问题描述
作为 为 maven 着色脚本 的核心,我运行了几个替换命令.sed
命令之一使用正则表达式,它可以在 shell 中找到作为 在这里讨论.可以在此处找到当前(不工作)的实现.
I run several substitution commands as the core of a colorize script for maven.One of the sed
commands uses a regular expression which works find in the shell as discussed here. The current (not working) implementation can be found here.
当我将命令的变体之一包含到脚本中时,会发生不同的行为:
When I include one of the variants of the command into the script different behavior occurs:
变体 1:
$ sed -re "s/([a-zA-Z0-9./\ :-]+)/1/g"
适应脚本:
-re "s/WARNING: ([a-zA-Z0-9./\ :-]+)/${warn}WARNING: 1${c_end}/g"
错误: shell 输出的信息与我输入 $ sed
的信息相同.奇怪!?
Error: The shell outputs the same information as if I would type $ sed
. Strange!?
变体 2:
$ sed -e "s/([a-zA-Z0-9./\ :-]+)/1/g"
适应脚本:
-e "s/WARNING: ([a-zA-Z0-9./\ :-]+)/${warn}WARNING: 1${c_end}/g"
错误:
sed: -e 表达式 #7, char 59: `s' 命令的 RHS 上的无效引用 1
推荐答案
您不需要实际捕获才能使其工作吗?即对于变体#2:
Don't you need to actually capture for that to work? i.e. for variant #2:
-r -e "s/WARNING: (([a-zA-Z0-9./\ :-]+))/${warn}WARNING: 1${c_end}/g"
(注:未经测试)
如果没有 -r
参数,反向引用(如 1
)将不起作用,除非每个括号都用 字符转义.
Without the -r
argument back-references (like 1
) won't work unless each parenthesis is escaped with a character.
使用 -r
,参数反向引用(如 1
)将不起作用,除非括号没有被转义.
With -r
, argument back-references (like 1
) won't work unless the parenthesis are NOT escaped.
这篇关于sed 错误:“'s' 命令的 RHS 上的引用无效 1"的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!