Shell Programming and Scripting

I still don't think you need an array:

Code:

grep ^MOVIE infile >/dev/null &&
  printf '%s\n' "there is a MOVIE without *' |
    mailx -s ... email@address

Notice that for some shells (like the Z-Shell) the ^ character is special and therefore you need to escape it.
Some implementations of grep support the -q/-s option to suppress the output, so if you use one of them, the command becomes:

Code:

grep -q ^MOVIE infile  &&
  printf '%s\n' "there is a MOVIE without *' |
    mailx -s ... email@address

Thank you for that code, I will try it.
Is there any good but short online manual for bash-shell?

---------- Post updated at 06:17 AM ---------- Previous update was at 04:51 AM ----------

The code is unfortunately not enough, because the string MOVIE should ONLY occur with a '*' as first literal. You did not check the case when it starts with an empty space for instance. Acctually the empty space is not a problem as long as a '*' follows, so let me state my problem elseway:
As long as the number occurances of the string MOVIE and '*' is the same, everything is right. So grepping the lines containing MOVIE is just the first step, in addition I hav to check whether ALL that lines have '*'

Could you post a sample data and elaborate further?

Yes, that was an example.
It's quite easy to modify the regular expression to match such a pattern.

As always, sample data and expected behavior (based on the sample data) will be useful.

Hi,
I need no array, but the postings so far do not solve my problem. Remember I need a file containing ALL lines from MySampleInfile containing the string 'MOVIE' but NOT beginning with '*'

I proceed now as follows, not needing an array anymore

Code:

grep MOVIE MySampleInfile >Output1

by this I have ALL lines containing Movie, no matter if they start with '*' or not

In a second step I want to filter all lines beginning with a space or another literal than '*'. So may statement has to be "Filter ALL lines NOT beginning with '*' "

Code:

grep not ^* Output1 >Output2

That does not work of course, but how can I write a NOT-Statement for the first literal?

If I understand correctly:

Code:

% cat infile 
* MOVIE A
bla bla
 MOVIE B
blubb blubb
xMOVIE C
% grep  '^[^*]*MOVIE' infile
 MOVIE B
xMOVIE C

this is already solved. im not sure what's the problem though.. perhaps you missed to escape *?

Code:

grep ^MOVIE MySampleInfile >Output1
grep ^\* MySampleInfile >Output2

I solved it without array and like this:

Code:

grep MOVIE $vINIFILE >>tmp_filter_1
grep ^[a-zA-Z0-9' '] tmp_filter_1 >>tmp_filter_2

So all lines NOT beginning with '*' are written into the file tmp_filter_2.
If the file tmp_filter_2 is NOT empty, I know for sure there is something wrong and I send my message.