Context;
After running the following command on my server:
zgrep "ResCode-5005" /loggers1/PCRF*/_01_03_2022 > analisis.txt
I get a text file with thousands of lines like this example:
loggers1/PCRF1_17868/PCRF12_01_03_2022_00_15_39.log:[C]|01-03-2022:00:18:20:183401|140404464875264|TRACKING: CCR processing Compleated for SubId-5281181XXXXX, REQNO-1, REQTYPE-3,
SId-mscp01.herpgwXX.epc.mncXXX.mccXXX.XXXXX.org;25b8510c;621dbaab;3341100102036XX-27cf0XXX,
RATTYPE-1004, ResCode-5005 |processCCR|ProcessingUnit.cpp|423
(X represents incrementing numbers)
Problem:
The output is filled with unnecessary data. The only string portions I need are the MSISDN,IMSI comma separated for each line, like this:
5281181XXXXX,3341100102036XX
Steps I tried
zgrep "ResCode-5005" /loggers1/PCRF*/_01_03_2022| grep -o -P
'(?<=SubId-).*?(?=, REQ)' > analisis1.txt
This gave me the first part of the solution
5281181XXXXX
However, when I tried to get the second string located between '334110' and "-"
zgrep "ResCode-5005" /loggers1/PCRF*/_01_03_2022| grep -o -P
'(?<=SubId-).?(?=, REQ)' | grep -o -P '(?<=334110).?(?=-)' >
analisis1.txt
it doesn't work.
Any input will be appreciated.
To get 5281181XXXXX or the second string located between '334110' and "-" you can use a pattern like:
\b(?:SubId-|334110)\K[^,\s-]+
The pattern matches:
\b A word boundary to prevent a partial word match
(?: Non capture group to match as a whole
SubId- Match literally
| Or
334110 Match literally
) Close the non capture group
\K Forget what is matched so far
[^,\s-]+ Match 1+ occurrences of any char except a whitespace char , or -
See the matches in this regex demo.
That will match:
5281181XXXXX
0102036XX
The command could look like
zgrep "ResCode-5005" /loggers1/PCRF*/_01_03_2022 | grep -oP '\b(?:SubId-|334110)\K[^,\s-]+' > analisis1.txt
Related
I have text that looks like this:
Name (OneData) [113C188D-5F70-44FE-A709-A07A5289B75D] (MoreData)
I want to use grep or some other way to get the ID inside [].
How to do it?
You can do something like this via bash (GNU grep required):
t="Name (OneData) [113C188D-5F70-44FE-A709-A07A5289B75D] (MoreData)"
echo "$t" | grep -Po "(?<=\[).*(?=\])"
The pattern will give you everything between the brackets, and uses a zero-width look-behind assertion (?<= ...) to eliminate the opening bracket and uses a zero-width look-ahead assertion (?= ...) to eliminate the closing bracket.
The -P flag activates perl-style regexes which can be useful not having too much to escape, then. The -o flag will give you only the wanted result (not the "non-capturing groups").
If you don't have GNU grep available, you can solve the problem in two steps (there are probably also other solutions):
Get the ID with the brackets (\[.*\])
Remove the brackets (] and [, here via sed, for example)
echo "$t" | grep -o "\[.*\]" | sed 's/[][]//g'
As Cyrus commented, you can also use the pattern grep -oE '[0-9A-F-]{36}' if you can ensure not having strings of length 36 or larger containing only the characters 0-9, A-F and - and if all the IDs have the length of 36 characters, of course. Then you can simply ignore the brackets.
I would like to grep for process path which has a variable. Example -
This is one of the proceses running.
/var/www/vhosts/rcsdfg/psd_folr/rcerr-m-deve-udf-172/bin/magt queue:consumers:start customer.import_proditns --single-thread --max-messages=1000
I would like to grep for "psd_folr/rcerr-m-deve-udf-172/bin/magt queue" from the running processes.
The catch is that the number 172 keeps changing, but it will be a 3 digit number only. Please suggest, I tried below but it is not returning any output.
sudo ps axu | grep "psd_folr/rcerr-m-deve-udf-'^[0-9]$'/bin/magt queue"
The most relevant section of your regular expression is -'^[0-9]$'/ which has following problems:
the apostrophes have no syntactical meaning to grep other than read an apostrophe
the caret ^ matches the beginning of a line, but there is no beginning of a line in ps's output at this place
the dollar $ matches the end of a line, but there is no end of a line in ps's output at this place
you want to read 3 digits but [0-9] will only match a single one
Thus, the part of your expression should be modified like this -[0-9]+/ to match any number of digits (+ matches the preceding character any number of times but at least once) or like this -[0-9]{3}/ to match exactly three times ({n} matches the preceding character exactly n times).
If you alter your command, give grep the -E flag so it uses extended regular expressions, otherwise you need to escape the plus or the braces:
sudo ps axu | grep -E "psd_folr/rcerr-m-deve-udf-[0-9]+/bin/magt queue"
I have two lists, one of which contains wildcards (in this case represented by *). I would like to compare the two lists and create an output of those that match, with each wildcard * representing a single character.
