Re: The string of characters in the special character by grep as a command

[Bob Proulx]

address@hidden wrote:
>    Now, I have four files in my directory:alarm.log
> ,alarm.log.20100802125915,alarm.log.wf and control.log; I will extract
> information from alarm.log.wf, orders for grep - E "FATAL|WARNING" |tail -n
> 1, at the command line -- I got the correct output WARNING: 08-04 20:41:11:
> alarm * 1126189408 close log_r.1126189408.wf. 

Your message was very difficult to read.  I cannot quite understand
it.  I will try.  I think you wanted to say:

  $ grep -E "FATAL|WARNING" alarm.log.wf | tail -n1
  WARNING: 08-04 20:41:11: alarm * 1126189408 close log_r.1126189408.wf

Okay.

> But use script: ss =`grep -E "FATAL|WARNING" alarm.log.wf|tail - n 1`, echo
> $ss, output is WARNING: 08-04 20:41:11: alarm alarm.log
> alarm.log.20100802125915 alarm.log.wf  control.log close
> log_r.1126189408.wf.It took "*" in the string as a special characters .I
> think this is a bug.

There is a space problem in your email making this very difficult to
read.  But I will guess you wanted to say:

  ss=`grep -E "FATAL|WARNING" alarm.log.wf | tail -n1`

I think that is what you meant to say.

> It took "*" in the string as a special characters .I think this is a
> bug.

Not in the assignment.  But if you do not quote it properly then when
printing it the '*' may be expanded as a file glob.

Case 1 (fail, insufficient quoting):

  $ echo $ss

Case 2:

  $ echo "$ss"

The first case will expand the variable on the command line.  Then
because it is not quoted it will do file glob expansion.  During the
file glob expansion the '*' will be expanded.  Therefore it is
necessary to quote it as in the second correct case.

I think the problem you experienced happened when you printed the
contents and not during the variable assignment.

Bob