如果行不以 # 开头，则重复删除标记 [英] remove token repeatedly if line does not start with #

问题描述

我想从我的文本文件中删除所有逗号，除非一行以 # 开头

I want to remove all commas from my text file unless a line starts with #

例如:

a, b, c
#a, b, c

应该转向:

a b c
#a, b, c

我不介意双重扫描文件，但我想用 sed 来做到这一点

I don't mind double scan the file but I want to do that with sed

推荐答案

你可以试试下面的 sed 命令，

You could try the below sed command,

$ sed '/^ *#/!s/,//g' file
a b c
#a, b, c

^ 断言我们在开始.所以上面的命令将匹配以零个或多个空格和 # 符号开头的行.然后下面的 ! 使 sed 反转选择，即它强制 sed 在不匹配的行上进行替换.s/,//g 用空字符串替换所有逗号.

^ asserts that we are at the start. So the above command will match the lines which starts with zero or more spaces and a # symbol. Then the following ! makes the sed to inverse the selections ie, it forces the sed to do the replacement on the lines which are not matched. s/,//g replaces all the commas with an empty string .

通过awk，

$ awk '!/^ *#/{gsub(/,/,"")}1' file
a b c
#a, b, c

! 在开头否定模式.同样，它只会在开头没有 # 的行上进行替换.

! at the start negates the patten. Likewise , it will do the replacement only on the lines which don't have # at the start.

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