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Wildcard in bash script

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# 1  
Old 03-01-2014
Wildcard in bash script
Hi there,
I am pretty new to linux scripting so ..

I am writing a script to loop through all my directories of sequence files in order to do stuff with them (trimming, normalizing, stuff that one would do with sequence files).

Here I need to pick out files that match each other. The files containing
R1??????R1 should be matched with R1??????R2, R2??????R1 with R2??????R2, and so on. The ‘?’ being wildcards. There are always six of them (barcodes).

So, I try the following test to pick a file:


Code:
if echo "$z" | grep -q "R1??????R1";



which does not work, most likely because the ? are treated as the text part I am searching for and not a wildcard.


Suggestion to solve this one would be greatly appreciated.


best regards
Last edited by Scrutinizer; 03-01-2014 at 06:09 AM.. Reason: code tags
# 2  
Old 03-01-2014
I didn't understand completely, can you try replacing those ? with . (period) if echo "$z" | grep -q "R1......R1";
This User Gave Thanks to michaelrozar17 For This Post:
jahndavik
# 3  
Old 03-01-2014
Grep uses regular expressions (regex), not pattern matching. It is a different syntax.

The equivalent regex of the pattern
Code:
R1??????R1

is
Code:
^R1......R1$

Last edited by Scrutinizer; 03-01-2014 at 06:18 AM..
# 4  
Old 03-01-2014
Code:
if [[ "$z" == R1??????R1 ]]
then
  ...
fi

# 5  
Old 03-02-2014
I used:
if echo "$z" | grep -q "R1......R1"; then

and it seem to work the way I want - pick files that contain a string with R1+sixletters+R1
in the filename.

Thank you for your replies.

jahn
# 6  
Old 03-02-2014
If you are using grep with regex as specified in post #2, you will need the opening ^ and the closing $ as well, otherwise unintended matches might occur.
# 7  
Old 03-02-2014
I tried that, but could not get it to work.
if echo "$z" | grep -q "^R1......R1$";

I may be doing something wrong here?
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