string with content of variables

chloraldo · 08-18-2006, 08:51 AM

Hello

I'm trying to write the last line in a file like this (colors are just for help in here):
8518082006154522000100000000

Code:

tempfile=tempf.txt
artikel=$(grep -c ^50 $tempfile)
echo 85$(date +%d%m%Y%H%M%S)$artikel000000>>$tempfile

$artikel has a value but doesn't add it to the line. What's wrong?
$artikel should also always be a 6-digit number. How can I do that?
The 000000 at the end doesn't appear...

Thanks for your help!

acid_kewpie · 08-18-2006, 08:58 AM

i normally wrap a sub command for that variable, e.g. echo $(date...)$(echo $variable)00000

jlliagre · 08-18-2006, 09:05 AM

or simply:

Code:

echo 85$(date +%d%m%Y%H%M%S)${artikel}000000>>$tempfile

otherwise, the shell will look for a variable named artikel000000.

chloraldo · 08-18-2006, 09:09 AM

Quote:

Originally Posted by jlliagre

or simply:

Code:

echo 85$(date +%d%m%Y%H%M%S)${artikel}000000>>$tempfile

otherwise, the shell will look for a variable named artikel000000.

Thanks that works....

Now I'm trying to get e.g. the number 6 to look like 000006 or 103 to look like 00103 etc. Any ideas?

soggycornflake · 08-18-2006, 09:34 AM

Quote:

Now I'm trying to get e.g. the number 6 to look like 000006 or 103 to look like 00103 etc. Any ideas?

Code:

printf '%06d' 6

(Note: if you try this on the command line, add a \n to the end of the string, or your prompt will clobber the output.)