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I have those strings

gcc-4.8.3-1.aix7.1.ppc.rpm
gcc-c++-4.8.3-1.aix7.1.ppc.rpm
gcc-cpp-4.8.3-1.aix7.1.ppc.rpm
gcc-gfortran-4.8.3-1.aix7.1.ppc.rpm
libgcc-4.8.3-1.aix7.1.ppc.rpm

I want to cut starting from ".aix7.1" to end to obtain

gcc-4.8.3-1
gcc-c++-4.8.3-1
gcc-gfortran-4.8.3-1
libgcc-4.8.3-1

Plese,how to do this with awk?

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6
  • I know a solution with sed,simply using s:4.3 etc etc::g,but i want to know awk Commented Apr 23, 2015 at 14:31
  • 1
    awk -F'.aix7.1' '{print $1}' But there is more appropriate solutons with sed and bash variables Commented Apr 23, 2015 at 14:31
  • @Costas, please post that as an answer Commented Apr 23, 2015 at 14:34
  • 1
    @Costas; the two dots (.) in the field separator should better be escaped \\. since the FS is a regexp. Commented Apr 23, 2015 at 14:34
  • @Janis aix7 is just enough to sure indicate where to separate the line, so for simplity OP can leave . as is (even if it can be treat as any symbol) Commented Apr 23, 2015 at 14:54

2 Answers 2

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5

I would split the line on dots and remove the last 4 fields

awk -F. -v OFS=. '{NF-=4}1'

But Costas's code in his comment more directly answers the requirements.

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  • tricky and nice! Commented Apr 23, 2015 at 14:40
1

You can do it like this:

root@localhost:~# cat test
gcc-4.8.3-1.aix7.1.ppc.rpm
gcc-c++-4.8.3-1.aix7.1.ppc.rpm
gcc-cpp-4.8.3-1.aix7.1.ppc.rpm
gcc-gfortran-4.8.3-1.aix7.1.ppc.rpm
libgcc-4.8.3-1.aix7.1.ppc.rpm
root@localhost:~# awk 'BEGIN{FS=".aix7.1"} {print $1}' test
gcc-4.8.3-1
gcc-c++-4.8.3-1
gcc-cpp-4.8.3-1
gcc-gfortran-4.8.3-1
libgcc-4.8.3-1
root@localhost:~#
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