UNIX programming linux - linux

Hi am trying to extract the part of the line after a decimal.
My input line is
A-b-2.0.2-c
I want to use sed to get the output as
2.0.2-c
Can some one help me out?

Not perfect, but working:
echo "A-b-2.0.2-c" | sed -r -n -e 's/^[^0-9]+([0-9.]+.*)$/\1/p'
The issue is that the sed command does not support non-greedy matching which is why you have to use an exclusion pattern for the leading part of the string. For a more precise matching pattern you'd have to use some other tool.

This is definetly an exercise sed is made for. With the information that you provide following would work:
$ sed 's/A-b-//' input.txt
There is also a bash way to do it:
#!/usr/bin/bash
input="A-b-2.0.2-c"
echo ${input/A-b-/}
would also extract the line after the decimal.

Related

How to make GNU sed remove certain characters from a line

I have a following line;
�5=?�#A00165:69:HKJ3YDMXX:1:1101:16812:7341 1:N:0:TCTTAAAG
and would like to remove characters, �5=?� in front of #. So the desired output looks as follows;
#A00165:69:HKJ3YDMXX:1:1101:16812:7341 1:N:0:TCTTAAAG
I used gnu sed (v4.8)with a following argument;
sed "s/.*#/#/"'
but this did not remove �5=?� thought it worked in the GNU sed live editor.
At this point, I really appreciate any help on this.
My system is 3.10.0-1160.71.1.el7.x86_64
Using sed, remove everything up to the first occurance of #
$ sed 's/^[^#]*//' input_file
#A00165:69:HKJ3YDMXX:1:1101:16812:7341 1:N:0:TCTTAAAG
This might work for you (GNU sed):
sed -E 's/(\o357\o277\o275)5=\?\1//g' file
This removes all occurrences of �5=?�.
N.B. To translate the octal strings use sed -n l file to display the file as is. The triplets \357\277\275 can be matched in the LHS of the substitute command by using \o357\o277\o275.

Conditional replace using sed

My question is probably rather simple. I'm trying to replace sequences of strings that are at the beginning of lines in a file. For example, I would like to replace any instance of the pattern "GN" with "N" or "WR" with "R", but only if they are the first 2 characters of that line. For example, if I had a file with the following content:
WRONG
RIGHT
GNOME
I would like to transform this file to give
RONG
RIGHT
NOME
I know i can use the following to replace any instance of the above example;
sed -i 's/GN/N/g' file.txt
sed -i 's/WR/R/g' file.txt
The issue is that I want this to happen only if the above patterns are the first 2 characters in any given line. Possibly an IF statement, although i'm not sure what the condition would look like. Any pointers in the right direction would be much appreciated, thanks.
just add the circumflex, remove g suffix (unnecessary, since you want at most one replacement), you can also combine them in one script.
sed -i 's/^GN/N/;s/^WR/R/' file.txt
Use the start-of-string regexp anchor ^:
sed -i 's/^GN/N/' file.txt
sed -i 's/^WR/R/' file.txt
Since sed is line-oriented, start-of-string == start-of-line.

Change variable evaluation method in all script from $VAR_NAME to ${VAR_NAME}

We have couple of scripts where we want to replace variable evaluation method from $VAR_NAME to ${VAR_NAME}
This is required so that scripts will have uniform method for variable evaluation
I am thinking of using sed for the same, I wrote sample command which looks like follows,
echo "\$VAR_NAME" | sed 's/^$[_a-zA-Z0-9]*/${&}/g'
output for the same is
${$VAR_NAME}
Now i don't want $ inside {}, how can i remove it?
Any better suggestions for accomplishing this task?
EDIT
Following command works
echo "\$VAR_NAME" | sed -r 's/\$([_a-zA-Z]+)/${\1}/g'
EDIT1
I used following command to do replacement in script file
sed -i -r 's:\$([_a-zA-Z0-9]+):${\1}:g' <ScriptName>
Since the first part of your sed command searches for the $ and VAR_NAME, the whole $VAR_NAME part will be put inside the ${} wrapper.
You could search for the $ part with a lookbehind in your regular expression, so that you end up ending the sed call with /{&}/g as the $ will be to the left of your matched expression.
http://www.regular-expressions.info/lookaround.html
http://www.perlmonks.org/?node_id=518444
I don't think sed supports this kind of regular expression, but you can make a command that begins perl -pe instead. I believe the following perl command may do what you want.
perl -p -e 's/(?<=\$)[_a-zA-Z0-9]*/{$&}/g'
PCRE Regex to SED

