){
        s.last.num++    
    }else{
        s.last = v;
        s.list.push(v)
    }
    return s
}, {list: [], last: {}})

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2021-11-27 08:59