[Linux 常見問題] bash script: get just filename from path

Source From Here
Question
How would I get just the filename without the extension and no path?

The following gives me no extension but I still have the path attached:
- test.sh

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1. #!/bin/sh  
2. source_file="/abc/123/test.txt"  
3. source_file_filename_no_ext=${source_file%.*}  
4. echo "$source_file_filename_no_ext"  

Execution result:

$ ./test.sh
/abc/123/test

How-To
Most UNIXes have a basename executable for just that purpose (strips directory information and suffixes from filenames.); another similar executable is dirname(strips the last part of a given filename, in effect outputting just the directory components of the pathname.).

For example, we can rewrite test.sh this way:

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1. #!/bin/sh  
2. source_file="/abc/123/test.txt"  
3. source_file_filename_no_ext=${source_file%.*}  
4.   
5. echo "Input: $source_file"  
6. source_file_filename_only=$(basename $source_file)  
7. source_file_dirname_only=$(dirname $source_file)  
8. source_file_filename_no_ext=${source_file_filename_only%.*}  
9. echo "    Filename=$source_file_filename_only"  
10. echo "    Dirname=$source_file_dirname_only"  
11. echo "    Filename without extension=$source_file_filename_no_ext"  

Execution result:

$ ./test.sh
Input: /abc/123/test.txt
Filename=test.txt
Dirname=/abc/123
Filename without extension=test

Supplement
* 鳥哥 - 認識與學習 BASH: 變數內容的刪除、取代與替換