I have to replace a field in one file with a field from other file.
I came across this awk command to replace a field with one string
nawk -F'|' -v OFS='|' '$2="replace"' temp2 > temp3
I need to have something like cut -f2 -d "|" temp1 (a field from other file) instead of 'replace'
Is... (8 Replies)
Howdy.
I know this is most likely possible using sed or awk or grep, most likely a combination of them together, but how would one go about running a grep like command on a file where you only try to match your pattern to the second field in a line, space delimited?
Example:
You are... (3 Replies)
hi
i have file as below , i want to add duplicate records like bell_bb to one record with valuve as 15 ( addition of both )
any oneline awk script to achive this ?
header 0
CAMPAIGN_NAME 1
Bell_BB 14
Bell_MONTHLY 803
SOLO_UNBEATABLE 644
Bell_BB 1
Bell_MONTHLY 25
SOLO_UNBEATABLE... (4 Replies)
i have something like this,
cat filename.txt
hui this si s"dfgdfg" omeone ipaddress="10.19.123.104" wel hope this works
i want to replace only 10.19.123.104 with different ip say 10.19.123.103
i tried this
sed -i "s/'ipaddress'/'ipaddress=10.19.123.103'/g" filename.txt
... (1 Reply)
Hi Friends,
Need Help. I have file1.txt as
File1.txt
|123|A|7267|Hyder|Cross|Sell|7801
|995|A|7051|2008|Lunar|New|Year|Promotion|7801
|996|A|7022|Q108|Targ|Prospect|&|SSCC|Savings|Promo|7801
|997|A|7182|Q1|Feb-Apr|08|Credit|ITA|PA|SBA|Campaign|7801
File2.txt... (7 Replies)
Hi,
I am trying with the below Perl command to print the first field when the second field matches the given pattern:
perl -lane 'open F, "< myfile"; for $i (<F>) {chomp $i; if ($F =~ /patt$/) {my $f = (split(" ", $i)); print "$f";}} close F' dummy_file
I know I can achieve the same with the... (7 Replies)
Hi all,
I have a requirement to replace a field with a character as per the length of the field.
Suppose i have a file where second field is of 20 character length. I want to replace second field with 20 stars (*). like ********************
As the field is not a fixed one, i want to do the... (2 Replies)
I have a .CSV file (file.csv) whose data are all enclosed in double quotes. Sample format of the file is as below:
column1,column2,column3,column4,column5,column6, column7, Column8, Column9, Column10
"12","B000QRIGJ4","4432","string with quotes, and with a comma, and colon: in... (3 Replies)
I am trying to remove lines in the target.txt file if $5 before the - in that file matches sorted_list. I have tried grep and awk. Thank you :).
grep
grep -v -F -f targets.bed sort_list
grep -vFf sort_list targets
awk
awk -F, '
> FILENAME == ARGV {to_remove=1; next}
> ! ($5 in... (2 Replies)
Discussion started by: cmccabe
2 Replies
LEARN ABOUT DEBIAN
partimaged-passwd
partimaged-passwd(8) Partition Image Server Configuration partimaged-passwd(8)NAME
partimaged-passwd - Manage partimaged user accounts
SYNTAX
partimaged-passwd [-Dhl] username password
partimaged-passwd [-Dhl] username
DESCRIPTION
partimaged can either authenticate against local user accounts (This needs access to /etc/shadow. As this is a potential security risk this
method is not recommended) or its own password database in /etc/partimaged/passwd.db. To simplify the management of the partimaged user
database this tool was written. It allows to easily add and remove users or list the users in the database. All users in this database are
allowed to access the partimaged server.
OPTIONS -D username
Delete the specified user from the password file.
-l List users in password file and exit.
-h Output help information and exit.
FILES
/etc/partimaged/passwd.db
AUTHORS
Michael Biebl <biebl@debian.org>
SEE ALSO partimaged(8), partimagedusers(5), partimage(1)Michael Biebl <biebl@teco.edu> 0.1 partimaged-passwd(8)