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date in awk

 
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# 8  
Old 08-06-2007
sorry madhan, i am not able to get the output. can you please provide your screenshot? when i tried its giving an error message as
awk: syntax error near line 1
awk: illegal statement near line 1
# 9  
Old 08-06-2007
Quote:
Originally Posted by jam_prasanna
sorry madhan, i am not able to get the output. can you please provide your screenshot? when i tried its giving an error message as
awk: syntax error near line 1
awk: illegal statement near line 1
Surprising,

Did you miss out any " ' " or any other punctuation marks ?

Which shell and which os are you using ?
# 10  
Old 08-06-2007
os: linux enterprise 4.0 version
shell : bash
# 11  
Old 08-06-2007
Actually date commands outputs a multiword string. So u have to quote the variable proprly.

v=`date`
echo 1 |awk -v var="$v" '{print var}'
# 12  
Old 08-06-2007
Quote:
Originally Posted by Sukhendu Naskar
Actually date commands outputs a multiword string. So u have to quote the variable proprly.

v=`date`
echo 1 |awk -v var="$v" '{print var}'
I don't think its necessary.

It works fine for me.

Code:
zsh
Fedora Core 5

# 13  
Old 08-06-2007
madhaban, actually i am using AIX V-5 and ksh.
On running: v=`date`
echo 1|awk -v var=$v '{print var}'
i got the following error msg:
awk: 0602-533 Cannot find or open file 6.
The source line number is 1.
would u plz expln. it
# 14  
Old 08-06-2007
Quote:
Originally Posted by Sukhendu Naskar
madhaban, actually i am using AIX V-5 and ksh.
On running: v=`date`
echo 1|awk -v var=$v '{print var}'
i got the following error msg:
awk: 0602-533 Cannot find or open file 6.
The source line number is 1.
would u plz expln. it
I see what you are saying Smilie

It doesn't work in ksh.

Might be there is default globbing of parameters in zsh Smilie
 
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