Linux Shell Programming. Implementing a Search, Find and Replace Technique

I have to implement an application in shell programming (Unix/Linux).
I have to search a word from a text file and replace that word with my given word. I have a knowledge on shell and still learning.
I am not expecting source code. Can anybody help me or suggest me or give me some similar solution....
cat abc.txt | grep "pattern" | sed 's/"pattern"/"new pattern"/g'
The above command should work
Thanks,
Regards,
Dheeraj Rampally
Say you are looking for pattern in a file (input.txt) and want to replace it with "new pattern" in another (output.txt)
Here is the main idea, without UUOC:
<input.txt sed 's/"pattern"/"new pattern"/g' >output.txt
todo
Now you need to embed this line in your program. You may want to make it interactive, or a command that you could use with 3 parameters.
edit
I tried to avoid the use of output.txt as a temporary file with this:
<input.txt sed 's/"pattern"/"new pattern"/g' >input.txt
but it empties input.txt for a reason I can't understand. So I tried with a subshell, so:
echo $(<input.txt sed 's/pattern/"new pattern"/g')>input.txt
... but the echo command removes line breaks... still looking.
edit2
From https://unix.stackexchange.com/questions/11067/is-there-a-way-to-modify-a-file-in-place , it looks like writing to the very same file at once it not easy at all. However, I could do what I wanted with sed -i for linux only:
sed -i 's/pattern/"new pattern"/g' input.txt
From sed -i + what the same option in SOLARIS , it looks like there's no alternative, and you must use a temporary file:
sed 's/pattern/"new pattern"/g' input.txt > input.tmp && mv input.tmp input.txt

Linux command line: split a string

I have long file with the following list:
/drivers/isdn/hardware/eicon/message.c//add_b1()
/drivers/media/video/saa7134/saa7134-dvb.c//dvb_init()
/sound/pci/ac97/ac97_codec.c//snd_ac97_mixer_build()
/drivers/s390/char/tape_34xx.c//tape_34xx_unit_check()
(PROBLEM)/drivers/video/sis/init301.c//SiS_GetCRT2Data301()
/drivers/scsi/sg.c//sg_ioctl()
/fs/ntfs/file.c//ntfs_prepare_pages_for_non_resident_write()
/drivers/net/tg3.c//tg3_reset_hw()
/arch/cris/arch-v32/drivers/cryptocop.c//cryptocop_setup_dma_list()
/drivers/media/video/pvrusb2/pvrusb2-v4l2.c//pvr2_v4l2_do_ioctl()
/drivers/video/aty/atyfb_base.c//aty_init()
/block/compat_ioctl.c//compat_blkdev_driver_ioctl()
....
It contains all the functions in the kernel code. The notation is file//function.
I want to copy some 100 files from the kernel directory to another directory, so I want to strip every line from the function name, leaving just the filename.
It's super-easy in python, any idea how to write a 1-liner in the bash prompt that does the trick?
Thanks,
Udi
cat "func_list" | sed "s#//.*##" > "file_list"
Didn't run it :)
You can use pure Bash:
while read -r line; do echo "${line%//*}"; done < funclist.txt
Edit:
The syntax of the echo command is doing the same thing as the sed command in Eugene's answer: deleting the "//" and everything that comes after.
Broken down:
"echo ${line}" is the same as "echo $line"
the "%" deletes the pattern that follows it if it matches the trailing portion of the parameter
"%" makes the shortest possible match, "%%" makes the longest possible
"//*" is the pattern to match, "*" is similar to sed's ".*"
See the Parameter Expansion section of the Bash man page for more information, including:
using ${parameter#word} for matching the beginning of a parameter
${parameter/pattern/string} to do sed-style replacements
${parameter:offset:length} to retrieve substrings
etc.
here's a one liner in (g)awk
awk -F"//" '{print $1}' file
Here's one using cut and rev
cat file | rev | cut -d'/' -f2-| rev

Resources