grep: Not a recognized flag: o












3














I am trying to extract the date and timestamps from my log file string ($Data in the below code) on AIX using a regex and loading it into a text file as below:



Data="Logs/2018-12-03/log.txt:3:2018-12-03 00:00:04,333 452621453 [blah blah blah"

echo "$data" | grep -o -n '[0-9]{4}-(0[1-9]|1[0-2])-(0[1-9]|[1-2][0-9]|3[0-1]) (2[0-3]|[01][0-9]):[0-5][0-9]' >  timestamps.txt


I get the below error most probably due to the unix version in my host being too old and it doesn't recognize the -o option. Is there any alternative method in which I can get the functionality of -o done?



grep: Not a recognized flag: o

Usage: grep [-r] [-R] [-H] [-L] [-E|-F] [-c|-l|-q] [-insvxbhwyu] [-p[parasep]] -e pattern_list...

    [-f pattern_file...] [file...]

Usage: grep [-r] [-R] [-H] [-L]  [-E|-F] [-c|-l|-q] [-insvxbhwyu] [-p[parasep]] [-e pattern_list...]

    -f pattern_file... [file...]

Usage: grep [-r] [-R] [-H] [-L] [-E|-F] [-c|-l|-q] [-insvxbhwyu] [-p[parasep]] pattern_list [file...]





grep variable aix







share|improve this question
























  • What operating system are you using?
    – terdon♦
    Dec 4 at 9:16










  • The command uname -s gives me 'AIX'
    – BlueNinja
    Dec 4 at 9:21






  • 1




    It's not that your Unix is "too old", it's that you are relying on non-standard functionality.
    – Kusalananda
    Dec 4 at 9:24










  • Is your Data variable a line from the output of grep? If it is, you can probably do whatever you are doing a fair bit more efficiently by running perl or awk over the original log files. Parsing the output of grep is not an ideal solution.
    – Kusalananda
    Dec 4 at 10:41


















3














I am trying to extract the date and timestamps from my log file string ($Data in the below code) on AIX using a regex and loading it into a text file as below:



Data="Logs/2018-12-03/log.txt:3:2018-12-03 00:00:04,333 452621453 [blah blah blah"

echo "$data" | grep -o -n '[0-9]{4}-(0[1-9]|1[0-2])-(0[1-9]|[1-2][0-9]|3[0-1]) (2[0-3]|[01][0-9]):[0-5][0-9]' >  timestamps.txt


I get the below error most probably due to the unix version in my host being too old and it doesn't recognize the -o option. Is there any alternative method in which I can get the functionality of -o done?



grep: Not a recognized flag: o

Usage: grep [-r] [-R] [-H] [-L] [-E|-F] [-c|-l|-q] [-insvxbhwyu] [-p[parasep]] -e pattern_list...

    [-f pattern_file...] [file...]

Usage: grep [-r] [-R] [-H] [-L]  [-E|-F] [-c|-l|-q] [-insvxbhwyu] [-p[parasep]] [-e pattern_list...]

    -f pattern_file... [file...]

Usage: grep [-r] [-R] [-H] [-L] [-E|-F] [-c|-l|-q] [-insvxbhwyu] [-p[parasep]] pattern_list [file...]





grep variable aix







share|improve this question
























  • What operating system are you using?
    – terdon♦
    Dec 4 at 9:16










  • The command uname -s gives me 'AIX'
    – BlueNinja
    Dec 4 at 9:21






  • 1




    It's not that your Unix is "too old", it's that you are relying on non-standard functionality.
    – Kusalananda
    Dec 4 at 9:24










  • Is your Data variable a line from the output of grep? If it is, you can probably do whatever you are doing a fair bit more efficiently by running perl or awk over the original log files. Parsing the output of grep is not an ideal solution.
    – Kusalananda
    Dec 4 at 10:41
















3












3








3







I am trying to extract the date and timestamps from my log file string ($Data in the below code) on AIX using a regex and loading it into a text file as below:



Data="Logs/2018-12-03/log.txt:3:2018-12-03 00:00:04,333 452621453 [blah blah blah"

echo "$data" | grep -o -n '[0-9]{4}-(0[1-9]|1[0-2])-(0[1-9]|[1-2][0-9]|3[0-1]) (2[0-3]|[01][0-9]):[0-5][0-9]' >  timestamps.txt


I get the below error most probably due to the unix version in my host being too old and it doesn't recognize the -o option. Is there any alternative method in which I can get the functionality of -o done?



grep: Not a recognized flag: o

Usage: grep [-r] [-R] [-H] [-L] [-E|-F] [-c|-l|-q] [-insvxbhwyu] [-p[parasep]] -e pattern_list...

    [-f pattern_file...] [file...]

Usage: grep [-r] [-R] [-H] [-L]  [-E|-F] [-c|-l|-q] [-insvxbhwyu] [-p[parasep]] [-e pattern_list...]

    -f pattern_file... [file...]

