[SOLVED] all in one line

mierdatuti · 08-05-2009, 12:56 PM

Hi,

I have one files where I need to extract the date.
I do:

Code:

grep -i start file | sed -e 's/.*start="\([^ ]*\) .*/\1/' > fileoutput

I get:
20090805030000
20090806070000
20090806090000
20090806093000
20090806100000
........

Know I would like to extract this character:

20090805030000 to 090805-0300
20090806070000 to 090806-0700
20090806090000 to 090806-0900

For this I use:

Code:

grep -i start ${ruta}la6 | sed -e 's/.*start="\([^ ]*\) .*/\1/' | cut -c 3-8 > temporalA
grep -i start ${ruta}la6 | sed -e 's/.*start="\([^ ]*\) .*/\1/' | cut -c 9-12 | sed "s/^/-/g" > temporalB

And then I paste them.

The problem is that I can't generete auxiliary files. Could you help me?

Many thanks and sorry for my english!

Hko · 08-05-2009, 02:17 PM

Code:

grep -i start file | sed 's/.*start="[0-9]\{2\}\([0-9]\{6\}\)\([0-9]\{4\}\).*/\1-\2/'

mierdatuti · 08-05-2009, 02:23 PM

Quote:

Originally Posted by Hko

Code:

grep -i start file | sed 's/.*start="[0-9]\{2\}\([0-9]\{6\}\)\([0-9]\{4\}\).*/\1-\2/'

Many thanks guru!!!!!!
It works.

PTrenholme · 08-05-2009, 03:40 PM

Have you looked at the gawk language? I think your whole task could be quite easily done with a simple gawk program.

If you could post a few lines from your input file, we could probably put together such a program for you. For example, your first script could be done like this:

Code:

BEGIN {IGNORECASE=1;FS=""}
/start *=/ {
   if (match($0,/([[:digit:]]+)/, Date)) {
      $0=Date[0]
      printf "%s-%s\n", $3 $4 $5 $6 $7 $8, $9 $10 $11 $12;
   }
}

For this data:

Code:

$ cat date.txt
Start=20090805030000
End=20090806070000
Start = 20090806090000
Start= 20090806093000
Start =20090806100000

I get this output:

Code:

$ gawk -f date.awk date.txt
090805-0300
090806-0900
090806-0900
090806-1000

Note that the above is a "quick-and-dirty" example: More elegant code should be used for a production application.