使用 .filter JS 删除行的问题

Question

var result_database = [
  { id: 49, absentName: 'man1', absentday: 1 },
  { id: 50, absentName: 'man2', absentday: 1 },
  { id: 49, absentName: 'man1', absentday: 2 },
  { id: 50, absentName: 'man2', absentday: 2 },
  { id: 49, absentName: 'man1', absentday: 3 },
  { id: 51, absentName: 'man3', absentday: 3 },
  { id: 51, absentName: 'man3', absentday: 4 },
  { id: 50, absentName: 'man2', absentday: 3 }
]

因此，absent_date 和 absent_days 是 return 当前日期值，星期一是 1。

const absent_date = new Date();
const absent_days = absent_date.getDay();

其余：

let absent_remove = result_database.filter(item => item.absentday != absent_days);
const absent_duplicate = [...new Map(absent_remove.map(item => [item.id, item])).values()]
console.log("Absent Result", absent_duplicate);

所以基本上，如果 absentday 匹配 absent_days，absent_remove 会删除数据库行。然后我得到一堆重复项，我用 absent_duplicate 解决了这些重复项，删除了重复项。我在下面留下了这些结果。

Absent Result [
  { id: 49, absentName: 'man1', absentday: 3 },
  { id: 50, absentName: 'man2', absentday: 3 },
  { id: 51, absentName: 'man3', absentday: 4 }
]

周一（1）Man1和Man2都缺席了，但是还在query中，问题是，怎么把他们拿出来？最后的结果应该是man3一个人。

Answer 1

试试这个：

res = [
  { id: 49, absentName: 'man1', absentday: 1 },
  { id: 50, absentName: 'man2', absentday: 1 },
  { id: 49, absentName: 'man1', absentday: 2 },
  { id: 50, absentName: 'man2', absentday: 2 },
  { id: 49, absentName: 'man1', absentday: 3 },
  { id: 51, absentName: 'man3', absentday: 3 },
  { id: 51, absentName: 'man3', absentday: 4 },
  { id: 50, absentName: 'man2', absentday: 3 }
]

answer = res.filter(x => !res.find(y => x.id == y.id && y.absentday == 1))
console.log(answer)

对于您结果中的每个条目，如果没有其他条目缺席日期为“1”，我们只会保留它。

使用 .filter JS 删除行的问题

Problem with removing rows with .filter JS

javascript

mysql