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segment of code from shell script 

 fname=$(sed 's/(.*//' <<< $p | awk '{ print $NF }')  
 if [[ $fname == *['!'@#\$%^\&*()_+]* ]]
 then
   flag1=0
 fi

where $p is one line from a text file .

p value is >= (short)BigBlock * DISK_SIZE + Size)

when above shell script is executed for p value , its gives garbage result for example this appear on terminal 

>= (short)BigBlock  file1 file2 file3 DISK_SIZE + Size)

where file1 file2 file3 are files in the folder

my guess is * is treated as ls command

what could be the possible solution

Kusalananda
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user143252
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    Please don't use abstracted values for the content of $p, you can't figure out regex issue without a real literal example. Show a real line from the file. – Lizardx Sep 21 '17 at 20:26
  • @Lizardx The OP is showing the value of $p. – Kusalananda Sep 22 '17 at 17:29
  • How do you know that? It's not at all clear from the example. If the string value is x, then how can $p >= x, that makes exactly no sense at all. I don't believe that's what the value is based on the example, but the person never responded, so I'm giving this a downvote as well. I believe that what the person is actually showing is the contents of a word problem assigned by the teacher, which isn't understood. – Lizardx Sep 22 '17 at 18:28

2 Answers2

1

Summary: You probably have echo $p somewhere in your code. You need to double-quote $p.

No, * is not treated as the ls command. It is, however, treated as a filename globbing pattern due to being used unquoted somewhere (not in code shown in the question)

You need to take care both when assigning and using the value of p and use appropriate quoting to protect the string.

Filename globbing patterns are not expanded in here-strings:

$ cat <<< *
*

So there is nothing wrong with the unquoted use of $p in the call to sed. However, it is almost always better to explicitly quote variable expansions. See also the unquoted use of $fname in your test. That should be double quoted for sure.

You say that the script outputs the text

>= (short)BigBlock  file1 file2 file3 DISK_SIZE + Size)

There is nothing that would provoke this string to be outputted in the script shown in the question. This is probably due to you doing

echo $p

somewhere else in the script.

Again, use double quotes around $p:

echo "$p"

This way, you stop the shell from performing file name globbing on its value (expanding *).

In general, use echo only for static strings, and use printf for variable data:

printf '$p is "%s"\n' "$p"

Related to that last point: Why is printf better than echo?

Kusalananda
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0

To start with, while awaiting the real $p value string: make sure to quote the $p.

fname=$(sed 's/(.*//' <<< "$p" | awk '{ print $NF }')

Also note, sed 's/(.*//'is going to match to the end of the string. 

f='(something) and more stuff';sed 's/(.*//' <<< "$f"

Results in of course, empty string. 

f='(something)-and more stuff';sed 's/([^)]*)//' <<< "$f"

Results in: -and more stuff

Lizardx
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