重命名文件并将扩展名保留在bash中

Question

I have to rename some files that I don't exactly know where they are and keep their extensions.

Ex: files system-2.1.3.war, system-2.1.3.ear, system-2.1.3.ejb to system.ear, system.war,system.ejb

So I wrote this.

find /DIR1 -name "*.ear" -o -name "*.war" -o -name "*.ejb" \ 
-exec bash -c 'export var1={}; cp $var1 NEW_NAME${var1: -4}' \;

The problem is: It works only for the last file in the "or list" of "find" command. So if the file is system-2.1.3.ejb, works, for system-2.1.3.war and system-2.1.3.ear don't.

If I change the find to

find /DIR1 -name "*.ejb" -o -name "*.war" -o -name "*.ear"

Notice that *.ear now is the last one, it will work for system-2.1.3.ear and not for the others and so on.

Please help me to fix this.

I know I can create a script to accomplish that but I want a "one line" code.

Answer 1

try this;

find /DIR1  \( -name "*.ear" -o -name "*.war"  -o -name "*.ejb" \) -exec bash -c 'export var1={}; cp $var1 NEW_NAME${var1: -4}' \;

or

find ./DIR1/  -regex  '.*\(.ear\|.war\|.ejb\)$' -exec bash -c 'export var1={}; cp $var1 NEW_NAME${var1: -4}' \;

Eg;

user@host $ ls -arlt DIR1/
total 76
-rw-rw-r--  1 user user     0 Oct 21 22:59 system-2.1.3.war
-rw-rw-r--  1 user user     0 Oct 21 22:59 system-2.1.3.ear
-rw-rw-r--  1 user user     0 Oct 21 22:59 system-2.1.3.ejb
drwxrwxr-x  2 user user  4096 Oct 21 22:59 .


user@host $ find . \( -name "*.ear" -o -name "*.war"  -o -name "*.ejb" \) -exec bash -c 'export var1={}; cp $var1 NEW_NAME${var1: -4}' \;

user@host $ ls -ralt
total 76
-rw-rw-r--  1 user user     0 Oct 21 22:59 system-2.1.3.war
-rw-rw-r--  1 user user     0 Oct 21 22:59 system-2.1.3.ear
-rw-rw-r--  1 user user     0 Oct 21 22:59 system-2.1.3.ejb
drwxrwxrwt 11 root   root   69632 Oct 21 23:10 ..
-rw-rw-r--  1 user user     0 Oct 21 23:10 NEW_NAME.war
-rw-rw-r--  1 user user     0 Oct 21 23:10 NEW_NAME.ear
-rw-rw-r--  1 user user     0 Oct 21 23:10 NEW_NAME.ejb
drwxrwxr-x  2 user user  4096 Oct 21 23:10 .

Answer 2

Rather than embedding the {} in the script, pass it as an argument:

find /DIR1 \( -name "*.ear" -o -name "*.war" -o -name "*.ejb" \) \ 
  -exec sh -c 'ext=${1##*.}; cp "$1" "NEW_NAME.$ext"' _ '{}' \;

Without the \\(...\\) grouping, -exec only applies to the primary it is implicitly "and"ed with, the previous one.

You can also limit the number of calls to the shell by looping over multiple arguments:

find /DIR1 \( ... \) -exec sh -c 'for f; do ext=${f##*.}; cp "$f" "NEW_NAME.$ext"; done' _ {} +

Answer 3

If you have rename utility then you can avoid forking BASH subprocess for each file and also make use of regex feature in find to avoid multiple -name options:

find /DIR1 -regextype awk -regex '.*\.([we]ar|ejb)$' \
    -exec rename 's/.*(\.[^.]+)$/system$1/' '{}' +

Answer 4

I'd use a while :

find . -name "*.war" -o -name "*.ejb" -o -name "*.ear" | while read file; do cp $file NEW_NAME${file: -4}; done

Keep in mind that both this and your example are copying the files in the current directory, so if you have more than one *.war , *.ejb or *.ear in your tree, only the last one(s) will be left in the target directory.