仅使用 bash 子字符串删除来去除前导零
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Stripping leading zeros using bash substring removal only
提问by user2969118
I am aware of other ways to solve this problem that do work (and I am currently using the 'expr' method), but it bugs me that I cannot figure out how to do it with the bash built-in functions only.
我知道解决这个问题的其他方法是有效的（我目前正在使用“expr”方法），但让我烦恼的是我无法弄清楚如何仅使用 bash 内置函数来解决这个问题。
When I try to strip the leading zeros from a variable using the ${variable##remove} construct I can either only get it to remove one zero or all numbers.
当我尝试使用 ${variable##remove} 构造从变量中去除前导零时，我只能让它删除一个零或所有数字。
string="00123456"
echo "${string##0}" // Only the first 0 is removed
echo "${string##0*0}" // The whole string is removed
echo "${string#0*0}" // Works
string="01230"
echo "${string##0}" // Works
echo "${string##0*0}" // The whole string is removed
echo "${string#0*0}" // Again, the whole string is removed
I read the bash manual twice to see if I am doing it correctly, but the official documentation is sparse at best. I am aware that enabling extglob may also solve this problem but it seems like overkill for a problem as simple as this.
我读了两次 bash 手册，看看我是否做得正确，但官方文档充其量是稀疏的。我知道启用 extglob 也可以解决这个问题，但对于像这样简单的问题来说似乎有点矫枉过正。
Am I missing something obvious or is it really that hard to remove one or more leading zeros from a string using bash functions only?
我是否遗漏了一些明显的东西，或者仅使用 bash 函数从字符串中删除一个或多个前导零真的很难吗？
回答by devnull
The following would remove all the leading 0s from the string:
以下内容将从0字符串中删除所有前导s：
$ string="000123456000"
$ echo "${string#"${string%%[!0]*}"}"
123456000
Saying "${string%%[!0]*}"would return the match after deleting the longest trailing portion of the string that satisifies [!0]*-- essentially returning the zeros at the beginning.
Saying"${string%%[!0]*}"将在删除满足的字符串的最长尾随部分后返回匹配[!0]*- 基本上返回开头的零。
"${string#"${string%%[!0]*}"}"would remove the part returned by the above from the beginning of the string.
"${string#"${string%%[!0]*}"}"将从字符串的开头删除上述返回的部分。
Alternatively, you could use shell arithmetic:
或者，您可以使用 shell 算法：
$ string="0000123456000"
$ echo $((10#$string))
123456000
回答by twalberg
Yet another way, although this requires the extglob functionality being enabled:
另一种方式，虽然这需要启用 extglob 功能：
echo ${string/#+(0)/}
回答by anubhava
This is also a pure BASH way of stripping leading zeroes in an one-liner:
这也是一种在单行中去除前导零的纯 BASH 方式：
string="00123456"
[[ "$string" =~ ^0*(.*)$ ]] && echo "${BASH_REMATCH[1]}"
123456
回答by LinuxGuru
Working with minutes and hours in a cron job generator script I found that leading zeros in front of 08 & 09 cause bash to think the value is an invalid octal amount. The script could not just strip leading zeros because a single zero is still valid. The solution was to pipe to bc.
在 cron 作业生成器脚本中使用分钟和小时，我发现 08 和 09 前面的前导零导致 bash 认为该值是无效的八进制数。该脚本不能仅仅去除前导零，因为单个零仍然有效。解决方案是通过管道连接到 bc。
HOUR=09
echo $HOUR | bc
9
HOUR=0
echo $HOUR | bc
0
HOUR=14
echo $HOUR | bc
14
回答by Robin A. Meade
Delete the longest substring from the beginning of the string matching the extended glob pattern +(0), that is, one or more zeros.
从匹配扩展glob模式的字符串的开头删除最长的子字符串+(0)，即一个或多个零。
shopt -s extglob
string="0000123456000"
echo ${string##+(0)}
prints:
印刷：
123456000
Note: this will reduce a string containing only zeros to an empty string.
注意：这会将仅包含零的字符串减少为空字符串。
shopt -s extglob
string="0000000000000"
echo ${string##+(0)}
prints an empty string.
打印一个空字符串。
If you want a string of zeros to reduce to a single zero, you can use arithmetic expansion together with the [base#]nform to force an arithmetic base of 10 to prevent octal interpretation:
如果您希望将一串零减少为单个零，您可以使用算术扩展和[base#]n形式来强制算术基数为 10 以防止八进制解释：
string="0000000000000"
echo $((10#$string))
prints:
印刷：
0
