How would I pipe the ls through grep to have an alternating character case, as output?
For example:
I have files (1) aAbBaBbA, (2) bAbBaA, (3) bbAb,(4) AAaBbAa,(5) BBBaaa, and (6) aBaB.
I want to find 1,2,6.
What command would I type?
(Edit): My example wasn't as specific as it should have been. I wanted to include alternating starting with lower and upper. (6) should be changed from aBaB to BaBa, but I want to keep the original.
try
ls -1 | grep -E '^[A-Z]?([a-z][A-Z])*[a-z]?$'
EDIT: as pointed very correctly by @mikeserv, this won't work for non-ASCII chars. And indeed they do happen quite often (e.g. music files with foreign titles for filename). So the more robust way is:
ls -1 | grep -E '^[[:upper:]]?([[:lower:]][[:upper:]])*[[:lower:]]?$'
In the following, I retain just [A-Z] for sake of readability.
Also, a caveat: this will match a single char (upper or lower). One could argue that "alternating case" is defined by no successive similar case in a sequence of zero or more chars... :-)
Test:
mkdir -p /tmp/junk
cd /tmp/junk
touch aAbBaBbA bAbBaA bbAb AAaBbAa BBBaaa aBaB
ls -1 | grep -E '^[A-Z]?([a-z][A-Z])*[a-z]?$'
# aAbBaBbA
# aBaB
# bAbBaA
But that's not enough. Some more tests:
touch aB
touch aBcD
touch aBcDeF
touch aBcDEf
touch Ab
touch AbCd
touch AbCdEf
touch AbCdeF
touch AbCdEF
ls -1 | grep -E '^[A-Z]?([a-z][A-Z])*[a-z]?$'
# aAbBaBbA
# aB
# Ab
# aBaB
# aBcD
# AbCd
# aBcDeF
# AbCdEf
# bAbBaA
[[:upper:]][[:lower:]] to avoid locale problems. Use a + rather than the Kleene * and you avoid the blank line thing.
Type this command:
ls |grep -E '^([[:upper:]][[:lower:]])+[[:upper:]]?$|^([[:lower:]][[:upper:]])+[[:lower:]]?$'
To break this down.
ls is the main command
It is piped to grep
The switch -P or -E are telling grep to search using regular expressions (regex)
Within the single quote ' and the parentheses () the pattern is layed out.
Basically it reads match any upper letter [[:upper:]] then any smaller letter [[:lower:]]
Or using the pipe |
Then match any smaller letter [[:lower:]] then match any upper letter [[:upper:]]
The + will match everything within the parentheses () and play it out through the list given.
$ tells when the pattern ends
grep are you using? Mine is GNU grep 2.20
set aAbBaBbA bAbBaA bbAb AAaBbAa BBBaaa aBaB
l= u=; printf %s\\n "$@" |
grep -E "^([${l:=[:lower:]}][${u:=[:upper:]}])+[$l]?$|^([$u][$l])+[$u]?$"
aAbBaBbA
bAbBaA
aBaB
But see Pierre's answer for the (obviously) better version of this.
Still, there is another way:
grep -vE '[^[:alpha:]]|[[:lower:]]{2}|[[:upper:]]{2}|^$'
...though that will match only a single char on a line. You can fix that by doing a .? between the ^ and $ at the end though.
for loop is just to get it all over to grep but actually, that's not the best way to do it....
[[:lower:]] and [[:upper:]] are the robust classes - so you might as well just do it robustly from the start.
The solution is
grep -E '^([a-z][A-Z])+$'
