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I need to parse out instances of +word+ line by line (replace +word+ with blank). I'm currently using the following (working) sed regex:
我需要逐行解析+word+实例(用空格替换+word+)。我目前正在使用以下(工作)sed regex:
newLine=$(echo "$line" | sed "s/+[a-Z]\++//g")
This violates "SC2001" according to "ShellCheck" validation;
根据“ShellCheck”验证，这违反了“SC2001”;
SC2001: See if you can use ${variable//search/replace} instead.
I've attempted several variations without success (The string "+word+" remains in the output):
我尝试过几种变体，但没有成功(字符串“+word+”仍然保留在输出中):
newLine=$(line//+[a-Z]+/)
newLine=$(line/+[a-Z]+//)
newLine=$(line/+[a-Z]\++/)
newLine=${line//+[a-Z]+/}
and more..
I've heard that in some cases sed is necessary, but I would like to use Bash's built in find and replace if possible.
我听说在某些情况下sed是必要的，但是如果可能的话，我希望使用Bash的build in find和replace。
1 个解决方案
#1
3
The substitution in parameter expansion doesn't use regular expressions, but patterns. To get closer to regular expressions, you can turn on extended patterns:
参数扩展中的替换不使用正则表达式，而是使用模式。为了更接近正则表达式，您可以打开扩展模式:
shopt -s extglob
new_line=${line//++([a-Z])+}
#1
3
The substitution in parameter expansion doesn't use regular expressions, but patterns. To get closer to regular expressions, you can turn on extended patterns:
参数扩展中的替换不使用正则表达式，而是使用模式。为了更接近正则表达式，您可以打开扩展模式:
shopt -s extglob
new_line=${line//++([a-Z])+}