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Q&A How to use grep to print only specific word from a string

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 simil...

posted 2y ago by celtschk‭

Answer
#1: Initial revision by user avatar celtschk‭ · 2022-07-21T09:44:19Z (over 2 years ago)
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:

 1. What regular expression matches exactly the part after the last slash.

 2. 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
```