bash 使用带通配符的 SED
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/9189120/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Using SED with wildcard
提问by mahmood
I want to replace a string with wildcard but it doesn't work.
我想用通配符替换字符串，但它不起作用。
The string looks like "some-string-8"
字符串看起来像“some-string-8”
I wrote
我写
sed -i 's/string-*/string-0/g' file.txt
but the output is
但输出是
some-string-08
回答by tom
The asterisk (*) means "zero or more of the previous item".
星号 (*) 表示“前一项的零个或多个”。
If you want to match any single character use
如果要匹配任何单个字符，请使用
sed -i 's/string-./string-0/g' file.txt
If you want to match any string (i.e. any single character zero or more times) use
如果要匹配任何字符串（即任何单个字符零次或多次），请使用
sed -i 's/string-.*/string-0/g' file.txt
回答by Michael Manoochehri
So, the concept of a "wildcard" in Regular Expressions works a bit differently. In order to match "any character" you would use "." The "*" modifier means, match any number of times.
因此，正则表达式中“通配符”的概念有点不同。为了匹配“任何字符”，您可以使用“。” “*”修饰符表示匹配任意次数。
