Wildcard in bash script

Shell Programming and Scripting
1

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:

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

2

I didn't understand completely, can you try replacing those ? with . (period) if echo "$z" | grep -q "R1......R1";

1 Like
3

Grep uses regular expressions (regex), not pattern matching. It is a different syntax.

The equivalent regex of the pattern

R1??????R1

is

^R1......R1$
4 
if [[ "$z" == R1??????R1 ]]
then
  ...
fi
5

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

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

I tried that, but could not get it to work.
if echo "$z" | grep -q "^R1......R1$";

I may be doing something wrong here?

8

Perhaps I misread, if the file names have names that exactly match R1??????R2 , for example R1abcdefR2 then you would need ^R1......R1$ .

If the file names contain more characters than these 10 then you would need a pattern like *R1??????R2* which would correspond to the regular expression R1......R1 .

9

Thank you.!
if echo "$z" | grep -q *R1??????R2*

works fine, as well!

To clarify: this string is just a part of the filename. There is stuff before and after.

jahn

10

In that case you could use patterns like this:

case $z in
   (*R1??????R1*)  echo "$z" 
esac

or

if [[ "$z" == *R1??????R1* ]]
then
  ...
fi

or if you want to use grep with regex like in post #2:

if echo "$z" | grep -q "R1......R1";
11

could? must!

grep -q *R1??????R2*

takes additional args as files to search in, and ignores stdin.