how to check missing files?

gholdbhurg · #1 (**permalink**) 04-26-2008

I have 50 files in this directory (/home/unixnewbie/wklyfiles) namely:
statistics.1
statistics.2
statistics.3
statistics.4
statistics.5
statistics.6
statistics.7
statistics.8
statistics.9
statistics.10
....
statistics.20
....
statistics.50

How can i determine if ever there will be some files missing in there?
And how to know which exact files are those? Like for example, the statistics.5, statistics.9, statistics.20 and statistics.35 files are missing that week.

Thanks in advance.

era · #2 (**permalink**) 04-26-2008

Create a list of all files which should be there, and compare.

Code:

perl -le 'for my $i (1..50) { print "missing: statistics.$i" unless -f "statistics.$i" }'

I meant for the Perl script to just generate the list, but it turned out to be so easy to do it all in Perl. Sorry, folks (-:

Just to illustrate my original proposal, here's another approach, using just simple shell commands:

Code:

yes . | head -50 | nl | sed -e 's/^ *\([1-9][0-9]*\) .*/statistics.\1/' >list
ls -rt statistics.* | diff list -

The file "list" is generated with the nl utility to have line numbers, and the file name prefix is added using sed. (This came out pretty tortured -- it would arguably be a lot easier with awk, but let's just say that using scripting languages would be cheating, okay?) Then we compare that list against the actual directory listing. This requires that your diff accepts "-" to mean "read the other file from standard input;" otherwise, you'll need to use two temporary files. (Don't forget to remove them when you're done.)

gholdbhurg · #3 (**permalink**) 04-26-2008

Wow great, thanks a lot era!

Hope you don't mind a follow-up question, what if i want to store in a file all those existing statistics files? And just just put spaces on those missing files?

Example
==>
statistics.1;statistics.2;statistics.3;statistics.4;<space>;statistics.6;......statistics.19; <space>;statistics.21;......statistics.34;<space>;....statistics.50

Hmmm..

era · #4 (**permalink**) 04-26-2008

Code:

perl -e 'for my $i (1..50) { my $f = "statistics.$i"; print ($i > 1 ? ";" : ""), (-f "$f" ? "$f" : " ") }
print "\n"'

The "if ? then : else" construct is a bit hard to sort out at first, but makes for pretty straightforward logic once you understand how it works.

ripat · #5 (**permalink**) 04-26-2008

Another bash solution:

Code:

#!/bin/bash

for nbr in $(seq 50); do
    if [ ! -e "statistics.$nbr" ]; then echo "statistics.$nbr is missing"; fi
done

For the one-liners fanatics:

Code:

for nbr in $(seq 50); do if [ ! -e "statistics.$nbr" ]; then echo "statistics.$nbr is missing"; fi; done

era · #6 (**permalink**) 04-26-2008

Proper fanatics would perhaps prefer

Code:

for n in $(seq 50); do [ -e statistics.$n ] || echo statistics.$n is missing; done

seq is not universal, that's why I started out with Perl; but if you have it, it's excellent for this type of job.

gholdbhurg · #7 (**permalink**) 04-27-2008

Unfortunately, the seq doesnt work

seq: command not found

Any other unix scripting lines possible?
Btw, i'm using ksh. Thanks!

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