UNIX for Dummies Questions & Answers

Hi!

I have thousands of sub-directories, and hundreds of thousands of files in them. What is the fast way to find out which files are older than a certain date? Is the "find" command the fastest? Or is there some other way?

Right now I have a C script that traverses through and checks each file. But it is taking forever to run! Can somebody tell me a better way?

Thanks

This is going to take a while no matter what you do. "find" should be able to do at close to the top possible speed. A custom C program should be able to beat "find", but not by much. I would not csh for anything.

Would grep be faster than find or no? I dont know much about any of it but just a thought. if you grepped for 10-2-04 or older *not exactly sure how to do that* but shouldnt turn up files fast?

Corrail -
He wants a -ctime or an -mtime older than a certain date. grep doesn't do file date/times

Ok thanks for clearing that up. I am working on teaching myself unix and grep/awk/sed and different things like that so i bring up questions on here of possible solutions to problems and then ppl can correct me and explain why and it helps ALOT. So thank you.

Hi Guys,

Thanks for your input...

From what I hear here, I guess that the way I've taken is the best. The C program is pretty fast, but then, with the hundreds of thousands of files, I'm looking for greased lightning. Every day about 20,000 files get added. I'm supposed to archive the old files (those older than 60 days). Sixty days means about 1.2 million files, with the rate growing as time goes by.

I guess one way to beat the large number of files is to go deeper down the file structure. But then, I'll have to run the application multiple times, one for each folder path that I select.

Just to complete the picture, the list of files older than 60 days is then fed to NetBackup (that archiving application from Symantec). NetBackup then moves the files to tape. I use Netbackup because it becomes easy to restore particular files whenever I need them.