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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 $CCSR | grep ????? "
But the main issue is that it can be anything. It is "branches/features" this time, though it will always be in this format, with only two forward slashes.
I'm not sure what to put here so that only the part I want is passed to the variable.
Could you please suggest any solutions for this?
3 answers
grep is not the right tool for your case. You can use basename:
basename a/b/c --> c
basename a/b/c/ --> c
or, in your case
basename branches/features/arm_and_musl --> arm_and_musl
which, within a sh script, you would use as
dirname=`basename "$CCSR"`
Note, that if you need to call sh explicitly, you have to give the -c command line argument to sh if you want to pass commands on the command line.
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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
grepto 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
I wouldn't use grep. As the other answers already said, it's not the right tool for this job.
Considering your specific case (fields separated by /), basename is the most straighforward way, as stated in Dirk's answer. I just want to provide some other alternatives.
cut
The cut command can separate the input in fields and allows you to get only the ones you need. If you know for sure that the inputs will always be branches/features/something, then you can use / as separator and get the third field:
echo $CCSR | cut -d / -f 3
Or, if you don't know how many fields there will be, there's a little trick to get the last one:
echo $CCSR | rev | cut -d / -f 1 | rev
rev reverses the input, so first I reverse it, get the first field and reverse it again. The result will be the last field.
awk
The awk command works in a similar way: we tell it what the separator is and can get specific fields. The difference is that it has a better way to get the last one, by using the predefined variable NF:
echo $CCSR | awk -F / '{print $NF}'
Although basename is the most straighforward solution for your case, please note thatcut and awk are more generic, as they can work with different separators and allows you to get specific fields (not only the last one).
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