问题
I'm using sed on Centos, bash.
I want to replace everything between \plain and }} with a space in the below line of text:
stuff here \plain \f2\fs20\cf2 4:21-23}} more stuff over here, could be anything.
The text between \plain and }} will vary (different numbers/numbers). How can I do a wildcard to include everything between \plain and }}.
I was hoping a simple * would grab everything between the two but the wildcard in shell doesn't seem to work like this:
s/\\plain *}}/ /g;
The answer may be something incorporating this? [a-zA-Z0-9.] but that doesn't account for the backslashes, colons, and dashes in the text.
回答1:
Just add dot before * to match everything.
i.e. s/\\plain .*}}/ /g should work.
回答2:
The following regex ...
^\\plain .*}}$
... will match lines beginning with \plain, having anything in the middle, and ending with }}.
If that's no use, instead of .* to match everything, use ^ (negation) to match everything that's not }}.
来源：https://stackoverflow.com/questions/5843870/sed-wildcard-search-replace