Usage: grep [-r] [-R] [-H] [-L] [-E|-F] [-c|-l|-q] [-insvxbhwyu] [-p[parasep]] pattern_list [file...]





grep variable aix







share|improve this question















I am trying to extract the date and timestamps from my log file string ($Data in the below code) on AIX using a regex and loading it into a text file as below:



Data="Logs/2018-12-03/log.txt:3:2018-12-03 00:00:04,333 452621453 [blah blah blah"

echo "$data" | grep -o -n '[0-9]{4}-(0[1-9]|1[0-2])-(0[1-9]|[1-2][0-9]|3[0-1]) (2[0-3]|[01][0-9]):[0-5][0-9]' >  timestamps.txt


I get the below error most probably due to the unix version in my host being too old and it doesn't recognize the -o option. Is there any alternative method in which I can get the functionality of -o done?



grep: Not a recognized flag: o

Usage: grep [-r] [-R] [-H] [-L] [-E|-F] [-c|-l|-q] [-insvxbhwyu] [-p[parasep]] -e pattern_list...

    [-f pattern_file...] [file...]

Usage: grep [-r] [-R] [-H] [-L]  [-E|-F] [-c|-l|-q] [-insvxbhwyu] [-p[parasep]] [-e pattern_list...]

    -f pattern_file... [file...]

Usage: grep [-r] [-R] [-H] [-L] [-E|-F] [-c|-l|-q] [-insvxbhwyu] [-p[parasep]] pattern_list [file...]





grep variable aix




grep variable aix






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Dec 9 at 14:29









Jeff Schaller

38.6k1053125




38.6k1053125










asked Dec 4 at 9:03









BlueNinja

213




213












  • What operating system are you using?
    – terdon♦
    Dec 4 at 9:16










  • The command uname -s gives me 'AIX'
    – BlueNinja
    Dec 4 at 9:21






  • 1




    It's not that your Unix is "too old", it's that you are relying on non-standard functionality.
    – Kusalananda
    Dec 4 at 9:24










  • Is your Data variable a line from the output of grep? If it is, you can probably do whatever you are doing a fair bit more efficiently by running perl or awk over the original log files. Parsing the output of grep is not an ideal solution.
    – Kusalananda
    Dec 4 at 10:41




















  • What operating system are you using?
    – terdon♦
    Dec 4 at 9:16










  • The command uname -s gives me 'AIX'
    – BlueNinja
    Dec 4 at 9:21






  • 1




    It's not that your Unix is "too old", it's that you are relying on non-standard functionality.
    – Kusalananda
    Dec 4 at 9:24










  • Is your Data variable a line from the output of grep? If it is, you can probably do whatever you are doing a fair bit more efficiently by running perl or awk over the original log files. Parsing the output of grep is not an ideal solution.
    – Kusalananda
    Dec 4 at 10:41


















What operating system are you using?
– terdon♦
Dec 4 at 9:16




What operating system are you using?
– terdon♦
Dec 4 at 9:16












The command uname -s gives me 'AIX'
– BlueNinja
Dec 4 at 9:21




The command uname -s gives me 'AIX'
– BlueNinja
Dec 4 at 9:21




1




1




It's not that your Unix is "too old", it's that you are relying on non-standard functionality.
– Kusalananda
Dec 4 at 9:24




It's not that your Unix is "too old", it's that you are relying on non-standard functionality.
– Kusalananda
Dec 4 at 9:24












Is your Data variable a line from the output of grep? If it is, you can probably do whatever you are doing a fair bit more efficiently by running perl or awk over the original log files. Parsing the output of grep is not an ideal solution.
– Kusalananda
Dec 4 at 10:41






Is your Data variable a line from the output of grep? If it is, you can probably do whatever you are doing a fair bit more efficiently by running perl or awk over the original log files. Parsing the output of grep is not an ideal solution.
– Kusalananda
Dec 4 at 10:41












1 Answer
1






active

oldest

votes


















4














Note that your regex wouldn't work even if -o was recognized by your grep implementation. You also need -E to enable extended regular expressions. Assuming you have perl, which you probably do, you can try:



$ Data="Logs/2018-12-03/log.txt:3:2018-12-03 00:00:04,333 452621453 [blah blah blah"

$ echo "$Data" | perl -lne '/[0-9]{4}-(0[1-9]|1[0-2])-(0[1-9]|[1-2][0-9]|3[0-1]) (2[0-3]|[01][0-9]):[0-5][0-9]/ && print "$.:$&"'

1:2018-12-03 00:00


But do you really need to make your regex that complicated? Isn't this enough?



$ echo "$Data" | perl -lne '/d{4}-d{2}-d{2} d{2}:d{2}/ && print "$.:$&"'

1:2018-12-03 00:00





share|improve this answer



















  • 1




    Note also that the data in the Data variable is a line produced by grep. This makes me think that the user is first running a grep of some kind to get the lines they want, and then iterating in a shell loop over these lines. This is quite an suboptimal approach to log file parsing.
    – Kusalananda
    Dec 4 at 10:43













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active

oldest

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4














Note that your regex wouldn't work even if -o was recognized by your grep implementation. You also need -E to enable extended regular expressions. Assuming you have perl, which you probably do, you can try:



$ Data="Logs/2018-12-03/log.txt:3:2018-12-03 00:00:04,333 452621453 [blah blah blah"

$ echo "$Data" | perl -lne '/[0-9]{4}-(0[1-9]|1[0-2])-(0[1-9]|[1-2][0-9]|3[0-1]) (2[0-3]|[01][0-9]):[0-5][0-9]/ && print "$.:$&"'

1:2018-12-03 00:00


But do you really need to make your regex that complicated? Isn't this enough?



