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how to suppress the status of fork in shell script

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# 1  
Old 02-29-2012
Java how to suppress the status of fork in shell script
When command is executed by forking, the console displays the status of that command. I want to suppress it.. how to do it ?
Example:
Code:
 var1=`date` &
echo "hello world";

output:
hello world
[1]+ Done var1=`date`

I want to suppress the second line "[1]+ Done var1=`date`".
I tried >/dev/null but of no use...
# 2  
Old 02-29-2012
The status messages are suppressed on my RHEL GNU Bash when I do this:

Code:
$ (x=`date` &); echo "hello world"
hello world
$

Check if its the same behaviour on your system too.
# 3  
Old 02-29-2012
it didnt help
This aint working in Solaris and SUSE.
Though the message is suppressed the value of X is null.
# 4  
Old 02-29-2012
Or if you run the statements in a script:

Code:
$ cat test1
var1=`date` &
echo "hello world";
wait
$ ./test1
hello world

The point is that these messages are then handled by a sub shell..

---------- Post updated at 08:07 ---------- Previous update was at 08:00 ----------

Quote:
Originally Posted by Arun_Linux
This aint working in Solaris and SUSE.
Though the message is suppressed the value of X is null.
The variable will be null anyway because you run it in a subshell in the background by using & and then X gets set in that shell, not the parent shell and when it returns X will still be null because it was never set in the current shell..
# 5  
Old 02-29-2012
Java is there anyway to export values between shells
understood.. Btw is there any workarounds to achieve this ?
# 6  
Old 02-29-2012
What do you want to achieve?
# 7  
Old 02-29-2012
Java
Code:
$ cat test1
var1=`date` &
echo "hello world";
wait
echo "value of var1 is  $var1";
$ ./test1
hello world

I want to get the value of "var1"
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