CODEWITHSUNDEEP
replacesedawk
Russo
Russo

Reputation: 341

Using awk to replace only once the delimiter between $4 and $5

I'm really new in awk language and I'm suprised of all the power it has. (I discovered this today), because I have a file in this format:

test1;test2;test3;test4;test5;test6;test7

I need to output this in a new file and get this as result:

test1;test2;test3;test4 test5;test6;test7

Basically add ' ' between $4 and $5. I know there is a lot of questions about this but I'm not able to do what I want.

I was testing this code that I found somewhere here:

   for (i=1;i<=3;i++)
      printf "%s;", $i

   n = split($0,tmp,/  +/)

   for (i=6;i>=8;i++)
      printf ";%s", tmp[n-i]

   print ""
}

But I get as output something like:

test1;test2;test3;test4;test5;test6;;;

Can you please tell my what I'm doing wrong? and, there is another simple method like one-line code in awk to do this? Thank you in advance

Upvotes: 2

Views: 923

Answers (3)

chooban
chooban

Reputation: 9256

I'd go with sed on this one. On the basis that the issue is "replace the 4th ; character" you can use this:

sed 's/;/ /4' <inputfile>

That'll output the result to stdout, or you can use "sed -i" to do it in place.

See http://www.gnu.org/software/sed/manual/html_node/The-_0022s_0022-Command.html

Upvotes: 5

Thor
Thor

Reputation: 47249

If you're using GNU awk you could use a similar solution to what chooban suggested:

 awk '{ $0=gensub(";", " ", "4") } 1'

The 1 at the end invokes the default block: { print $0 }.

Upvotes: 2

Gilles Qu&#233;not
Gilles Qu&#233;not

Reputation: 185881

Try this :

echo "test1;test2;test3;test4;test5;test6;test7" | awk -F';' '{
   for (i=1; i<=NF; i++)
        printf "%s%s", $i, (i == 4) ? " " : ";"
}'

Output

test1;test2;test3;test4 test5;test6;test7

Upvotes: 3

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