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Old 04-20-2004, 08:24 AM   #1
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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!
Old 04-20-2004, 09:58 AM   #2
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I'm not any good with sed, but here's a couple of C programs to do it easily:

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;
int main(int argc, char **argv)
  if(argc != 2 || strlen(argv[1]) != 14)
    return 1;
  argv[1][8] = '\0';
  return 0;
Old 04-20-2004, 10:33 AM   #3
Registered: Aug 2003
Location: Akron, OH
Distribution: Fedora Core 3
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echo 20040419204537 |
while read a; do
hr=`echo $a | cut -c9-10`
dt=`echo $a | cut -c1-8`
echo $hr $dt
Old 04-20-2004, 10:37 AM   #4
jim mcnamara
Registered: May 2002
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day=`expr substr $d 9 2 `
echo "date=$d  day=$day"
Old 04-20-2004, 12:01 PM   #5
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Thank you all very much. Such a quick response and good examples. This is a great site!


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