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Shell Programming and Scripting
Grep: show only first match per line?
Post 303000827 by Corona688 on Thursday 20th of July 2017 12:11:06 PM
07-20-2017
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1. Shell Programming and Scripting
I'm having problems since few days ago, and i'm not able to make it works with a simple awk+grep script (or other way to do this).
For example, i have a input file1.txt:
cat inputfile1.txt
218299910417
1172051195
1172070231
1172073514
1183135117
1183135118
1183135119
1281440202
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Hello
I need some help with this job.
file.txt
----- cut ----
TARGET
13/11/08
20:43:21
POINT 1
MOVE 8
772102y64312417771
TARGET
13/11/08
21:10:01
POINT 2
MOVE 5
731623jjd12njhd
----- cut ----
this is the example.
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Hello
I have a silly question. I need to grep a match in text file
and then print 5 lines after it.
grep -A 5 .... do it.
OK
The next thing I can not handle is I need each output to be on 1 line
match line2 line3 line4 line5
match line2 line3 line4 line5
etc..
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Hi
i have a file which have a pattern like this
Nov 10 session closed
Nov 10 Nov 9 08:14:27 EST5EDT 2010 on tty .
Nov 10 Oct 19 02:14:21 EST5EDT 2010 on pts/tk .
Nov 10 afrtetryytr
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Dec 8 Nov 10 05:03:02 EST5EDT 2010
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Hi All
I have a file with tha same line multiple times and its easy to grep out those lines using grep "pattern" filealthough I need to know exactly what the next line after those lines are
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Hello,
Here I have some grep command which is not working correctly:
cat file1.txt:
apples
Date: Sun, 24 Feb 2013 8:14:06 -0800
peaches melons
cherry sky cloud
green purple
yellow
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apples
Date
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Hi Guru's,
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Hello,
one step in a shell script i am writing, involves Grep command to search a regular expression in a line an only print the string after the match
an example line is below
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