Question

我的数据库中的记录如下所示：

From: 2012-01-16 06:20:00   To: 2012-01-16 06:30:00
From: 2012-01-16 06:30:00   To: 2012-01-16 06:40:00
From: 2012-01-16 06:40:00   To: 2012-01-16 06:50:00

在我的应用程序中，我必须将它们组合成1个记录，如下所示：

From: 2012-01-16 06:20:00   To: 2012-01-16 06:50:00

有时记录不是任何顺序。他们可能看起来像这样：

From: 2012-01-16 06:20:00   To: 2012-01-16 06:30:00
From: 2012-01-16 06:40:00   To: 2012-01-16 06:50:00
From: 2012-01-16 06:30:00   To: 2012-01-16 06:40:00

Answer 1

如果这是时间戳，您可以使用MIN（从）和MAX（到）。

Answer 2

可能不是最有效的方式，但我会这样做：

let timeRanges = [];

timeRanges.push({
  from:  new Date(Date.parse('2012-01-16 06:20:00')),
  to: new Date(Date.parse('2012-01-16 06:30:00'))
});

timeRanges.push({
  from:  new Date(Date.parse('2012-01-16 06:40:00')),
  to: new Date(Date.parse('2012-01-16 06:50:00'))
});

timeRanges.push({
  from:  new Date(Date.parse('2012-01-16 06:30:00')),
  to: new Date(Date.parse('2012-01-16 06:40:00'))
});

let combinedTimeRange = {
  from: null,
  to: null
};

for(let timeRange of timeRanges) {
  if(combinedTimeRange.from === null 
    || combinedTimeRange.from.getTime() > timeRange.from.getTime()) {
      combinedTimeRange.from = timeRange.from;
  }

  if(combinedTimeRange.to === null 
    || combinedTimeRange.to.getTime() < timeRange.to.getTime()) {
      combinedTimeRange.to = timeRange.to;
  }
}

// just a test ..
console.log(combinedTimeRange.from.getMinutes());
console.log(combinedTimeRange.to.getMinutes());

结合日期的最有效方法是什么？

2 个答案: