计算 App Inventor 2 项目中使用的源代码行数或块数？

Question

有没有什么方法可以计算 App Inventor 2 项目中使用的源代码行 (SLOC) 或块的数量？

Answer 1

我编写了一个 Python 函数 aia_count_blocks(aia_filename)，通过计算 bky 文件中字符串 <block 的出现次数来计算给定 AIA 文件中的块数：

from __future__ import print_function
from __future__ import division

import glob
import ntpath
import os
import shutil
import zipfile

def unzip(source_filename, dest_dir):
    '''
    Source: http://stackoverflow.com/questions/12886768/how-to-unzip-file-in-python-on-all-oses
    '''
    with zipfile.ZipFile(source_filename) as zf:
        for member in zf.infolist():
            # Path traversal defense copied from
            # http://hg.python.org/cpython/file/tip/Lib/http/server.py#l789
            words = member.filename.split('/')
            path = dest_dir
            for word in words[:-1]:
                drive, word = os.path.splitdrive(word)
                head, word = os.path.split(word)
                if word in (os.curdir, os.pardir, ''): continue
                path = os.path.join(path, word)
            zf.extract(member, path)

def list_subdirectories(dir):
    '''
    Source: http://stackoverflow.com/questions/800197/get-all-of-the-immediate-subdirectories-in-python
    '''
    return filter(os.path.isdir, [os.path.join(dir,f) for f in os.listdir(dir)])


def path_leaf(path):
    '''
    Source: http://stackoverflow.com/questions/8384737/python-extract-file-name-from-path-no-matter-what-the-os-path-format
    '''
    head, tail = ntpath.split(path)
    return tail or ntpath.basename(head)

def bky_count_blocks(bky_filename):
    return open(bky_filename).read().count('<block ')

def aia_count_blocks(aia_filename):
    '''
    Count blocks in an AIA project 
    '''

    # unzip
    temp_folder = 'temp_aia'
    unzip(aia_filename, temp_folder)

    # Build path to .bky files, which contains all blocks for each screen of the AI2 project
    bky_files_path = os.path.join(temp_folder, 'src', 'appinventor', )
    for i in range(6): bky_files_path = list_subdirectories(bky_files_path)[0] # walk inside...
    #print(bky_files_path)    

    # Count
    total_blocks_count = 0
    bky_filenames = glob.glob(os.path.join(bky_files_path, '*.bky'))
    for bky_filename in bky_filenames :
        bky_block_count = bky_count_blocks(bky_filename)
        print('Screen {0} contains {1} blocks'.format(path_leaf(bky_filename), bky_block_count))
        total_blocks_count += bky_block_count
    print('The AIA project {0} contains {1} blocks spread across {2} screens.'.format(aia_filename, total_blocks_count, len(bky_filenames)))

    # Clean temp files
    shutil.rmtree(temp_folder)

def main():
    '''
    This is the main function
    '''
    aia_filename = 'test.aia'
    aia_count_blocks(aia_filename)

if __name__ == "__main__":
    main()
    #cProfile.run('main()') # if you want to do some profiling

输出：

Screen address.bky contains 42 blocks
Screen edit.bky contains 265 blocks
Screen list.bky contains 233 blocks
Screen logic.bky contains 954 blocks
Screen plan.bky contains 70 blocks
Screen table1.bky contains 206 blocks
Screen table2.bky contains 157 blocks
Screen Screen1.bky contains 16 blocks
The AIA project test.aia contains 1943 blocks spread across 8 screens.

可能会有很多改进，例如避免计算禁用块、按块类型计数等。

Answer 2

对于碰巧安装 Python 版本 3+ 而不是 2+ 的其他人

找到这个

return filter(os.path.isdir, [os.path.join(dir,f) for f in os.listdir(dir)])

用这个替换

return list(filter(os.path.isdir, [os.path.join(dir,f) for f in os.listdir(dir)]))