Grep: Not a recognized flag: o

aixgrepvariable

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...]

Best 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

Related Solutions

awk grep – Extract Lines Containing Terms from One File to Another Using Grep/Awk

To extract the lines from data.txt with the genes listed in genelist.txt:

grep -w -F -f genelist.txt data.txt > newdata.txt

grep options used:

-w tells grep to match whole words only (i.e. so ABC123 won't also match ABC1234).
-F search for fixed strings (plain text) rather than regular expressions
-f genelist.txt read search patterns from the file

If you want the header (Sample 1, Sample 2, etc) line as well:

grep -w -F -f genelist.txt -e Sample data.txt > newdata.txt

-e Sample also search for "Sample"

To find lines in genelist.txt that aren't in newdata.txt:

grep -v -w -F -f <(sed -E -e 's/(\t|  +).*//' newdata.txt) genelist.txt

-v invert the search, print non-matching lines.

The rest of the grep options are the same, but instead of using a file with the -f option, it's using something called Process Substitution (See also), which allows you to use a command in place of an actual file. Whatever output the command creates is treated as the "file"'s contents.

In this case, we're using the command sed -E -e 's/(\t| +).*//' newdata.txt, which outputs each line of newdata.txt after first deleting everything from either the first TAB character or the first pair of spaces it sees. In other words, the first field (e.g. "Gene A"). I had to use TAB or double space because a) i wasn't sure if your data was space-separated or TAB separated and b) the first fields in your example contained spaces.

sed options used:

-E use extended regular expressions, so we can use plain (, ), and + which are more readable than having to escape them with \ as $, $, \+.
-e 's/(\t| +).*//' specifies the sed script to apply against the input (newdata.txt)

Running that command on your sample data.txt would produce the following output:

$ sed -E -e 's/(\t|  +).*//' data.txt

Gene A
Gene B
Gene C
Gene D

Anyway, the output of that sed command is used as the list of search patterns by the grep command.

Shell – grep: Not a recognized flag in AIX

Looks like $LINE contains a value which starts with a dash.

You can protect against this with

grep -q -e "$LINE"

More generally, most Unix commands allow -- to mark the end of options, and so any argument after this "end of options" option will be taken as a literal, non-option argument.

echo will have a problem, too; the portable solution is to switch to printf, which works fine with arguments which start with dashes, as long as it's not the first argument, which is a format string.

You should also avoid using uppercase variable names; these are reserved for system use.

Finally, scripts should almost never need to explicitly examine $? - this is already done by if, while and other control constructs.

if grep -q -e "$line" log.log; then
    printf '%s\n' "$line"
fi

As an optimization, placing the redirection outside the loop will make things a lot quicker.

while read -r line; do
    grep -q -e "$line" log.log && printf '%s\n' "$line"
done <results.txt >notfound.txt

Best Answer

Related Solutions

awk grep – Extract Lines Containing Terms from One File to Another Using Grep/Awk

Shell – grep: Not a recognized flag in AIX

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