How do I get awk to print a " in it's print part?

#1 (**permalink**) 06-27-2006

The line is simple, use "[" as a field separator for the bits. What I need to do is put a " between $2 and $3.

I can't seem to figure out the escape character needed to get awk to actually print out the " and not think it's part of a bigger double ".

Any ideas?

awk -F[ '{ print $1"["$2$3 }' file.txt| awk -F] '{ print $1"]"$2"\"$3THE " NEEDS TO GO HERE$4 }'

I've tried \", "\, ^" and '"" but none of it works. What am I missing? Putting in the [ between $1 and $2 works fine, I just need to do the same with a ".

Thanks.

x96riley3 · #2 (**permalink**) 06-27-2006

I'm not sure this is exactly what you wanted.

cat file | awk -F: '{print $1, $2, "\"", $3}'

sybqad0l 1706 " 546
sybbev0x 6218 " 542
wpati0k 8145 " 253
wembi02 4621 " 267
pimdp01 223 " 133
abainn1 7098 " 1141
abaibm1 7094 " 1141

-X

Ygor · #3 (**permalink**) 06-27-2006

Try...

Code:

awk '{print $3 "\042" $4}'

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