sed用括号外的逗号替换

Question

I have this partial SQL string.

select ID,to_char(ts2date(created_t),'DD-MM-YYYY'),name,segment_code from sometable

Using sed, I tried to replace any comma that reside outside the outmost parentheses with string char '~'.

The desired result would be:-

select ID~to_char(ts2date(created_t),'DD-MM-YYYY')~name~segment_code from sometable

Here is what I tried:-

sed '
:a
s/[,]\(.*(\)/~\1/g
s/\().*\)[,]/\1~/g
ta

But the result become:-

select ID~to_char(ts2date(created_t)~'DD-MM-YYYY')~name~segment_code from sometable

How can I ignore the comma inside the outmost parentheses?

TQ for any answer .. :)

Answer 1

It's not possible to reach such goal with pure sed regex. Correct/incorrect bracketing and its depth cannot be recognized by regular automatas (and therefore it cannot be recognized by regular expressions).

If you want to reach this with a "regex", you might want to use perl and its look-ahead/look-behind features. Or write a simple loop that checks the bracketing.

Answer 2

sed is an excellent tool for simple substitutions on a single line. For any other text manipulation, just use awk:

$ awk '{
    match($0,/\(.*\)/)

    head = substr($0,1,RSTART-1)
    tail = substr($0,RSTART+RLENGTH)

    gsub(/,/,"~",head)
    gsub(/,/,"~",tail)

    print head substr($0,RSTART,RLENGTH) tail
}' file
select ID~to_char(ts2date(created_t),'DD-MM-YYYY')~name~segment_code from sometable

Couldn't be much more straight forward...