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Comments on How to use grep to print only specific word from a string
Parent
How to use grep to print only specific word from a string
I have a variable that contains a string:
$CCSR = "branches/features/arm_and_musl"
I want to get only the part after the last /
. In this case it's "arm_and_musl" but it can be anything.
So something like:
def dirname= sh " echo $CCSR | grep ????? "
But the main issue is that it can be anything. It is "branches/features" this time, though it will always be in this format, with only two forward slashes.
I'm not sure what to put here so that only the part I want is passed to the variable.
Could you please suggest any solutions for this?
Post
While I agree with Dirk Herrmann that basename
is the right tool for the job, I think it is still worthwhile to know how to do it with grep
, because you might one day encounter a sufficiently similar situation that is not covered by a specialised tool.
The question can be broken into two questions:
-
What regular expression matches exactly the part after the last slash.
-
How to get
grep
to only output the part that matches, instead of the whole line.
For the first question, note that the part after the last slash is characterized by the fact that it doesn't contain a slash (otherwise the previous slash would not have been the last one) and extends to the end of the line.
To match a character that's anything but a slash, you can use [^/]
.
You want to match any number of them, so put a star afterwards: [^/]*
. But you only want to match it if it extends to the end of line (so you do not match e.g. the branches
in your example), this is done by terminating the regular expression with a dollar sign. So the complete regular expression you need is [^/]*$
For the second question, the option -o
does exactly what you need. From the man page:
-o, --only-matching
Print only the matched (non-empty) parts of a matching line,
with each such part on a separate output line.
So putting it all together, we get the desired behaviour:
$ echo "branches/features/arm_and_musl" | grep -o '[^/]*$'
arm_and_musl
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