bash脚本，非法数字：08 [英] Bash script, Illegal number: 08

问题描述

我运行在Ubuntu一个pretty简单的bash脚本，但已经提出了一个问题。
如果需要的话我会后整个脚本，但我已经缩小的问题。
基本上，我想运行一些$ C $每15秒C，所以我开始用这样的：

I'm running a pretty simple bash script in ubuntu but have come up with a problem. If needed I'll post the whole script, but I've narrowed down the problem. Basically, I want to run some code every 15 seconds, so I started with this:

time=`date +%S`

time2=$((time%15))

if [ $time2 -eq 0 ]

then

等等，等等，等等...

etc, etc, etc....

问题出现时，时间应该为：08秒。该脚本数目非法终止：08。
添加了这个，使用时：

The problem comes up when time is 08 seconds. The script terminates with Illegal number: 08. Adding to that, when using:

time2=$(($time%15))

而不是非法编号的错误，它会算术的前pression终止：期待EOF：08％15

instead of the illegal number error, it would terminate with Arithmetic expression: expecting EOF: "08%15"

我猜测08是不是PTED为数字间$ P $。还是有一些基本的问题，喜欢它认为它在八进制或东西。任何帮助吗？

I'm guessing 08 isn't interpreted as a number. Or there's some base issue, like it thinks it's in octal or something. Any help?

推荐答案

最短的解决方案：

time2=$(( ${time#0} % 15 ))

$ {VAR＃水珠} 办法 $ VAR 与水珠从一开始就删除，如果present。

${var#glob} means "$var with glob removed from the beginning if present".

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