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(/...$/,"",$2)+1' file
It removes the last 3 chars from column 2.
update with sed one liner
If you are a fan of sed:
sed -r 's/(;[^;]*)...;/\1;/' file
With sed:
sed -r 's/^([^;]+;[^;]+)...;/\1;/' file
(Or)
sed -r 's/^([^;]+;[0-9]{2}:[0-9]{2}:[0-9]{2})...;/\1;/' file
It also can be something like sed 's/(.*)([0-9]{2}\:){2}([0-9]{3})[0-9]*\;(.*)/\1\2\3\4/g'
It is not very clean, but at least is more clear for me.
Regards
I'd use perl for this:
perl -pe 's/(?<=:\d\d)\d+(?=;)//' file
That removes any digits between "colon-digit-digit" and the semicolon (first match only, not globally in the line).
If you want to edit the file in-place: perl -i -pe ...
With sed:
sed -E 's/(:[0-9]{2})[0-9]{3}/\1/' file
or perl:
perl -pe's/:\d\d\B\K...//' file
Related
I have the followiing input file and I need to remove all the characters from the strings that appear after the last '/'. I'll also show my expected output below.
input:
/start/one/two/stopone.js
/start/one/two/three/stoptwo.js
/start/one/stopxyz.js
expected output:
/start/one/two/
/start/one/two/three/
/start/one/
I have tried to use sed but with no luck so far.
You could simply use good old grep:
grep -o '.*/' file.txt
This simple expression takes advantage of the fact that grep is matching greedy. Meaning it will consume as much characters as possible, including /, until the last / in path.
Original Answer:
You can use dirname:
while read line ; do
echo dirname "$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/' ${t}
done
select string starting with text and change digit value is present
Using sed:
sed '/^text-/ s/-[0-9.]*-/-5.0-/' file
I want to replace the
/fdasatavol/ankit
to
/fdasatavol_sata/ankit
Can anyone help me out in this?
to write to a new file (without modifying file1):
sed 's/fdasatavol/fdasatavol_sata/g' file1 > file2
or to replace in the original file:
sed -i 's/fdasatavol/fdasatavol_sata/g' file1
This will replace each occurrence of fdasatavol with fdasatavol_sata:
sed 's/fdasatavol/&_sata/g'
If your input has occurrence of fdasatavol that are not in /fdasatavol/ankit and you don't want to substitute these then use:
sed 's#/fdasatavol/ankit#/fdastatavol_sata/ankit#g'
Note: you can use any character as sed's delimilter to aviod the confusion with the parrtern contiaing /. sed prints to stdout by default, if you are happy with the changes produced by sed you can use the -i option to store back to the file.
sed -i 's/fdasatavol/&_stat/g' file
I want to delete first and last line from the file
file1 code :
H|ACCT|XEC|1|TEMP|20130215035845|
849002|48|1208004|1
849007|28|1208004|1
T|2
After delete the output should be
849002|48|1208004|1
849007|28|1208004|1
I have tried below method but has to run it 2 times, I want one liner solution to remove both in one go!
sed '1,1d' file1.txt >> file1.out
sed '$d' file1.out >> file2
Please suggest one liner code....
You could use ;
sed '1d; $d' file
Use Command Separator
In sed, you can separate commands using a semicolon. For example:
sed '1d; $d' /path/to/file
How about:
sed '$d' < file1.txt | sed "1d"
Try sed -i '1d;$d' /path/to/file
awk 'NR>2{print v}{v=$0}'
Starting with line 3, print the previous line each time. This means the first and last lines will not be printed.
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