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need date in awk (Part II)

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# 8  
Old 09-21-2004
Perderabo...I tried your suggestion. I even copied/pasted your syntax. I got the following error message:

awk: syntax error at source line 1
context is
{print09/21/04 >>> 12: <<<
awk: illegal statement at source line 1



-cd
# 9  
Old 09-21-2004
Farzolito:

I tried your suggestion also, copying/pasting your syntax into my script. I received the following error when I ran the script:

awk: syntax error at source line 1
context is
today=09/21/04 >>> 12: <<<
awk: bailing out at source line 1


-cd
# 10  
Old 09-21-2004
One more try...we need a space after that print on any version of awk....

cat $LOG | grep SWI100 | sort | uniq -c | awk '{print '"$today "' "MSP\t" "Rich\t" $3 $12 "\t" $1}'
# 11  
Old 09-21-2004
Perderabo:

Still did not work. I get the following error:

awk: syntax error at source line 1
context is
{print 09/21/04 >>> 14: <<<
awk: illegal statement at source line 1

-cd
# 12  
Old 09-21-2004
How about...

cat $LOG | grep SWI100 | sort | uniq -c | awk '{print '\"$today \"' "MSP\t" "Rich\t" $3 $12 "\t" $1}'
# 13  
Old 09-21-2004
I think it is the space caracter between date and hour in your today's variable which cause error in the awk.

try this perhaps it will be works:

use this variable :
today=`date +'%m/%d/%y_%T'

cat $LOG | grep SWI100 | sort | uniq -c | awk -v today="$today" '{printf today" MSP\t" "Rich\t" $3 $12 "\t" $1}' | sed 's/_/ /g'
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