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#1
07-29-2010
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Replace a string after n semicolon every line
I have a file that is formatted in this way. Code:
a1;b2;c33;d4;e5;e;f;f;f;s d;ds;d;a;v;b;g;gr;r;rt;fdf s1;s2;s2;s3;s4; b1;f2;g3;h4;a3c4e;xcsd;fds; sd2;fs4;fs2;sdf3; I want to replace the value just before the 4th semicolon to empty string, regardless the value, such that it looks like this for every block of data separated by a line Code:
a1;b2;c33;;e5;e;f;f;f;s d;ds;d;a;v;b;g;gr;r;rt;fdf s1;s2;s2;s3;s4; b1;f2;g3;;a3c4e;xcsd;fds; sd2;fs4;fs2;sdf3; I tried using Code:
sed 's/[^*];//4' but to no avail. Any ideas? Last edited by alienated; 07-29-2010 at 11:23 AM.. |
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#2
07-29-2010
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Code:
echo 'a1;b2;c33;d4;e5' | nawk -F';' '{$4=""}1' OFS=';'
OR
echo 'a1;b2;c33;d4;e5' | sed 's/[^;][^;]*;/;/4'Last edited by vgersh99; 07-29-2010 at 11:38 AM.. Reason: regex fix |
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alienated (07-29-2010) | ||
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#4
07-30-2010
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Code:
# echo 'a1;b2;c33;d4;e5' | sed 's/[^;]*;/;/4' a1;b2;c33;;e5 Code:
# echo 'a1;b2;c33;d4;e5' | sed 's/\(.*\)[a-z][0-9]*;\([a-z0-9]*\)*$/\1;\2/' a1;b2;c33;;e5 |
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