run script error

I try to run a new script on the RH linux shell , it always pop the
sign "##4" to each line , but I can run the same script on other shell
, what is wrong in my system ? please suggest , thx in advance.

#!/bin/bash
who -u awk '{ print $6":"$7 }' grep -v '\.'

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Andy wrote:
> I try to run a new script on the RH linux shell , it always pop the
> sign "##4" to each line , but I can run the same script on other shell
> , what is wrong in my system ? please suggest , thx in advance.
>
> #!/bin/bash
> who -u awk '{ print $6":"$7 }' grep -v '\.'

Well, first off, if this is the script you run, then you have a very
non-standard version of the who(1) binary. On my system (admittedly, Slackware
Linux 9.0), who(1) doesn't take arguments like "awk" or "grep". So, is this
truely the script you run, or are there typos in the above?

Assuming that there are typos in the script you presented, and you /meant/ to
present us with something like
#!/bin/bash
who -u | awk '{ print $6":"$7 }' | grep -v '\.'
then your version of who(1) still doesn't match up with what I have. When I
run 'who -u', I get 6 fields back, but your script is explicitly looking for
seven fields (the awk(1) portion uses $6 and $7, the sixth and seventh field
respectively).

Could you show us the results of:
1) who -u
2) who -u | awk '{ print $6":"$7 }'
3) who -u | awk '{ print $6":"$7 }' | grep -v '\.'
This might help determine what it is you see, and how it gets derived.

- --
Lew Pitcher

Master Codewright & JOAT-in-training | GPG public key available on request
Registered Linux User #112576 (http://counter.li.org/)
Slackware - Because I know what I'm doing.
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