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codeman
Asked: 2022-05-13 12:27:38 +0800 CST 2022-05-13 12:27:38 +0800 CST

Leave only the last 4 Bash characters

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I have the text SAR_aPP.D7HG32689 , and I only want to get the last 4 characters, in this case it would be 2689 , but it can be something else, since the text is random and the total number of characters can also vary.

How can I get this with sed?

I tried with:

$ sed 's/SAR_aPP.//g'

To remove the first part, but what follows I have no idea.

I tried with: cut -c, but I could only see that it strips characters from front to back, or it won't use it correctly.

linux

1 Answers

  1. Best Answer

    You can use the following regular expression: (.{4})$, which encloses in a group those strings that have four characters and that are before reaching the end of the line.

    Wearingsed

    Its use with sedis:

    $ sed -r 's/.*(.{4})$/\1/g

    Where it -renables the use of extended regular expressions, and the metacharacter \1indicates that the match (the four-digit string) is to be displayed; the rest not.

    Example generating 2 random numbers of 5 digits:

    $ shuf -i 10000-90000 -n 2 | xargs -I {} echo 'SAR_aPP.D7HG{}' | sed -r 's/.*(.{4})$/\1/g'
8607
7169

    Wearinggrep

    But you can also with grep -Po '.{4}$':

    $ shuf -i 10000-90000 -n 2 | xargs -I {} echo 'SAR_aPP.D7HG{}' | grep -Po '.*\K(.{4})$'
2071
2481

    Wearingcut

    You can use a workaround with cut:

    $ shuf -i 10000-90000 -n 2 | xargs -I {} echo 'SAR_aPP.D7HG{}' | rev | cut -c 1-4 | rev
1817
9935

    Here I used revin each row to reverse the order. Thus, cuthe can now take the first four (which were the last), and then he invests those four again with another rev.

    with the shell

    You can also use the shell tools:

    In Bash or in Zsh:

    $ cadena=SAR_aPP.D7HG32689
$ echo ${cadena: -4}
2689
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