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 BSD
zgrep
ZGREP(1) General Commands Manual ZGREP(1)NAME
zgrep - search possibly compressed files for a regular expression
SYNOPSIS
zgrep [ grep_options ] [ -e ] pattern filename...
DESCRIPTION
Zgrep invokes grep on compressed or gzipped files. These grep options will cause zgrep to terminate with an error code:
(-[drRzZ]|--di*|--exc*|--inc*|--rec*|--nu*). All other options specified are passed directly to grep. If no file is specified, then the
standard input is decompressed if necessary and fed to grep. Otherwise the given files are uncompressed if necessary and fed to grep.
If the GREP environment variable is set, zgrep uses it as the grep program to be invoked.
EXIT CODE
2 - An option that is not supported was specified.
AUTHOR
Charles Levert (charles@comm.polymtl.ca)
SEE ALSO grep(1), gzexe(1), gzip(1), zdiff(1), zforce(1), zmore(1), znew(1)ZGREP(1)