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nausicaa3000

04-20-2004 07:24 AM

using sed to grab two characters?

I am new to sed, but it seems to me this may be able to do what I want. I need to be able to capture two integers from a variable. I have a variable holding a date and time in the form YYYYMMDDHHMMSS.

For example if I have a variable holding 20040419204537, and I want to grab the hour (20) and place it in a new variable.

Also, if possible, I would like to be able to capture the first 8 characters and put them into a new variable (20040419)

Thanks for any help you can give!

itsme86

04-20-2004 08:58 AM

I'm not any good with sed, but here's a couple of C programs to do it easily:

Code:

gethour.c

int main(int argc, char **argv)

{

  if(argc != 2 || strlen(argv[1]) != 14)

    return 1;

  printf("%c%c\n", argv[1][8], argv[1][9]);

  return 0;

}

Code:

getdate.c

int main(int argc, char **argv)

{

  if(argc != 2 || strlen(argv[1]) != 14)

    return 1;

  argv[1][8] = '\0';

  puts(argv[1]);

  return 0;

}

mfeat

04-20-2004 09:33 AM

echo 20040419204537 |
while read a; do
hr=`echo $a | cut -c9-10`
dt=`echo $a | cut -c1-8`
echo $hr $dt
done

jim mcnamara

04-20-2004 09:37 AM

Code:

d="20040419204537"

day=`expr substr $d 9 2 `

echo "date=$d  day=$day"

nausicaa3000

04-20-2004 11:01 AM

Thank you all very much. Such a quick response and good examples. This is a great site!

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