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#1
02-08-2013
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Finding directories with expression
Hi All,
I need your help in finding pattern of directories. need to search for all pattern have "mypatern" from base directory folder. example ------- server1 - base directory 100 server1/ab_123456_1/mypattern 100 server1/ab_123456_2/mypattern 200 server1/ab_123457_1/mypattern 400 server1/ab_123457_3/mypattern If the pattern "mypattern" is found in server1/ab_123456_1 and if another pattern is found ab_123456_2 then i want the to du -k mypattern and sum and the result saved as shown. result.txt --------- 200k 123456/mypattern 600k 123457/mypattern Appreciate your help in Advance. Thanks lxdorney |
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#2
02-08-2013
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Code:
find * -type d -name mypattern | xargs -r du -sk | while read k n
do
$(( sum += k ))
done
echo $sum |
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#3
02-09-2013
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very appreciate here thanks Code:
find * -type d -name mypattern | xargs -r du -sk | while read k n do $(( sum += k )) done echo $sum executed the above command got error message: Code:
./test.sh: line 3: 8: command not found ./test.sh: line 3: 12: command not found ./test.sh: line 3: 16: command not found ./test.sh: line 3: 24: command not found ./test.sh: line 3: 28: command not found but when execute this Code:
find * -type d -name mypattern | xargs -r du -sk output: Code:
8 server/ab_1234568_2/mypattern 4 server/ab_1234567_2/mypattern 4 server/ab_1234569_2/mypattern 8 server/ab_1234567_1/mypattern 4 server/ab_1234569_3/mypattern is there any chances that can group the same sequence of directory filename? 8 server/ab_1234567_1/mypattern 4 server/ab_1234567_2/mypattern 4 server/ab_1234569_2/mypattern 4 server/ab_1234569_3/mypattern the output would be Code:
8 ab_1234569/mypattern 12 ab_1234567/mypattern Last edited by Scrutinizer; 02-09-2013 at 02:43 AM.. Reason: Additional code tags |
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#4
02-09-2013
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Code:
awk -F"[ _]" '$0~p {a[$3]+=$1} END {for (i in a) print a[i]"k "i"/"p}' p="mypattern" server1 > result.txt
cat result.txt
200k 123456/mypattern
600k 123457/mypattern |
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#5
02-09-2013
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Code:
awk -F"[ _]" '$0~p {a[$3]+=$1} END {for (i in a) print a[i]"k "i"/"p}' p="mypattern" server1> result.txt
awk: cmd. line:1: fatal: cannot open file `server1' for reading (Inappropriate ioctl for device)my apologies, server1 is a directory not a file my process goes like this: Code:
mkdir -p server1/ab_1234568_2/mypattern mkdir -p server1/ab_1234567_2/mypattern mkdir -p server1/ab_1234569_2/mypattern mkdir -p server1/ab_1234567_1/mypattern mkdir -p server1/ab_1234569_3/mypattern create some files Code:
cd server1 ls /etc/ -l > ab_1234567_1/mypattern/etc.txt ls /etc/ -l > ab_1234567_2/mypattern/etc.txt ls /etc/ -l > ab_1234569_3/mypattern/etc.txt ls /etc/ -l > ab_1234569_2/mypattern/etc.txt Code:
cd server1 du -k 4 ./ab_1234568_2/mypattern 8 ./ab_1234568_2 16 ./ab_1234567_2/mypattern 20 ./ab_1234567_2 16 ./ab_1234569_2/mypattern 20 ./ab_1234569_2 16 ./ab_1234567_1/mypattern 20 ./ab_1234567_1 16 ./ab_1234569_3/mypattern 20 ./ab_1234569_3 92 . is there any chances can get the below result something like this? 32 1234567/mypattern 4 1234568/mypattern 36 1234569/mypattern sum of all color red sum of all color green sum of all color blue really appreciate your help Last edited by lxdorney; 02-09-2013 at 07:35 AM.. |
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#6
02-09-2013
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I did assume that you have the information stored in file server1 You can use this Code:
du -k | awk -F"[ _]+" '$0~p {a[$3]+=$1} END {for (i in a) print a[i]"k "i"/"p}' p="mypattern" > result.txtCode:
cat result.txt 32k 1234567/mypattern 4k 1234568/mypattern 32k 1234569/mypattern Also added a + since du seems to have many space as separator. |
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#7
02-09-2013
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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.txtCode:
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.txtCode:
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 12:19 AM.. |
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