date -d 'May 20, 2010' +%d%m in awk


 
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# 1  
Old 08-17-2010
date -d 'May 20, 2010' +%d%m in awk

Hi,

I am working on Redhat GNU linux

Code:
#!/usr/bin/ksh
awk '
{
        dial_date="May 26, 2010"
        print dial_date
        aa ='`date -d '"dial_date"' +%d%m%Y`'
        print aa
} END {}' < "input"

Error output:
Code:
date: invalid date `"dial_date"'
awk: cmd. line:5:       aa =
awk: cmd. line:5:           ^ unexpected newline or end of string

The variable dial_date will have a value like "May 26, 2010". I need the output as dd/mm/yyyy. This logic should run within an awk loop. I am not able to represent the variable 'dial_date' inside the date syntax correctly. Please help.

Last edited by Scott; 08-17-2010 at 12:30 PM.. Reason: Please use code tags
# 2  
Old 08-17-2010
Code:
"date -d "dial_date " +%d%m%Y" | getline aa

# 3  
Old 08-17-2010
I tried :
`"date -d "dial_date " +%d%m%Y"` | getline aa
I got an error : sh -c
# 4  
Old 08-17-2010
Dont add backticks. Just replace your code by mine and run it
# 5  
Old 08-18-2010
#!/usr/bin/ksh
awk '
{
dial_date="May 26, 2010"
print dial_date
"date -d "dial_date " +%d%m%Y" | getline aa
print aa
} END {}' < "input"
exit 0


Messge
/tenant1> test.ksh
May 26, 2010
sh: +%d%m%Y: command not found
May 26, 2010
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