使用 sed 或 awk 替换匹配后的所有行

Question

I have the following txt file:

Col1,,Col2,,Col 3,,Session,,Time
Mike,,Rg,,Tx,,32658723,,2:00
,,,,,,,,
,,,,,,23623623,,
,,,,,,,,
Joe,,Tx,,Rg,,47235623,,1:00
,,,,,,,,
Peter,,Un,,Xs,,6523,,1:00
,,,,,,,,
Nick,,Xe,,Lk,,67286734,,3:00
,,,,,,,,
,,,,,,,,
,,,,,,32623,,
,,,,,,,,
Bob Li,,Yh,,Xa,,2362,,3:00
,,,,,,,,
,,,,,,,,
,,,,,,,,
,,,,,,,,
,,,,,,323,,
,,,,,,,,
,,,,,,,,
,,,,,,,,
,,,,,,,,
Lin Xu,,Rw,,NB,,1352362,,1:00
,,,,,,,,

The most important value on this file is what is on column 7th. I would like to fill all the empty lines right below the first line which has a value for column 7th. Leaving that value untouched.

I have been trying some send commands like:

sed -n '/^,/{g;};h;p'

But it is replacing all the empty lines, even if they do have an expected value.

What I would like the above file is to become like this:

Col1,,Col2,,Col 3,,Session,,Time
Mike,,Rg,,Tx,,32658723,,2:00
Mike,,Rg,,Tx,,32658723,,2:00
Mike,,Rg,,Tx,,23623623,,2:00
Mike,,Rg,,Tx,,23623623,,2:00
Joe,,Tx,,Rg,,47235623,,1:00
Joe,,Tx,,Rg,,47235623,,1:00
Peter,,Un,,Xs,,6523,,1:00
Peter,,Un,,Xs,,6523,,1:00
Nick,,Xe,,Lk,,67286734,,3:00
Nick,,Xe,,Lk,,67286734,,3:00
Nick,,Xe,,Lk,,67286734,,3:00
Nick,,Xe,,Lk,,32623,,3:00
Nick,,Xe,,Lk,,32623,,3:00
Bob Li,,Yh,,Xa,,2362,,3:00
Bob Li,,Yh,,Xa,,2362,,3:00
Bob Li,,Yh,,Xa,,323,,3:00
Bob Li,,Yh,,Xa,,323,,3:00
Bob Li,,Yh,,Xa,,323,,3:00
Bob Li,,Yh,,Xa,,323,,3:00
Bob Li,,Yh,,Xa,,323,,3:00
Lin Xu,,Rw,,NB,,1352362,,1:00
Lin Xu,,Rw,,NB,,1352362,,1:00

Answer 1

This might work for you (GNU sed):

sed -E '1b
        /^,{8}$/{g;b}
        /^,{6}\S+,,$/{G;s/^,{6}(\S+),,\n(([^,]*,){6})[^,]*/\2\1/}
        h' file

Ignore the header line.

If the line contains empty fields, replace it by the previous good line and break.

If the line contains all empty fields bar the seventh, append the last good line and replace its seventh field by that of the current lines.

This line is now good, make a copy.

All lines are printed by default.