There is a directory containing the following files:
.
├── bla-bla-bla1.tar.7z
├── bla-bla-bla2.tar.7z
├── bla-bla-bla3.tar.7z
└── _bla-bla-bla_foo.tar.7z
I need to find and delete all of the files that in the format of *.7z except _*.7z
I've tried:
find /backups/ -name "*.7z" -type f -mtime +180 -delete
How I can do it?
Another approach is to use an additional, negated primary with find:
find /backups/ -name "*.7z" ! -name '_*.7z' -type f -mtime +180 -delete
The simple regex in the other answers is better for your use case, but this demonstrates a more general approach using the ! operator available to find.
'_*.7z', which will exclude the 7z files starting with "_", in the current form will exclude only a file named (exactly) "_.7z".
In regular expressions, the ^ operator means "any character except for". Thus [^_] means "any character except for _". E.g.:
"[^_]*.7z"
So, if your intention is to exclude files starting with _, your full command line would be:
find /backups/ -name "[^_]*.7z" -type f -mtime +180 -delete
If you'd like to exclude any occerance of _, you can use the and and not operators of find, like:
find . -name "*.7z" -and -not -name "*_*"
It should be
find . -name "*[^_].7z"
A quick way given you have bash 4.2.25, is to simply use bash pattern matching to remove all .7z, but the ones having _.7z, like this:
touch a.7z b.7z c.7z d_.7z e_.7z f.txt
rm *[^_].7z
