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09-04-2008
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Get date from cal function
Cal
Sun Mon Tue Wed Thu Fri Sat 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 cal | awk '{print $7}' Sat 13 20 27 In the above output i am missing 6 because the first column is empty so it is shfting. How can i print all the dates under sat. I need the output like below. Sat 6 13 20 27 |
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09-04-2008
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Use cut to cut out the column by character count.
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09-04-2008
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A very ugly way to do this...But the solution suggested by radoulov must be the best way.. Code:
cal | sed -n '/\(.*\) \(.*\) \(.*\) \(.*\) \(.*\) \(.*\) \(.*\)/p' |sed 's/\(.*\) \(.*\)/\2/' |
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