Hi -
I am guessing this is fairly simple for someone .. but I can not quite figure it out. I need a sed command to print just parts of lines from a file.
e.g. filea.txt
4710451 : Success : MODIFY : cn=user1,dc=org,dc=uk
Message log started
Message log ended
4710452 : Success : MODIFY :... (7 Replies)
Hey, I found a way to print the lines which is just before a regular expression, not including the expression.
sed -n '/regexp/{n;p;}' myfile
Now I'm looking for a way to print all lines, exept the regular expression and also the line before the same regular expression.
Use code tags. (1 Reply)
I am trying to print those line which has no # in the begining of the line.
The sed I used for this purpose as shown below is not giving the required output.
echo 'PDE 5600' | sed -n 's/^\!#/&/p'
Where lies the problem:confused: (3 Replies)
I have read many threads, but I still didn't find the right answer. May be i didn't find the right thread, though are so many threads for the same question.
Basically the situation is - find date in a file and replace it with another date. (its not homework, its part of lot of a big processing,... (10 Replies)
I would like to print 3 lines after a regular expression is found in the logfile. I'm using the following code:
grep -n "$reg_exp" file.txt |while read LINE ;do i=$(echo $LINE |cut -d':' -f1 ) ;sed -n "$i,$(($i+3))p" file.txt ;done
The above code things works fine,but sometimes gives erroneous... (3 Replies)
Hello All,
I have something like below
LDC100/rel/prod/libinactrl.a
LAA2000/rel/prod/libinactrl.a
I want to remove till first forward slash that is outputshould be as below
rel/prod/libinactrl.a
rel/prod/libinactrl.a
How can I do that ??? (8 Replies)
I am trying to extract a table of data (mysql query output) from a log file. I need to print everything below the header and not past the end of the table. I have spent many hours searching with little progress. I am matching the regexp +-\{99\} with no problem. I just can't figure out how to print... (5 Replies)
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
/logs/GRAS/LGT/applogs/lgt-2016-08-24/2016-08-24.8.log.zip:2016-08-24 19:12:48,602 ERROR... (9 Replies)
Discussion started by: Ramneekgupta91
9 Replies
LEARN ABOUT PLAN9
grep
GREP(1) General Commands Manual GREP(1)NAME
grep - search a file for a pattern
SYNOPSIS
grep [ option ... ] pattern [ file ... ]
DESCRIPTION
Grep searches the input files (standard input default) for lines (with newlines excluded) that match the pattern, a regular expression as
defined in regexp(6). Normally, each line matching the pattern is `selected', and each selected line is copied to the standard output.
The options are
-c Print only a count of matching lines.
-h Do not print file name tags (headers) with output lines.
-i Ignore alphabetic case distinctions. The implementation folds into lower case all letters in the pattern and input before interpre-
tation. Matched lines are printed in their original form.
-l (ell) Print the names of files with selected lines; don't print the lines.
-L Print the names of files with no selected lines; the converse of -l.
-n Mark each printed line with its line number counted in its file.
-s Produce no output, but return status.
-v Reverse: print lines that do not match the pattern.
Output lines are tagged by file name when there is more than one input file. (To force this tagging, include /dev/null as a file name
argument.)
Care should be taken when using the shell metacharacters $*[^|()= and newline in pattern; it is safest to enclose the entire expression in
single quotes '...'.
SOURCE
/sys/src/cmd/grep.c
SEE ALSO ed(1), awk(1), sed(1), sam(1), regexp(6)DIAGNOSTICS
Exit status is null if any lines are selected, or non-null when no lines are selected or an error occurs.
GREP(1)