delimiter file need to insert characters, awk problem.

fantasy1215 · 07-09-2012, 01:35 AM

I have space delimitered data file, data file as follows: The time format is not specified, may be 09:46:34 or 094634 or 09/46/34

Code:

data.txt
4367424271020235424  000210584060118   09:46:34  000000050000  110 3002
7424271020235424     00330210584060118 09:46:52                    0000

I need to insert fix time( ) to field 3, to make it look like this:

Code:

4367424271020235424  000210584060118   time(09:46:34)  000000050000  110 3002
7424271020235424     00330210584060118 time(09:46:52)                    0000

I think I may use awk to deal with such delimitered file, but don't know how to insert.
Thanks in advance!

druuna · 07-09-2012, 02:02 AM

Have a look at this sed solution:

Code:

sed 's/\([0-2][0-9]:[0-5][0-9]:[0-5][0-9]\)/time(\1)/' input

I personally think that sed is the easier solution.

fantasy1215 · 07-09-2012, 02:07 AM

Quote:

Originally Posted by druuna

Have a look at this sed solution:

Code:

sed 's/\([0-2][0-9]:[0-5][0-9]:[0-5][0-9]\)/time(\1)/' input

I personally think that sed is the easier solution.

But what if the time format is 20120709094634 instead of 09:46:34?
still Thanks for your quick reply.

druuna · 07-09-2012, 02:33 AM

Try this:

Code:

awk '{ printf("%-19.19d  %-17.17d  %-20s  %-12.12d  %-3.3d  %-4.4d\n", $1, $2, "time("$3")", $4, $5, $6 ) }' input

Example run:

Code:

$ cat input
4367424271020235424  000210584060118   09:46:34  000000050000  110 3002
7424271020235424     00330210584060118 09:46:52                    0000
7424271020235424     00330210584060118 20120709094634              0000

$ awk '{ printf("%-19.19d  %-17.17d  %-20s  %-12.12d  %-3.3d  %-4.4d\n", $1, $2, "time("$3")", $4, $5, $6 ) }' input 
4367424271020235264  00000210584060118  time(09:46:34)        000000050000  110  3002
0007424271020235424  00330210584060118  time(09:46:52)        000000000000  000  0000
0007424271020235424  00330210584060118  time(20120709094634)  000000000000  000  0000

One side effect: fields that aren't there are printed as zero's (italic parts in the above output). This could be fixed with a more elaborate awk script.

EDIT: I just noticed you edited your original post after an answer has been given, please don't do that. The answer(s) given might look out of place if you do so.

grail · 07-09-2012, 03:28 AM

You did not specifically mention formatting, so the obvious solution if not important would be:

Code:

awk '$3 = "time("$3")"' file

If the formatting is important:

Code:

awk 'sub($3,"time("$3")")' file

fantasy1215 · 07-09-2012, 04:44 AM

Quote:

Originally Posted by grail

You did not specifically mention formatting, so the obvious solution if not important would be:

Code:

awk '$3 = "time("$3")"' file

If the formatting is important:

Code:

awk 'sub($3,"time("$3")")' file

Thanks so much,
awk 'sub($3,"time("$3")")' file
really solves my problem. Thanks. The formatting problem is just the problem.