Another Solaris question.
This is my file.
/abc/123/gfh/hello/what/is/up <THIS WOULD BE WHERE A NEW LINE STARTS>
bhn/fda/fds/hello/the/sky/is/blue <THIS WOULD BE WHERE A NEW LINE STARTS>
...etc
I need to delete everything before "hello" include the forward slash "/" infront of it for everyline in the file...
I'm stuck -> I had used a sed -E command but Solaris doesn't recognize the "-E". sigh
I think you can grep this:
grep -o hello.*
This should delete everything up to the slash before "hello":
sed -e 's|^.*hello/|hello|' <inputfile >outputfile
That should do it:
sed -e 's/.hello(.)/hello\1/'
#user4815162342: No need for "^" in the sed solution, ".*" would suffice.
Since there was an awk tag too, here's the equivalent awk solution:
awk '{sub(/.*hello\//,"hello")}1'
Related
27211;18:05:03479;20161025;0;0;0;0;10991;0;10991;000;0;0;000;1000000;0;0;000;0;0;0;82
Second string after ; is time. gg:mm:sssss:. I just want to be gg:mm:ss:
Like so:
27211;18:05:03;20161025;0;0;0;0;10991;0;10991;000;0;0;000;1000000;0;0;000;0;0;0;82
I tried with cut but it deletes everything after n'th occurance of character, and for now I am stuck, please help.
give this one liner a try:
awk -F';' -v OFS=";" 'sub(/...$/,"",$2)+1' file
It removes the last 3 chars from column 2.
update with sed one liner
If you are a fan of sed:
sed -r 's/(;[^;]*)...;/\1;/' file
With sed:
sed -r 's/^([^;]+;[^;]+)...;/\1;/' file
(Or)
sed -r 's/^([^;]+;[0-9]{2}:[0-9]{2}:[0-9]{2})...;/\1;/' file
It also can be something like sed 's/(.*)([0-9]{2}\:){2}([0-9]{3})[0-9]*\;(.*)/\1\2\3\4/g'
It is not very clean, but at least is more clear for me.
Regards
I'd use perl for this:
perl -pe 's/(?<=:\d\d)\d+(?=;)//' file
That removes any digits between "colon-digit-digit" and the semicolon (first match only, not globally in the line).
If you want to edit the file in-place: perl -i -pe ...
With sed:
sed -E 's/(:[0-9]{2})[0-9]{3}/\1/' file
or perl:
perl -pe's/:\d\d\B\K...//' file
I have the followiing input file and I need to remove all the characters from the strings that appear after the last '/'. I'll also show my expected output below.
input:
/start/one/two/stopone.js
/start/one/two/three/stoptwo.js
/start/one/stopxyz.js
expected output:
/start/one/two/
/start/one/two/three/
/start/one/
I have tried to use sed but with no luck so far.
You could simply use good old grep:
grep -o '.*/' file.txt
This simple expression takes advantage of the fact that grep is matching greedy. Meaning it will consume as much characters as possible, including /, until the last / in path.
Original Answer:
You can use dirname:
while read line ; do
echo dirname "$line"
done < file.txt
or sed:
sed 's~\(.*/\).*~\1~' file.txt
perl -lne 'print $1 if(/(.*)\//)' your_file
Try this GNU sed command,
$ sed -r 's~^(.*\/).*$~\1~g' file
/start/one/two/
/start/one/two/three/
/start/one/
Through awk,
awk -F/ '{sub(/.*/,"",$NF); print}' OFS="/" file
I have a file data.txt with the following strings:
text-common-1.1.1-SNAPSHOT.jar
text-special-common-2.1.2-SNAPSHOT.jar
some-text-variant-1.1.1-SNAPSHOT.jar
text-another-variant-text-3.3.3-SNAPSHOT.jar
I want to change all of the text-something-digits-something.jar to text-something-5.0.jar.
Here is my script with sed (GNU sed version 4.2.1
), but it doesn't work, I don't know why:
#!/bin/bash
for t in ./data.txt
do
sed -i "s/\(text-[a-z]*-(\d|\.)*\).*\(.jar\)/\15.0\2/" ${t}
done
What is wrong with my sed usage?
How about this awk
awk '/^text/ {sub(/[0-9].*\./,"5.0.")}1'
text-common-5.0.jar
text-special-common-5.0.jar
some-text-variant-1.1.1-SNAPSHOT.jar
text-another-variant-text-5.0.jar
text-something-digits-something.jar to text-something-5.0.jar
equal change digits-someting to 5.0
It also takes care of changing line only starting with text
I think a simpler approach might be enough: sed -r -e 's/(text-(.*-)?common-)([0-9\.]+)(-.*\.jar)/\15.0\4/' < your_data.
