Invalid max count in grep function

Question

If I do a find . -mmin -1 I get 'myfile' which was modified in last one minute.

But when I want to search a particular string in that file by doing

grep 'myString' myfile -mmin -1 

I get the error invalid max count

I also tried

find . -name "myfile" -exec grep 'myString' myfile -mmin -5

I get the error find: missing argument to -exec

So my question is How do I grep or cat only the changes within a file which happened in last 5 mins. Now that I know the file name which was modified during that period.

Suggestions? Thanks in adv.

Solution 1

Grep doesn't have an mmin argument as far as I can see. It does have a -m argument with a number parameter. grep 'myString' myfile -m3 will stop after 3 lines containing myString. So, the error message means that 'min' in -mmin is not a valid maximum count.

Solution 2

grep something *

Error:

grep: invalid max count

Verify that you have a file with a leading dash in the name in the current directory. The file name might be taken for an option.

For example:

grep something // okay

touch -- -mmin

**grep something **

grep: invalid max count

Workaround:

**grep -- something **

From man getopt:

Each parameter after a -- parameter is always interpreted as a non-option parameter.

Comments

[Niels Bech Nielsen]

I think with this one, you came closest: find . -name "myfile" -exec grep 'myString' myfile -mmin -5, however, if you need to execute the grep from find, it must be find . -mmin -1 -exec grep -H "mystring" {} \; where the {} are where to insert the filename, and \; terminates the grep command. In this way, find will search for new files and execute grep on each.