I have a log and it has data like :
ProcessID='3940', Key='1', Number='5547', TotalNumberOfInputMessages='1', TotalElapsedTime='1332',
there are many other such info in log, but I am interested in particularly printing only TotalNumberOfInputMessages='1' occurrences ... There are many such occurrences in the log file with this value changing for TotalNumberOfInputMessages.
I want output like :
TotalNumberOfInputMessages='1'
TotalNumberOfInputMessages='diff value'
TotalNumberOfInputMessages='diff value'
TotalNumberOfInputMessages='diff value'
How can i achieve it by cut, sed or grep ?
You can use grep -Eo:
grep -Eo "\bTotalNumberOfInputMessages='[^']*'" file
TotalNumberOfInputMessages='1'
Related
I want to find the number of 8 letter words that do not contain the letter "e" in a number of text files (*.txt). In the process I ran into two issues: my lack of understanding in quantifiers and how to exclude characters.
I'm quite new to the Unix terminal, but this is what I have tried:
cat *.txt | grep -Eo "\w+" | grep -i ".*[^e].*"
I need to include the cat command because it otherwise includes the names of the text files in the pipe. The second pipe is to have all the words in a list, and it works, but the last pipe was meant to find all the words that do not have the letter "e" in them, but doesn't seem to work. (I thought "." for no or any number of any character, followed by a character that is not an "e", and followed by another "." for no or any number of any character.)
cat *.txt | grep -Eo "\w+" | grep -wi "[a-z][a-z][a-z][a-z][a-z][a-z][a-z][a-z]"
This command works to find the words that contain 8 characters, but it is quite ineffective, because I have to repeat "[a-z]" 8 times. I thought it could also be "[a-z]{8}", but that doesn't seem to work.
cat *.txt | grep -Eo "\w+" | grep -wi "[a-z][a-z][a-z][a-z][a-z][a-z][a-z][a-z]" | grep -i ".*[^e].*"
So finally, this would be my best guess, however, the third pipe is ineffective and the last pipe doesn't work.
You may use this grep:
grep -hEiwo '[a-df-z]{8}' *.txt
Here:
[a-df-z]{8}: Matches all letters except e
-h: Don't print filename in output
-i: Ignore case search
-o: Print matches only
-w: Match complete words
In case you are ok with GNU awk and assuming that you want to print only the exact words and could be multiple matches in a line if this is the case one could try following.
awk -v IGNORECASE="1" '{for(i=1;i<=NF;i++){if($i~/^[a-df-z]{8}$/){print $i}}}' *.txt
OR without the use of IGNORCASE one could try:
awk '{for(i=1;i<=NF;i++){if(tolower($i)~/^[a-df-z]{8}$/){print $i}}}' *.txt
NOTE: Considering that you want exact matches of 8 letters only in lines. 8 letter words followed by a punctuation mark will be excluded.
Here is a crazy thought with GNU awk:
awk 'BEGIN{FPAT="\\<\\w{8}\\>"}{c+=NF}END{print c}' file
Or if you want to make it work only on a select set of characters:
awk 'BEGIN{FPAT="\\<[a-df-z]{8}\\>"}{c+=NF}END{print c}' file
What this does is, it defines the fields, to be a set of 8 characters (\w as a word-constituent or [a-df-z] as a selected set) which is enclosed by word-boundaries (\< and \>). This is done with FPAT (note the Gory details about escaping).
Sometimes you might also have words which contain diatrics, so you have to expand. Then this might be the best solution:
awk 'BEGIN{FPAT="\\<\\w{8}\\>"}{for(i=1;i<=NF;++i) if($i !~ /e/) c++}END{print c}' file
I have a csv file with a list of workers and I wanna make an script for modify their work group given their ID's. Lines in CSV files are like this:
Before:
ID TAG GROUP
niub16677500;B00;AB0
After:
ID TAG GROUP
niub16677500;B00;BC0
How I can make this?
I'm working with awk and sed commands but I couldn't get anything at the moment.
With awk:
awk -F';' -v OFS=';' -v id="niub16677500" -v new_group="BC0" '{if($1==id)$3=new_group}1' input.csv
ID;TAG;GROUP
niub16677500;B00;BC0
Redirect the output to a file and note that the csv header should use the same field separator as the body.
Explanations:
-F';' to have input field separator as ;
-v OFS=';' same for the output FS
-v id="niub16677500" -v new_group="BC0" define the variables that you are going to use in the awk commands
'{if($1==id)$3=new_group}1' when the first column is equal to the value contained in variable id the overwrite the 3rd field and print the line
With sed:
id="niub16677500"; new_group="BC0"; sed "/^$id/s/;[^;]*$/;$new_group/" input.csv
ID;TAG;GROUP
niub16677500;B00;BC0
You can either do an inline change using -i.bak option, or redirect the output to a file.
Explanations:
Store the values in 2 variables
/^$id/ when you reach a line that starts with the ID store in the variable id, run sed search and replace
s/;[^;]*$/;$new_group/ search and replace command that will replace the last field by the new value
Sed can do it,
echo 'niub16677500;B00;AB0' | sed 's/\(^niub16677500;...;\)\(...\)$/\1BC0/'
will replace the AB0 group in your example with BC0, by matching the user name, semicolon, whatever 3 characters and another semicolon, and then matching the remaining 3 characters. Then as an output it repeats the first match with \1 and adds BC0.
