The grep command argument list is too long? We have find, Xargs, and grep.

First, the problem of the current directory in the specified sub-directory specified in the server for the specified time range of all containing "meritorious acquisition" of the log content 1, the specified subdirectory 1.2.3.25 .... 1.2.3.83
2, specify the server the server's log file name format is: Sceneserver.log. (the latter two digits of the year, two digits) (month, not enough with 0 complement, two digits) (Day, not enough with 0 complement, two digits)-(hours, not enough with 0 complement, two digits)
For example, January 22, 2014 16 o'clock, the log file name is: sceneserver.log.140122-163, the specified time range from October 2014 to 22-2014 December 9

Second, grep's helplessness undoubtedly, directly think of is certainly powerful grep, write command:

grep ' exploits get ' 1.2.3. {2[5-9],[3-7][0-9],8[0-3]}/sceneserver.log.14{10{2[2-9],3},11,120[1-9]}* > Tmp.log

The results will come out. Tips:

Hint: command argument list is too long

(The matching files were found to be close to 3w later)
Okay, so here's the question: besides grep, what are the Linux commands that support regular expression search files?

Three, a strong combination to find the file is definitely the Find command, man, found that find is actually supporting the regular expression. Command:

Find. -regextype "Egrep"-regex "\./1\.2\.3\. (2[5-9]| [3-7] [0-9]|8[0-3])/sceneserver\.log\.14 (2[2-9]|3) |11|120[1-9]). *-print0 | xargs-0 grep ' meritorious acquisition ' >> Tmp.log

Note that there are: 1. Note the difference between-print and-print0 in the Find command parameter, and the difference between the Xargs parameter and the 0 parameter and no-0 parameter 2, you can use the following command to view the number of matching files

Find. -regextype "Egrep"-regex "\./1\.2\.3\. (2[5-9]| [3-7] [0-9]|8[0-3])/sceneserver\.log\.14 (2[2-9]|3) |11|120[1-9]). *-print | Wc-l

The grep command argument list is too long? We have find, Xargs, and grep.