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looping in awk

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# 1  
Old 05-16-2011
looping in awk
How do I remove last comma?
Code:
echo "xx yy zz" | awk 'BEGIN{FS=" "}{for (i=1; i<=NF; i++) printf "%s,", $i}'

output:
Code:
xx,yy,zz,

required output:
Code:
xx,yy,zz

or (ideally!):
Code:
xx, yy & zz

many thanks in advance!
# 2  
Old 05-16-2011
Code:
echo "xx yy zz" | awk '{for (i=1; i<=NF; i++) printf "%s%s", $i,(i==NF)?RS:(i==NF-1)?" & ":OFS}' OFS=,

# 3  
Old 05-16-2011
Try this

Code:
echo "xx yy zz" | awk '{for (i=1; i<NF-1; i++) printf "%s, ", $i; printf "%s & %s",$(NF-1),$NF}'

regards,
Ahamed
This User Gave Thanks to ahamed101 For This Post:
euval
# 4  
Old 05-16-2011
How about this:

Code:
 
echo "xx yy zz" | awk 'BEGIN{FS=" "}{ for (i=1; i<=(NF-1); i++) { printf("%s,", $i); } printf("%s\n", $NF); }'

This User Gave Thanks to TRB For This Post:
euval
# 5  
Old 05-16-2011
that does exactly what I need - many thanks indeed!
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