Question

    ubu@bruntu:/home/wideDB1$ echo $(($(date +%s%N)/1000000))
1441637184683
ubu@bruntu:/home/wideDB1$ echo $(($(date +%N)/1000000))
730
ubu@bruntu:/home/wideDB1$ echo $(($(date +%N)/1000))
bash: 088253206: value too great for base (error token is "088253206")
ubu@bruntu:/home/wideDB1$ echo $(($(date +%N)/1000))
bash: 085061725: value too great for base (error token is "085061725")
ubu@bruntu:/home/wideDB1$ echo $(($(date +%N)/10000))
59078
ubu@bruntu:/home/wideDB1$ echo $(($(date +%N)/1000))
672523
ubu@bruntu:/home/wideDB1$ echo $(($(date +%N)/10))
12079796
ubu@bruntu:/home/wideDB1$ echo $(($(date +%N)/10000))
74644
ubu@bruntu:/home/wideDB1$ echo $(($(date +%N)/100000))
1403
ubu@bruntu:/home/wideDB1$ echo $(($(date +%N)/1000000))
5
ubu@bruntu:/home/wideDB1$ echo $(($(date +%N)/100000))
9977
ubu@bruntu:/home/wideDB1$ echo $(($(date +%N)/1000000))
360
ubu@bruntu:/home/wideDB1$ echo $(($(date +%N)/100000))
6663
ubu@bruntu:/home/wideDB1$ echo $(($(date +%N)/10))
10325856
ubu@bruntu:/home/wideDB1$ echo $(($(date +%N)/100))
8020128
ubu@bruntu:/home/wideDB1$ echo $(($(date +%N)/1000))
9461
ubu@bruntu:/home/wideDB1$ echo $(($(date +%N)/10000))
81272
ubu@bruntu:/home/wideDB1$ echo $(($(date +%N)/10000))
bash: 046340492: value too great for base (error token is "046340492")

我必须准确地在bash脚本中使用此函数来同步任务之间的毫秒数，但这里非常不稳定。任何人都可以帮我找出这项工作的最佳解决方案吗？

干杯！的Alessandro

Answer 1

默认情况下，以0开头的数字字符串被视为八进制数字;您只需告诉bash在所有情况下将date +%N中捕获的字符串视为十进制数字，前缀为10#。

$ echo $(( 10#$(date +%N)/1000000))

Answer 2

$((...))内的数字字符串被解释为数字，但以0开头的数字被解释为八进制数字，即基数为8。

您可以通过参数扩展轻松删除前导零：

#!/bin/bash
shopt -s extglob          # Enable extended globbing -- in this case, +(0)

for i in {1..20} ; do
    nanos=$(date +%N)
    echo -n "$nanos -> "
    nanos=${nanos##+(0)}  # Remove as many 0's from the left as possible.
    echo $nanos
done

定期“对于基数来说太大了”在bash中对日期输出进行数学运算

2 个答案: