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# 1  
Old 07-16-2004
printf format
hi, i would like to extract the header and put it in a variable, then use printf to output the variable, but i keep on getting errors...please tell me if my format is incorrect.
Code:
HDR = "`ps -e -o user,pid,ppid,pcpu,stime,etime,time,comm | head -n 1`"

printf (%s, $HDR);

thanks!
# 2  
Old 07-16-2004
I figured out why i got the message for "invalid format character" for the string i'm trying to print. The string procuded by the command contains "%CPU", and it doesn't seem to like the "%" in the middle of the string. Is there any way to make the whole string prints including "%" using printf?

Thank you much.
# 3  
Old 07-16-2004
You can escape the % by using %% instead (to print a literal %).

Change your code to
Code:
HDR=`ps -e -o user,pid,ppid,pcpu,stime,etime,time,comm | head -n 1 | sed 's/%/%%/g'`
printf "%s" HDR

This code is for bash/ksh

Cheers
ZB
# 4  
Old 07-16-2004
Thank you thank you thank you!

happy friday! Smilie
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