UNIX for Dummies Questions & Answers

I have the following file:

Code:

Loc1,20080102,230100,0.5617,0.5617
Loc1,20080102,230400,0.5616,0.5616
Loc1,20080102,230700,0.5615,0.5615

Then I use the following code to turn the third column into something which to mysql has the time datatype:

Code:

sed -i '' -e "s#,\(..\)\(..\)\(..\),#,\1:\2:\3,#" file.dat

It correctly transforms the third column into times:

Code:

Loc1,20080102,23:01:00,0.5617,0.5617
Loc1,20080102,23:04:00,0.5616,0.5616
Loc1,20080102,23:07:00,0.5615,0.5615

What I am wondering is why the fourth column is unaffected, since it also has 6 characters and has a comma (,) on both sides. Is because of the decimal point (.)?

It is because your sed statement is executed only once per line, because, there is no flag after the last # (like e.g. g). In this case even if there were a flag g, then there would still only be one match, since the last comma was already part of the first match, so it cannot be part of the second match..

If you left out one of the commas for example then it would happen everywhere with the g flag.:

Code:

$ echo "Loc1,20080102,230100,0.5617,0.5617" | sed  -e "s#,\(..\)\(..\)\(..\)#,\1:\2:\3#g"
Loc1,20:08:0102,23:01:00,0.:56:17,0.:56:17

Makes sense, thanks.