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Help with Removing extra characters in Filename

 
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Top Forums UNIX for Dummies Questions & Answers Help with Removing extra characters in Filename
# 1  
Old 11-04-2010
Computer Help with Removing extra characters in Filename
Hi,

It's my first time here... anyways, I have a simple problem with these filenames. This is probably too easy for you guys:
Code:
ABC_20101.2A.2010_01
ABD_20103.2E.2010_04
ABE_20107.2R.2010_08


Expected Output:
Code:
ABC_20101
ABD_20103
ABE_20107

The only pattern available are the ff:
1) All filenames start with AB.
2) Characters before the first dot have fixed length of 9.

I tried using the ${i#...} function, but ended up ruining the entire filenames, and I don't want to do that again... Smilie On my attempt, I ended up renaming files as:
Code:
20101
20103
20107

which is terribly wrong...

can anyone please help?

(btw, I am using Tera Term 3.1.3 via Unix -- pretty old version huh Smilie)

Moderator's Comments:
Mod Comment Use code tags when posting code, data or logs to preserve formatting and enhance readability, thanks
Last edited by zaxxon; 11-04-2010 at 12:32 PM.. Reason: updated details
# 2  
Old 11-04-2010
Code:
$> while read LINE; do echo ${LINE%%.*}; done < infile
ABC_20101
ABD_20103
ABE_20107

If the input comes from a directory instead of a text file it could look like this:
Code:
$> for file in AB*; do echo ${file%%.*}; done
ABC_20101
ABD_20103
ABE_20107

# 3  
Old 11-04-2010
Hi zaxxon,

Wow, many thanks for the immediate answer.

I jsut have one more question though, isn't the 'echo' command just going to display the output? but what I need is to rename the files permanently with this expected output.
# 4  
Old 11-04-2010
As I noticed you seem not to process a text file with those names, I added a 2nd block below, working on directory entries ie. files. You could just convert the echo into a mv like:
Code:
$> for file in AB*; do mv $file ${file%%.*}; done

This User Gave Thanks to zaxxon For This Post:
Joule
# 5  
Old 11-04-2010
Code:
bash-3.2$ cat test4
ABC_20101.2A.2010_01
ABD_20103.2E.2010_04
ABE_20107.2R.2010_08
bash-3.2$ sed 'p;s/\..*$//g' test4 |xargs -n2 echo mv
mv ABC_20101.2A.2010_01 ABC_20101
mv ABD_20103.2E.2010_04 ABD_20103
mv ABE_20107.2R.2010_08 ABE_20107
bash-3.2$ sed 'p;s/\..*$//g' test4 |xargs -n2 echo mv|sh

hope this helps u!!
This User Gave Thanks to phoenix_nebula For This Post:
Joule
# 6  
Old 11-04-2010
Hi again, thanks for quick reply.

I got an error of files being identical. I could have posted the entire list of file names if I had known it will make sense to do so. Anyways, to illustrate, here are the existing filenames:
Code:
ABC_20101.2A.2010_01
ABC_20102.2C.2010_06
ABD_20103.2E.2010_04
ABD_20105.2J.2010_09
ABE_20107.2R.2010_08
ABE_20109.2B.2010_02

Expected Output:
Code:
ABC_20101
ABC_20102
ABD_20103
ABD_20105
ABE_20107
ABE_20107

when I tried to run the code, it gave me the error that pairs of AB* (two ABC, ABD, and ABE) are identical. Hence, it could not proceed with renaming these files.

any ideas please?
# 7  
Old 11-04-2010
Sorry, I can't see the problem or maybe I misunderstood:
Code:
$> ls -la
total 8
drwxr-xr-x 2 root root 4096  4. Nov 17:06 .
drwxr-x--- 3 isau isau 4096  4. Nov 17:06 ..
-rw-r--r-- 1 root root    0  4. Nov 17:06 ABC_20101.2A.2010_01
-rw-r--r-- 1 root root    0  4. Nov 17:06 ABC_20102.2C.2010_06
-rw-r--r-- 1 root root    0  4. Nov 17:06 ABD_20103.2E.2010_04
-rw-r--r-- 1 root root    0  4. Nov 17:06 ABD_20105.2J.2010_09
-rw-r--r-- 1 root root    0  4. Nov 17:06 ABE_20107.2R.2010_08
-rw-r--r-- 1 root root    0  4. Nov 17:06 ABE_20109.2B.2010_02
$> for file in AB*; do mv $file ${file%%.*}; done
$> ls -la
total 8
drwxr-xr-x 2 root root 4096  4. Nov 17:07 .
drwxr-x--- 3 isau isau 4096  4. Nov 17:06 ..
-rw-r--r-- 1 root root    0  4. Nov 17:06 ABC_20101
-rw-r--r-- 1 root root    0  4. Nov 17:06 ABC_20102
-rw-r--r-- 1 root root    0  4. Nov 17:06 ABD_20103
-rw-r--r-- 1 root root    0  4. Nov 17:06 ABD_20105
-rw-r--r-- 1 root root    0  4. Nov 17:06 ABE_20107
-rw-r--r-- 1 root root    0  4. Nov 17:06 ABE_20109

 
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