Unix command to remove dot at end of each line in file.
Sample rec in file
11234567 0.
23456789 5569.
34567810 1.
10162056 0.
Just use sed:
sed 's/\.$//' yourfile
Escape the special character . using \.
Put an achor $ to only remove it from the end.
To make infile changes use -i option of sed.
sed -e 's/\.$//'
done. (padding to make answer long enough. grumble)
sed -i 's/\.$//' /path/to/file
This will match a literal period that is anchored to the end of the line, and replace it with nothing. The -i tells sed to make the change inline.
Related
In zabbix-agent.conf I have lines:
# Example: Server=127.0.0.1,192.168.1.0/24,::1,2001:db8::/32,zabbix.example.com
Server=127.0.0.1
I want to replace line
Server=127.0.0.1
with my
Server=zabbix.mydomain.com
But if I do
sed -i -e 's/Server=127.0.0.1/Server=zabbix.mydomain.com/g' /etc/zabbix/zabbix_agentd.conf
it found also line with commented example and replace string in it. I get:
#<----->Example: Server=zabbix.mydomain.com,192.168.1.0/24,::1,2001:db8::/32,zabbix.example.com
Server=zabbix.mydomain.com
How to replace only one line?
You need to
Match the text at the start of string
Escape the dots
Remove g flag since the match will only be found at the string start.
Also, you do not need the -e option, you can use
sed -i 's/^Server=127\.0\.0\.1/Server=zabbix.mydomain.com/' /etc/zabbix/zabbix_agentd.conf
See the online demo:
#!/bin/bash
s='# Example: Server=127.0.0.1,192.168.1.0/24,::1,2001:db8::/32,zabbix.example.com
Server=127.0.0.1'
sed 's/^Server=127\.0\.0\.1/Server=zabbix.mydomain.com/g' <<< "$s"
Output:
Example: Server=127.0.0.1,192.168.1.0/24,::1,2001:db8::/32,zabbix.example.com
Server=zabbix.mydomain.com
This might work for you (GNU sed):
sed -i '/^Server=127\.0\.0\.1/cServer=zabbix.mydomain.com' file
Change line beginning Server=127.0.0.1 to Server=zabbix.mydomain.com.
The other answers are almost fine but they would also replace a line like:
Server=127.0.0.10
A complete solution with any sed could be:
sed -i 's/^Server=127\.0\.0\.1$/Server=zabbix.mydomain.com/' /etc/zabbix/zabbix_agentd.conf
^ and $ anchor the string to the beginning and end of line, respectively. Dots need backslash escape, else they stand for any character.
I have a file file.txt and it has the lines below. I want the queuename to be converted to uppercase, like this: queuename=SP00245B
# Queue name
#
queuename=sp00245b
awk '$1 == "queuename" {$2 = toupper($2)}1' FS== OFS== input-file
Note that this will fail if there are 2 = in the line, and only the values between the first 2 = will be uppercased. If that's an issue, it's an easy fix (left as an exercise for the reader).
A simple Perl solution:
perl -i -pe 's/^\s*queuename=\K(.*)/\U$1/' file.txt
(Remove -i if you don't want to modify the file in place.)
With GNU sed:
sed -i 's/\(^[[:blank:]]*queuename=\)\(.*\)/\1\U\2/' file.txt
This uses two captures groups and the \U sequence to toggle uppercase substitution for the second group.
You can also use the sed conversion \U to convert the portions of the matched pattern with the substitution command to uppercase. To covert everything following the '=' sign you could use, e.g.
sed '/^queuename=/s/=.*$/\U&/' filename
To edit the file in-place, include the -i option, e.g.
sed -i '/^queuename=/s/=.*$/\U&/' filename
Example Use/Output
$ echo "queuename=sp00245b" | sed '/^queuename=/s/=.*$/\U&/'
queuename=SP00245B
I have a text file and I would like to only delete the first character of the text file, is there a way to do this in shell script?
I'm new to writing scripts so I really don't know where to start. I understand that the main command most people use is "sed" but I can only find how to use that as a find and replace tool.
All help is appreciated.
You can use the tail command, telling it to start from character 2:
tail -c +2 infile > outfile
You can use sed
sed '1s/^.//' startfile > endfile
1s means match line 1, in substitution mode (s)
^. means at the beginning of the line (^), match any character (.)
There's nothing between the last slashes, which means substitute with nothing (remove)
I used to use cut command to do this.
For example:
cat file|cut -c2-80
Will show characters from column 2 to 80 only.
In your case you can use:
cat file|cut -c2-10000 > newfile
I hope this help you.
[]s
You can also use the 0,addr2 address-range to limit replacements to the first substitution, e.g.
sed '0,/./s/^.//' file
That will remove the 1st character of the file and the sed expression will be at the end of its range -- effectively replacing only the 1st occurrence.
To edit the file in place, use the -i option, e.g.
sed -i '0,/./s/^.//' file
or simply redirect the output to a new file:
sed '0,/./s/^.//' file > newfile
A few other ideas:
awk '{print (NR == 1 ? substr($0,2) : $0)}' file
perl -0777 -pe 's/.//' file
perl -pe 's/.// unless $done; $done = 1' file
ed file <<END
1s/.//
w
q
END
dd allows you to specify an offset at which to start reading:
dd ibs=1 seek=1 if="$input" of="$output"
(where the variables are set to point to your input and output files, respectively)
"The Unix Programming Environment" states that '$' used in regular expression in sed means end-of-the line which is fine for me, because
cat file | sed 's/$/\n/'
is interpretted as "add newline at the end of each line".
The question arises, when I try to use the command:
cat file | sed '$d'
Shouldn't this line remove each line instead of the last one? In this context, dollar sign means end of the LAST line. What am I getting wrong?
$ is treated as regex anchor when used in pattern in s command e.g.
s/$/\n
However in $d, $ is not a regex anchor, it is address notation that means the last line of the input, which is deleted using the d command.
Also note that cat is unnecessary in your last command. It can be used as:
sed '$d' file
In the second usage, there is no regular expression. The $ there is an address, meaning the last line.
Note that regex in sed must be inside the delimiters(;,:, ~, etc) other than quotes.
/regex/
ex:
sed '/foo/s/bar/bux/g' file
or
~regex~
ex:
sed 's~dd~s~' file
but not 'regex'. So $ in '$d' won't be considered as regex by sed. '$d' acts like an address which points out the last line.
I am trying to use sed to append the "^A" control character at the end of each line:
sed -i 's/$/^A/' testfile
I want the '\001' special character instead of the "^A" literal string.
Please suggest how to achieve this.
You can use \x01 for ^A i.e.:
sed -i.bak 's/$/\x01/' file
sed -i 's/$/^A/' file
when you press the ^A, just press in this way: Ctrl+(v+a together)
sed 's/$/'$'\01''/' testfile
Explanation
s/$/ — start the sed argument
$'\01' — let bash expand this
'/' — end the sed argument
(I recommend you start off without the rather dangerous -i flag and only add it once you're sure it does what you want.)