如何将时间戳转为秒
通常的做法
public static long toSecondMethod1(long timestamp) {
return timestamp / 1000;
}
直接使用TimeUnit工具箱中的方法
一个更友好点的方法
public static long toSecondMethod2(long timestamp) {
return TimeUnit.MILLISECONDS.toSeconds(timestamp);
}
测试代码
public static void main(String[] args) {
long timestamp = System.currentTimeMillis();
long method1 = toSecondMethod1(timestamp);
long method2= toSecondMethod2(timestamp);
System.out.println("timestamp:" + timestamp);
System.out.println("method1:" + method1);
System.out.println("method2:" + method2);
}
输出结果
timestamp:1618100030482
method1:1618100030
method2:1618100030
可以看到结果是一样,我们可以进入TimeUnit(java.util.concurrent)中,看一下具体的实现
/**
* Time unit representing one thousandth of a second
*/
MILLISECONDS {
public long toNanos(long d) { return x(d, C2/C0, MAX/(C2/C0)); }
public long toMicros(long d) { return x(d, C2/C1, MAX/(C2/C1)); }
public long toMillis(long d) { return d; }
public long toSeconds(long d) { return d/(C3/C2); }
public long toMinutes(long d) { return d/(C4/C2); }
public long toHours(long d) { return d/(C5/C2); }
public long toDays(long d) { return d/(C6/C2); }
public long convert(long d, TimeUnit u) { return u.toMillis(d); }
int excessNanos(long d, long m) { return 0; }
},
// Handy constants for conversion methods
static final long C0 = 1L;
static final long C1 = C0 * 1000L;
static final long C2 = C1 * 1000L;
static final long C3 = C2 * 1000L;
static final long C4 = C3 * 60L;
static final long C5 = C4 * 60L;
static final long C6 = C5 * 24L;
通过以上的两段代码,可以看出,内部实现和我们的实现一样的。
使用TimeUnit有什么好处呢?
-
如果我们除了将时间戳转为秒的需求,还可能将时间戳转为小时,转为分钟等,那该工具箱提供了toHours等方法使用。
-
如果我们的源头不止时间戳,可能是将秒转为小时等,同样的也会有TimeUnit.SECONDS对应的方法使用。
SECONDS { public long toNanos(long d) { return x(d, C3/C0, MAX/(C3/C0)); } public long toMicros(long d) { return x(d, C3/C1, MAX/(C3/C1)); } public long toMillis(long d) { return x(d, C3/C2, MAX/(C3/C2)); } public long toSeconds(long d) { return d; } public long toMinutes(long d) { return d/(C4/C3); } public long toHours(long d) { return d/(C5/C3); } public long toDays(long d) { return d/(C6/C3); } public long convert(long d, TimeUnit u) { return u.toSeconds(d); } int excessNanos(long d, long m) { return 0; } }
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