For example:
File 1
123456|Jane|Johnson|Pharmacist|janejohnson#gmail.com
09876579|Frank|Roberts|Butcher|frankie1#hotmail.com
092362936|Joe|Jordan|Joiner|joe#joesjoinery.com
928|Bob|Horton|Farmer|bhorton#farmernews.co.uk
File 2
1***6|Jane|Johnson|Pharmacist|janejohnson#gmail.com
09876579|Frank|Roberts|Butcher|f**1#hotmail.com
092362936|Joe|Jordan|J*****|joe#joesjoinery.com
928|Bob|Horton|Farmer|b*****n#f*********.co.uk
Output
092362936|Joe|Jordan|Joiner|joe#joesjoinery.com
928|Bob|Horton|Farmer|bhorton#farmernews.co.uk
Explanation
The first two lines are not considered matches because the number of *s is not equal to the number of characters shown in the first file. The latter two are, so they are added to output.
I have tried to reason out ways to do this in AWK and using Join, but I don't know any way to even start trying to achieve this. Any help would be greatly appreciated.
$ cat tst.awk
NR==FNR {
file1[$0]
next
}
{
# Make every non-* char literal (see https://stackoverflow.com/a/29613573/1745001):
gsub(/[^^*]/,"[&]") # Convert every char X to [X] except ^ and *
gsub(/\^/,"\\^") # Convert every ^ to \^
# Convert every * to .:
gsub(/\*/,".")
# Add line start/end anchors
$0 = "^" $0 "$"
# See if the current file2 line matches any line from file1
# and if so print that line from file1:
for ( line in file1 ) {
if ( line ~ $0 ) {
print line
}
}
}
$ awk -f tst.awk file1 file2
092362936|Joe|Jordan|Joiner|joe#joesjoinery.com
928|Bob|Horton|Farmer|bhorton#farmernews.co.uk
sed 's/\./\\./g; s/\*/./g' file2 | xargs -I{} grep {} file1
Explanation:
I'd take advantage of regular expression matching. To do that, we need to turn every asterisk * into a dot ., which represents any character in regular expressions. As a side effect of enabling regular expressions, we need to escape all special characters, particularly the ., in order for them to be taken literally. In a regular expression, we need to use \. to represent a dot (as opposed to any character).
The first step is perform these substitutions with sed, the second is passing every resulting line as a search pattern to grep, and search file1 for that pattern. The glue that allows to do this is xargs, where a {} is a placeholder representing a single line from the results of the sed command.
Note:
This is not a general, safe solution you can simply copy and paste: you should watch out for any characters, in your file containing the asterisks, that are considered special in grep regular expressions.
Update:
jhnc extends the escaping to any of the following characters: .\^$[], thus accounting for almost all sorts of email addresses. He/she then avoids the use of xargs by employing -f - to pass the results of sed as search expressions to grep:
sed 's/[.\\^$[]/\\&/g; s/[*]/./g' file2 | grep -f - file1
This solution is both more general and more efficient, see comment below.
I'm trying to do a grep command that finds all lines in a file whos first word begins "as" and whos first word also ends with "ng"
How would I go about doing this using grep?
This should just about do it:
$ grep '^as\w*ng\b' file
Regexplanation:
^ # Matches start of the line
as # Matches literal string as
\w # Matches characters in word class
* # Quantifies \w to match either zero or more
ng # Matches literal string ng
\b # Matches word boundary
May have missed the odd corner case.
If you only want to print the words that match and not the whole lines then use the -o option:
$ grep -o '^as\w*ng\b' file
Read man grep for all information on the available options.
I am pretty sure this should work:
grep "^as[a-zA-Z]*ng\b" <filename>
hard to say without seeing samples from the actual input file.
sudo has already covered it well, but I wanted to throw out one more simple one:
grep -i '^as[^ ]*ng\b' <file>
-i to make grep case-insensitive
[^ ]* matches zero or more of any character, except a space
^ finds the 'first character in a line', so you can search for that with:
grep '^as' [file]
\w matches a word character, so \w* would match any number of word characters:
grep '^as\w*' [file]
\b means 'a boundary between a word and whitespace' which you can use to ensure that you're matching the 'ng' letters at the end of the word, instead of just somewhere in the middle:
grep '^as\w*ng\b' [file]
If you choose to omit the [file], simply pipe your files into it:
cat [file] | grep '^as\w*ng\b'
or
echo [some text here] | grep '^as\w*ng\b'
Is that what you're looking for?
grep (GNU grep) 2.14
Hello,
I have a log file that I want to filter on a selected word. However, it tends to filter on many for example.
tail -f gateway-* | grep "P_SIP:N_iptB1T1"
This will also find words like this:
"P_SIP:N_iptB1T10"
"P_SIP:N_iptB1T11"
"P_SIP:N_iptB1T12"
etc
However, I don't want to display anything after the 1. grep is picking up 11, 12, 13, etc.
Many thanks for any suggestions,
You can restrict the word to end at 1:
tail -f gateway-* | grep "P_SIP:N_iptB1T1\>"
This will work assuming that you have a matching case which is only "P_SIP:N_iptB1T1".
But if you want to extract from P_SIP:N_iptB1T1x, and display only once, then you need to restrict to show only first match.
grep -o "P_SIP:N_iptB1T1"
-o, --only-matching show only the part of a line matching PATTERN
More info
At least two approaches can be tried:
grep -w pattern matches for full words. Seems to work for this case too, even though the pattern has punctuation.
grep pattern -m 1 to restrict the output to first match. (Also doable with grep xxx | head -1)
If the lines contains the quotes as in your example, just use the -E option in grep and match the closing quote with \". For example:
grep -E "P_SIP:N_iptB1T1\"" file
If these quotes aren't in the text file, and there's blank spaces or endlines after the word, you can match these too:
# The word is followed by one or more blanks
grep -E "P_SIP:N_iptB1T1\s+" file
# Match lines ending with the interesting word
grep -E "P_SIP:N_iptB1T1$" file