You can use :
sed 's/\(^niub16677500;...;\)\(...\)$/\1BC0/' <old_file >new_file
to make a new_file with this change.
https://www.grymoire.com/Unix/Sed.html is a great resource, you should take a look at it.
I have a log file and I am trying to extract data between 2 words of that log file.
username=#$^#$^&###%^&==&employeeid
There is data before and after these words but I am only interested in the data between them. Thus the expected output is (just the value between the username= and &employeeid
#$^#$^&###%^&==
I want to grep the file first and then search using sed in that file. Something like below. This is not working for me exactly..
grep "e553bb57-b94b-cb0f-f4ba-eb9a02ab0050" /path/abc/logfile.txt | sed -n '/username=/{s/.*username=//;s/\S*=.*//;p}'
How about
echo 'username=#$^#$^&###%^&==&employeeid' | sed 's/username=\(.*\)==&employeeid/\1/'
The output is
#$^#$^&###%^&
The matched part would be in \1.
I have a log which got like this :
.....client connection.....remote=/xxx.xxx.xxx.xxx]].......
I need to extract all lines in the log which contain the above,and print just the ip after remote=.. This would be something in the pattern :
grep "client connection" xxx.log | sed -e ....
Using grep:
grep -oP '(?<=remote=/)[^\]]+' file
o is to extract only the pattern, instead of entire line.
P is to match perl like regex. In this case, we are using "negative look behind". It will try to match set of characters which is not "]" which is preceeded by remote=/
grep -oP 'client connection.*remote=/\K.*?(?=])' input
Prints anything between remote=/ and closest ] on the lines which contain client connection.
Or by using sed back referencing: Here the line is divided into three parts/groups which are later referred by \1 \2 or \3. Each group is enclosed by ( and ). Here IP address belongs to 2nd group, so whole line is replaced by 2nd group which is IP address.
sed -r '/client connection/ s_(^.*remote=/)(.*?)]](.*)_\2_g' input
Or using awk :
awk -F'/|]]' '/client connection/{print $2}' input
Try this:
grep 'client connection' test.txt | awk -F'[/\\]]' '{print $2}'
Test case
test.txt
---------
abcd
.....client connection.....remote=/10.20.30.40]].......
abcs
.....client connection.....remote=/11.20.30.40]].......
.....client connection.....remote=/12.20.30.40]].......
Result
10.20.30.40
11.20.30.40
12.20.30.40
Explanation
grep will shortlist the results to only lines matching client connection. awk uses -F flag for delimiter to split text. We ask awk to use / and ] delimiters to split text. In order to use more than one delimiter, we place the delimiters in [ and ]. For example, to split text by = and :, we'd do [=:].
However, in our case, one of the delimiters is ] since my intent is to extract IP specifically from /x.x.x.x] by spitting the text with / and ]. So we escape it ]. The IP is the 2nd item from the splitting.
A more robust way, improved over this answer would be to also use GNU grep in PCRE mode with -P for perl style regEx match, but matching both the patterns as suggested in the question.
grep -oP "client connection.*remote=/\K(\d{1,3}\.){3}\d{1,3}" file
10.20.30.40
11.20.30.40
12.20.30.40
Here, client connection.*remote matches both the patterns in the lines and extracts IP from the file. The \K is a PCRE syntax to ignore strings up to that point and print only the capture group following it.
(\d{1,3}\.){3}\d{1,3}
To match the IP i.e. 3 groups of digits separated by dots of length from 1 to 3 followed by 4th octet.
For example suppose I have the following piece of data
ABC,3,4
,,ExtraInfo
,,MoreInfo
XYZ,6,7
,,XyzInfo
,,MoreXyz
ABC,1,2
,,ABCInfo
,,MoreABC
It's trivial to get grep to extract the ABC lines. However if I want to also grab the following lines to produce this output
ABC,3,4
,,ExtraInfo
,,MoreInfo
ABC,1,2
,,ABCInfo
,,MoreABC
Can this be done using grep and standard shell scripting?
Edit: Just to clarify there could be a variable number of lines in between. The logic would be to keep printing while the first column of the CSV is empty.
grep -A 2 {Your regex} will output the two lines following the found strings.
Update:
Since you specified that it could be any number of lines, this would not be possible as grep focuses on matching on a single line see the following questions:
How can I search for a multiline pattern in a file?
Regex (grep) for multi-line search needed
Why can't i match the pattern in this case?
Selecting text spanning multiple lines using grep and regular expressions
You can use this, although a bit hackity due to the grep at the end of the pipeline to mute out anything that does not start with 'A' or ',':
$ sed -n '/^ABC/,/^[^,]/p' yourfile.txt| grep -v '^[^A,]'
Edit: A less hackity way is to use awk:
$ awk '/^ABC/ { want = 1 } !/^ABC/ && !/^,/ { want = 0 } { if (want) print }' f.txt
You can understand what it does if you read out loud the pattern and the thing in the braces.
The manpage has explanations for the options, of which you want to look at -A under Context Line Control.