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Old 03-14-2010, 01:47 PM   #1
flea89
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Registered: Mar 2010
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Sort in descending order according to a column of an array


Hello ! I have the following bash script:

Code:
#!/bin/bash


echo "The script starts now."

echo "Hi, $USER!"


echo "Count of logins for each user:"
#!/bin/sh
last > logs.txt
awk '{count[$1]++}END{for(j in count) print j, "("count[j]"logins)"}' logs.txt
What it does is count logins to my network and print on screen the total
of logins each user has.
How can I have my results in descending order according to the loggins of each user?
For example the result should be like:
ampaltas (80logins)
amaxilat (61logins)
altanis (70logins)
alkalai (70logins)
alexopo (20logins)
alexiou (18logins)
alexandz (17logins)
alavernt (11logins)
akribopo (11logins)
akretsis (10logins)
akis (2logins)
akatis (1logins)
 
Old 03-14-2010, 01:57 PM   #2
AlucardZero
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Registered: May 2006
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Distribution: Debian
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sort -r -n -k2

see: man sort
 
Old 03-14-2010, 02:02 PM   #3
flea89
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Registered: Mar 2010
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oh thanks ! i had already read man sort but i hadnt understood how i would choose the column via k2... thanks a lot
i suppose this will go after awk, like awk '{...}' | sort -r -n -k2 ?
 
Old 03-14-2010, 07:00 PM   #4
ArfaSmif
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Registered: Oct 2008
Location: Brisbane Australia
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This will work (you can ignore the first line of the output)

last | awk '{count[$1]++}END{for(j in count) print count[j] " logins" , j}' | sort -g
 
  


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