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($0,i,L)}'
Related
27211;18:05:03479;20161025;0;0;0;0;10991;0;10991;000;0;0;000;1000000;0;0;000;0;0;0;82
Second string after ; is time. gg:mm:sssss:. I just want to be gg:mm:ss:
Like so:
27211;18:05:03;20161025;0;0;0;0;10991;0;10991;000;0;0;000;1000000;0;0;000;0;0;0;82
I tried with cut but it deletes everything after n'th occurance of character, and for now I am stuck, please help.
give this one liner a try:
awk -F';' -v OFS=";" 'sub(/...$/,"",$2)+1' file
It removes the last 3 chars from column 2.
update with sed one liner
If you are a fan of sed:
sed -r 's/(;[^;]*)...;/\1;/' file
With sed:
sed -r 's/^([^;]+;[^;]+)...;/\1;/' file
(Or)
sed -r 's/^([^;]+;[0-9]{2}:[0-9]{2}:[0-9]{2})...;/\1;/' file
It also can be something like sed 's/(.*)([0-9]{2}\:){2}([0-9]{3})[0-9]*\;(.*)/\1\2\3\4/g'
It is not very clean, but at least is more clear for me.
Regards
I'd use perl for this:
perl -pe 's/(?<=:\d\d)\d+(?=;)//' file
That removes any digits between "colon-digit-digit" and the semicolon (first match only, not globally in the line).
If you want to edit the file in-place: perl -i -pe ...
With sed:
sed -E 's/(:[0-9]{2})[0-9]{3}/\1/' file
or perl:
perl -pe's/:\d\d\B\K...//' file
Say I have a script where I want to change several lines for a single line.
For example, I got a new function that can summarize several commands, so that I can replace in my script as follows:
Original
some_code
command1
command2
command3
some_more_code
Edited
some_code
foo()
some_more_code
How would you do that using sed?
sed '/some_code/,/command3/ !b
/some_code/ b
/command3/ a\
foo()
d' YourFile
be carrefull about meta character ( like &\\^$[]{}().) in any of the pattern (except your foo() line)
I am answering my own question here.
I couldn't figure out a way to do it in one go, so I split the problem into two parts.
Part 1: replace the first line
sed -e 's/command1/foo()/g' file1 > file2
Part 2: remove the rest of the lines
sed -e '/command2/,+1d/' file2 > file3
I'd prefer a more elegant way though, where I can be flexible in the number of lines that I am replacing, possibly matching the last command in the block. Any ideas?
Just use awk:
$ awk -v RS='^$' -v ORS= '{sub(/command1\ncommand2\ncommand3/,"foo()")}1' file
some_code
foo()
some_more_code
The above uses GNU awk for multi-char RS.
This might work for you (GNU sed):
sed '/command1/,/command3/c\foo()' 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
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'
I need a one liner using sed, awk or perl to remove blank lines from my data file. The data in my file looks like this -
Aamir
Ravi
Arun
Rampaul
Pankaj
Amit
Bianca
These blanks are at random and appear anywhere in my data file. Can someone suggest a one-liner to remove these blank lines from my dataset.
it can be done in many ways.
e.g with awk:
awk '$0' yourFile
or sed:
sed '/^$/d' yourFile
or grep:
grep -v '^$' yourFile
A Perl solution. From the command line.
$ perl -i.bak -n -e'print if /\S/' INPUT_FILE
Edits the file in-place and creates a backup of the original file.
AWK Solution:
Here we loop through the input file to check if they have any field set. NF is AWK's in-built variable that is set to th number of fields. If the line is empty then NF is not set. In this one liner we test if NF is true, i.e set to a value. If it is then we print the line, which is implicit in AWK when the pattern is true.
awk 'NF' INPUT_FILE
SED Solution:
This solution is similar to the ones mentioned as the answer. As the syntax show we are not printing any lines that are blank.
sed -n '/^$/!p' INPUT_FILE
You can do:
sed -i.bak '/^$/d' file
A Perl solution:
perl -ni.old -e 'print unless /^\s*$/' file
...which create as backup copy of the original file, suffixed with '.old'
for perl it is as easier as sed,awk, or grep.
$ cat tmp/tmpfile
Aamir
Ravi
Arun
Rampaul
Pankaj
Amit
Bianca
$ perl -i -pe 's{^\s*\n$}{}' tmp/tmpfile
$ cat tmp/tmpfile
Aamir
Ravi
Arun
Rampaul
Pankaj
Amit
Bianca