How extract a field from file with awk without separators in Unix

Question

I've got a file calling "file" where there are some fields separated with a separator. The structure of the file is:

@field1@field2@field3@field4

With awk i've extracted the fields without separators.

vr=$file;

sep_mx=`echo $vr | awk '{
    n=split($0,x,"@");
    print n
}'` 
echo $sep_mx
## here the number of substring produced is calc.

while ((i<=$max)); 
do
    # would print individual substring one at a
    # time as the counter increases.
    echo $vr | awk -v er=$i '{
        n=split($0,x,"@"); print x[er]
    }'
    ((i+=1))
done

The output is:

field1
field2
field3
field4

If i want to extract only the second field from the code i've just posted how can i do it? Thanks

Answer 1

With bash:

$ while IFS=@ read -r col1 col2 col3 col4; do 
      echo $col3
  done <<< '@field1@field2@field3@field4'
field2

Another way with awk

$ awk '{split($0,ary,/@/); print ary[3]}' <<< '@field1@field2@field3@field4'
field2

Answer 2

This should make it:

awk -F@ '{print $3}' file

In general, try to use -F. It means "field separator" and makes your life easier, with no needs to split, substr and things like that.

Test

$ awk -F@ '{print $3}' file
field2

Answer 3

The GNU sed invocation

sed -r 's/\W\w+\W(\w+).*/\1/' file

outputs

field2

Answer 4

Your code doesn't make sense. If want to convert

@field1@field2@field3@field4

to

field1
field2
field3
field4

then all you need is this (or any of various other simple solutions):

$ cat file
@field1@field2@field3@field4
$ 
$ gawk -v RS='[@\n]' 'NR>1' file
field1
field2
field3
field4

If you want to print text "field3" then it's simply:

$ gawk -v RS='[@\n]' 'NR==4' file 
field3

The record number is 4 instead of 3 because your file starts with "@" so there's an empty record before "field1".

If you tell us what you're really trying to do we can help you, so far you seem to be going down the wrong path.