Another way of saying the same thing with perl: perl -pe 's/(text-(?:(.*-))*common-)([\d\.]+)(-.*\.jar)/${1}1.5${4}/' < your_data.
#!/bin/bash
for t in ./data.txt
do
sed -i '/^text-/ s/[.0-9]\{1,\}-something\(\.jar\)$/5.0\2/' ${t}
# for "any" something
#sed -i '/^text-/ s/[.0-9]\{1,\}-[^?]\{1,\}\(\.jar\)$/5.0\2/' ${t}
done
select string starting with text and change digit value is present
Using sed:
sed '/^text-/ s/-[0-9.]*-/-5.0-/' file
I have a large file which is slightly corrupted. The new lines have disappeared. There should have been a new line at every 250th character. How can I fix that?
Thanks in advance.
How about
sed 's/.\{250\}/&\n/g'
The .\{250\} captures 250 of any type of character. The characters are replaced by themselves, plus a newline.
try this:
sed -r 's/.{250}/&\n/g'
gawk:
awk -v FPAT='.{1,25}' -v OFS='\n' '$1=$1'
There is a command in coreutils that can wrap lines, it is called fold:
fold -w 250
sed 's/^.\{250\}/&\
/;P;D' YourFile
Could be faster on huge file
An awk version
awk '{L=250;for (i=1;i<=length($0);i+=L) print substr($0,i,L)}'
I have this line inside a file:
ULNET-PA,client_sgcib,broker_keplersecurities
,KEPLER
I try to get rid of that ^M (carriage return) character so I used:
sed 's/^M//g'
However this does remove everything after ^M:
[root#localhost tmp]# vi test
ULNET-PA,client_sgcib,broker_keplersecurities^M,KEPLER
[root#localhost tmp]# sed 's/^M//g' test
ULNET-PA,client_sgcib,broker_keplersecurities
What I want to obtain is:
[root#localhost tmp]# vi test
ULNET-PA,client_sgcib,broker_keplersecurities,KEPLER
Use tr:
tr -d '^M' < inputfile
(Note that the ^M character can be input using Ctrl+VCtrl+M)
EDIT: As suggested by Glenn Jackman, if you're using bash, you could also say:
tr -d $'\r' < inputfile
still the same line:
sed -i 's/^M//g' file
when you type the command, for ^M you type Ctrl+VCtrl+M
actually if you have already opened the file in vim, you can just in vim do:
:%s/^M//g
same, ^M you type Ctrl-V Ctrl-M
You can simply use dos2unix which is available in most Unix/Linux systems. However I found the following sed command to be better as it removed ^M where dos2unix couldn't:
sed 's/\r//g' < input.txt > output.txt
Hope that helps.
Note: ^M is actually carriage return character which is represented in code as \r
What dos2unix does is most likely equivalent to:
sed 's/\r\n/\n/g' < input.txt > output.txt
It doesn't remove \r when it is not immediately followed by \n and replaces both with just \n. This fails with certain types of files like one I just tested with.
alias dos2unix="sed -i -e 's/'\"\$(printf '\015')\"'//g' "
Usage:
dos2unix file
If Perl is an option:
perl -i -pe 's/\r\n$/\n/g' file
-i makes a .bak version of the input file
\r = carriage return
\n = linefeed
$ = end of line
s/foo/bar/g = globally substitute "foo" with "bar"
In awk:
sub(/\r/,"")
If it is in the end of record, sub(/\r/,"",$NF) should suffice. No need to scan the whole record.
This is the better way to achieve
tr -d '\015' < inputfile_name > outputfile_name
Later rename the file to original file name.
I agree with #twalberg (see accepted answer comments, above), dos2unix on Mac OSX covers this, quoting man dos2unix:
To run in Mac mode use the command-line option "-c mac" or use the
commands "mac2unix" or "unix2mac"
I settled on 'mac2unix', which got rid of my less-cmd-visible '^M' entries, introduced by an Apple 'Messages' transfer of a bash script between 2 Yosemite (OSX 10.10) Macs!
I installed 'dos2unix', trivially, on Mac OSX using the popular Homebrew package installer, I highly recommend it and it's companion command, Cask.
This is clean and simple and it works:
sed -i 's/\r//g' file
where \r of course is the equivalent for ^M.
Simply run the following command:
sed -i -e 's/\r$//' input.file
I verified this as valid in Mac OSX Monterey.
remove any \r :
nawk 'NF+=OFS=_' FS='\r'
gawk 3 ORS= RS='\r'
remove end of line \r :
mawk2 8 RS='\r?\n'
mawk -F'\r$' NF=1