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Full Discussion: Finding directories with expression
Top Forums Shell Programming and Scripting Finding directories with expression Post 302768747 by lxdorney on Saturday 9th of February 2013 11:24:38 AM
Old 02-09-2013
Thank you so much.
after execute the script, notice is different from your output.
also adding ab_1234567_9 directory, supposed they added to the sum of 36k which is 2/mypattern/mypattern +16 and will become 52 and so on but is not.

steps:
d_path=/root/server1
pattern=mypattern
1. find to $d_path with exact $pattern to all directory expression something like this 'ab_[0-9]\{6,\}\_[0-9]\{0,\}'
2. grep my $pattern to all directory found my expression. ei ab_1234567_9 or ab_12345673334_96
3. du -k to all $pattern that was found in expression and sum them all $pattern belong to same expression.
4. slice the expression something like ab_1234567_9, ab_1234567_19 and ab_1234567_20 and so on become 1234567 and the final output would be:

total expression/pattern
32k 1234567/mypattern
4k 1234568/mypattern
32k 1234569/mypattern

 wish you could elaborate or explain then is nice.

Code:
ls
result.txt  server1  server2  test.sh

Code:
pwd
/root

Code:
du -k | awk -F"[ _]+" '$0~p {a[$3]+=$1} END {for (i in  a) print a[i]"k "i"/"p}' p="mypattern" > result.txt

Code:
cat  result.txt
36k 2/mypattern/mypattern
16k 3/mypattern/mypattern
16k 1/mypattern/mypattern

Code:
cp -r ab_1234567_2 ab_1234567_9

Code:
ls
ab_1234567_1  ab_1234567_2  ab_1234567_9  ab_1234568_2  ab_1234569_2   ab_1234569_3  result.txt

Code:
du -k | awk -F"[ _]+" '$0~p  {a[$3]+=$1} END {for (i in a) print a[i]"k "i"/"p}' p="mypattern" >  result.txt

Code:
cat result.txt
36k 2/mypattern/mypattern
16k 3/mypattern/mypattern
16k 1/mypattern/mypattern
16k 9/mypattern/mypattern

Last edited by lxdorney; 02-10-2013 at 01:19 AM..
 

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