Question

我目前有以下变量，想知道以秒为单位的时间量。所以CURRENT_TIME-TEST_START

TEST_START=$(date '+%d/%m/%Y %H:%M:%S')
sleep 4
CURRENT_TIME=$(date '+%d/%m/%Y %H:%M:%S')

我在下面尝试过，但仍继续受到非法行为

START_IN_SECONDS=$(date --date "$(TEST_START)" +%s)

这是在Mac中。

Answer 1

#!/bin/bash

# Time Arithmetic

TIME1=05:36:11
TIME2=06:46:00

# Convert the times to seconds from the Epoch
SEC1=`date +%s -d ${TIME1}`
SEC2=`date +%s -d ${TIME2}`

# Use expr to do the math, let's say TIME1 was the start and TIME2 was the finish
DIFFSEC=`expr ${SEC2} - ${SEC1}`

echo Start ${TIME1}
echo Finish ${TIME2}

echo Took ${DIFFSEC} seconds.

# And use date to convert the seconds back to something more meaningful
echo Took `date +%H:%M:%S -ud @${DIFFSEC}`

输出：

Start 05:36:11
Finish 06:46:00
Took 4189 seconds.
Took 01:09:49

Answer 2

首先，用于扩展变量的语法为$variable或${variable}，而不是$(variable)。

第二，date不知道如何解析%d/%m/%Y格式的日期。当日期使用斜杠时，它期望它是%m/%d/%Y。因此，请更改您的前两行以使用正确的格式。

TEST_START=$(date '+%m/%d/%Y %H:%M:%S')
sleep 4
CURRENT_TIME=$(date '+%m/%d/%Y %H:%M:%S')
START_IN_SECONDS=$(date --date "$TEST_START" +%s)
CURRENT_IN_SECONDS=$(date --date "$CURRENT_TIME" +%s)
diff=$((CURRENT_IN_SECONDS - START_IN_SECONDS))
echo $diff

当然，您可以直接使用xxx_IN_SECONDS设置+%s变量，而不用解析格式化的日期：

START_IN_SECONDS=$(date +%s)
sleep 4
CURRENT_IN_SECONDS=$(date +%s)

如何找到bash中两个日期之间以秒为单位的时差？

2 个答案: