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print argv inside awk

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# 1  
Old 06-18-2009
print argv inside awk
Code:
 
 mode=$1
  psg telnetd | awk current=`date +%M`'{
               printf ("mode is %s",mode)
               printf ("mode is %s",ARGV[1])
                }'


at command prompt when i run the script along with the argument i get only--
'mode is '
argument is not printed.(If the argument is printed outside the awk it gets printed!!) why is it so? . what can be done to make it print in the awk action block.
# 2  
Old 06-18-2009
You have to do like this...
e.g.
Code:
awk '{print variable}' variable=$argv[1] <filename>

# 3  
Old 06-18-2009
Use the -v option for shell variables:

Code:
awk -v mode=$1 'BEGIN{
  printf ("mode is %s\n",mode)
}'

Have a read of one of the awk tutorial here:

https://www.unix.com/answers-frequent...tutorials.html
# 4  
Old 06-18-2009
Quote:
Originally Posted by Franklin52
Use the -v option for shell variables:

Code:
awk -v mode=$1 'BEGIN{
  printf ("mode is %s\n",mode)
}'

Have a read of one of the awk tutorial here:

https://www.unix.com/answers-frequent...tutorials.html
yep..thanks..

from the code removed BEGIN and its working fine.
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