[英]Bash - replacing underscores with spaces, except leading/trailing ones
I want underscores between words to be replaced with spaces, and leading and trailing underscores left alone. 我希望将单词之间的下划线替换为空格,并单独保留前导和尾随下划线。 For example:
例如:
__hello_world_a_b___
hello___world
should become 应该成为
__hello world a b___
hello world
Using Bash with its regular expression support: 使用Bash及其正则表达式支持:
string='__hello_world_a_b___'
[[ $string =~ ^(_*)(.*[^_])(_*)$ ]]
echo "${BASH_REMATCH[1]}${BASH_REMATCH[2]//_/ }${BASH_REMATCH[3]}"
To check that it works, let's make a script that will take the string as argument: 为了检查它是否有效,让我们创建一个脚本,该脚本将字符串作为参数:
#!/bin/bash
string=$1
[[ $string =~ ^(_*)(.*[^_])(_*)$ ]]
echo "${BASH_REMATCH[1]}${BASH_REMATCH[2]//_/ }${BASH_REMATCH[3]}"
Call this script banana
, chmod +x banana
and let's go: 将此脚本
banana
, chmod +x banana
,让我们开始:
$ ./banana '__hello_world_a_b___'
__hello world a b___
$ ./banana '__hello_world_a_b'
__hello world a b
$ ./banana 'hello_world_a_b___'
hello world a b___
$ ./banana 'hello_world_a_b'
hello world a b
$ ./banana '___'
$ # the previous output is empty
$ ./banana $'___hello_world_with\na_newline___'
___hello world with
a newline___
$ ./banana 'hello___world'
hello world
You could simply use the below Perl command which uses the PCRE verb (*SKIP)(*F)
. 您可以简单地使用下面的Perl命令,该命令使用PCRE动词
(*SKIP)(*F)
。
$ echo "hello___world" | perl -pe 's/(?:^_+|_+$)(*SKIP)(*F)|_/ /g'
hello world
$ echo "__hello_world_a_b___" | perl -pe 's/(?:^_+|_+$)(*SKIP)(*F)|_/ /g'
__hello world a b___
The above regex would match all the _
except the leading and trailing ones. 上面的正则表达式将匹配所有
_
除了前导和尾随的。
Another Pure Bash possibility that doesn't use regular expression but extended globs, in a very pedestrian way: 另一个不使用正则表达式而是扩展glob的Pure Bash可能性,以一种非常行人的方式:
#!/bin/bash
shopt -s extglob
string=$1
wo_leading=${string##+(_)}
wo_underscore=${wo_leading%%+(_)}
printf -v leading '%*s' "$((${#string}-${#wo_leading}))"
printf -v trailing '%*s' "$((${#wo_leading}-${#wo_underscore}))"
echo "${leading// /_}${wo_underscore//_/ }${trailing// /_}"
The variables wo_leading
will contain the string without leading underscores, and the variable wo_underscore
will contain the string without leading and trailing underscores. 变量
wo_leading
将包含不带下划线的字符串,变量wo_underscore
将包含不带下划线和结尾的字符串。 From here, it's easy to get the number of leading and trailing underscore, to replace underscores by spaces in wo_underscore
and put back everything together. 从这里,很容易获得前导和尾随下划线的数目,用
wo_underscore
的空格替换下划线并将所有内容放回去。
Another Perl answer: Perl的另一个答案:
perl -pe 's/(?<=[^\W_])(_+)(?=[^\W_])/ " " x length($1) /ge' <<END
__hello_world_a_b___
hello___world
END
__hello world a b___
hello world
That is: a sequence of underscores preceded by a character that is a word character except underscore, and followed by a character that is a word character except underscore. 即:一个下划线序列,其后是一个字符,该字符是除下划线之外的单词字符,然后是一个字符,该字符是除下划线之外的单词字符。
如果您有GNU awk,则可以使用
awk '{match($0,"^(_*)(.*[^_])(_*)$",arr); print arr[1] gensub("_"," ","g",arr[2]) arr[3]}'
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