$ echo "$Data" | perl -lne '/d{4}-d{2}-d{2} d{2}:d{2}/ && print "$.:$&"'

1:2018-12-03 00:00





share|improve this answer



















  • 1




    Note also that the data in the Data variable is a line produced by grep. This makes me think that the user is first running a grep of some kind to get the lines they want, and then iterating in a shell loop over these lines. This is quite an suboptimal approach to log file parsing.
    – Kusalananda
    Dec 4 at 10:43


















4














Note that your regex wouldn't work even if -o was recognized by your grep implementation. You also need -E to enable extended regular expressions. Assuming you have perl, which you probably do, you can try:



$ Data="Logs/2018-12-03/log.txt:3:2018-12-03 00:00:04,333 452621453 [blah blah blah"

$ echo "$Data" | perl -lne '/[0-9]{4}-(0[1-9]|1[0-2])-(0[1-9]|[1-2][0-9]|3[0-1]) (2[0-3]|[01][0-9]):[0-5][0-9]/ && print "$.:$&"'

1:2018-12-03 00:00


But do you really need to make your regex that complicated? Isn't this enough?



$ echo "$Data" | perl -lne '/d{4}-d{2}-d{2} d{2}:d{2}/ && print "$.:$&"'

1:2018-12-03 00:00





share|improve this answer



















  • 1




    Note also that the data in the Data variable is a line produced by grep. This makes me think that the user is first running a grep of some kind to get the lines they want, and then iterating in a shell loop over these lines. This is quite an suboptimal approach to log file parsing.
    – Kusalananda
    Dec 4 at 10:43
















4












4








4






Note that your regex wouldn't work even if -o was recognized by your grep implementation. You also need -E to enable extended regular expressions. Assuming you have perl, which you probably do, you can try:



$ Data="Logs/2018-12-03/log.txt:3:2018-12-03 00:00:04,333 452621453 [blah blah blah"

$ echo "$Data" | perl -lne '/[0-9]{4}-(0[1-9]|1[0-2])-(0[1-9]|[1-2][0-9]|3[0-1]) (2[0-3]|[01][0-9]):[0-5][0-9]/ && print "$.:$&"'

1:2018-12-03 00:00


But do you really need to make your regex that complicated? Isn't this enough?



$ echo "$Data" | perl -lne '/d{4}-d{2}-d{2} d{2}:d{2}/ && print "$.:$&"'

1:2018-12-03 00:00





share|improve this answer














Note that your regex wouldn't work even if -o was recognized by your grep implementation. You also need -E to enable extended regular expressions. Assuming you have perl, which you probably do, you can try:



$ Data="Logs/2018-12-03/log.txt:3:2018-12-03 00:00:04,333 452621453 [blah blah blah"

$ echo "$Data" | perl -lne '/[0-9]{4}-(0[1-9]|1[0-2])-(0[1-9]|[1-2][0-9]|3[0-1]) (2[0-3]|[01][0-9]):[0-5][0-9]/ && print "$.:$&"'

1:2018-12-03 00:00


But do you really need to make your regex that complicated? Isn't this enough?



$ echo "$Data" | perl -lne '/d{4}-d{2}-d{2} d{2}:d{2}/ && print "$.:$&"'

1:2018-12-03 00:00






share|improve this answer














share|improve this answer



share|improve this answer








edited Dec 4 at 9:28

























answered Dec 4 at 9:20









terdon♦

128k31247423




128k31247423








  • 1




    Note also that the data in the Data variable is a line produced by grep. This makes me think that the user is first running a grep of some kind to get the lines they want, and then iterating in a shell loop over these lines. This is quite an suboptimal approach to log file parsing.
    – Kusalananda
    Dec 4 at 10:43
















  • 1




    Note also that the data in the Data variable is a line produced by grep. This makes me think that the user is first running a grep of some kind to get the lines they want, and then iterating in a shell loop over these lines. This is quite an suboptimal approach to log file parsing.
    – Kusalananda
    Dec 4 at 10:43










1




1




Note also that the data in the Data variable is a line produced by grep. This makes me think that the user is first running a grep of some kind to get the lines they want, and then iterating in a shell loop over these lines. This is quite an suboptimal approach to log file parsing.
– Kusalananda
Dec 4 at 10:43






Note also that the data in the Data variable is a line produced by grep. This makes me think that the user is first running a grep of some kind to get the lines they want, and then iterating in a shell loop over these lines. This is quite an suboptimal approach to log file parsing.
– Kusalananda
Dec 4 at 10:43




















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