I try to replace this samples:
http_proxy="http://QQ@127.0.0.1:8080/"
http://test1:test2@127.0.0.1:8080/
http_proxy="http://QQ@127.0.0.1:8080/
http://@127.0.0.1:8080/"
with this regex (^.+?\/\/).+?(@.*$)
to get it like this
http_proxy="http://user:pass@127.0.0.1:8080/"
http://user:pass@127.0.0.1:8080/
http_proxy="http://user:pass@127.0.0.1:8080/"
http://user:pass@127.0.0.1:8080/
According to https://regex101.com/r/AE3Wxi/3 the regex seems to be working for the first 3 lines.
But when i try it with
echo http_proxy=\"http://QQ@127.0.0.1:8080/\" | sed 's/\(^.+?\/\/\).+?\(@.*$\)/\1user:pass\2/g'
It has this output:
http_proxy="http://QQ@127.0.0.1:8080/"
>Solution :
You have to escape the plus \+ and sed does not support non greedy quantifiers like .\+?
If you also want a match for the last example http://@127.0.0.1:8080/" the quantifier after the double forward slash should be * instead of +
You could write the command as:
echo http_proxy=\"http://QQ@127.0.0.1:8080/\" | sed 's/\(^.\+\/\/\).*\(@.*$\)/\1user:pass\2/'
Output
http_proxy="http://user:pass@127.0.0.1:8080/"
If you only want to replace the first occurrence in the line, you might shorten it to
echo http_proxy=\"http://QQ@127.0.0.1:8080/\" | sed 's~\(//\)[^@]*\(@\)~\1user:pass\2~'
See a regex demo and here for the captured group values.