python - 无法打开资源 : OS Error

Question

OSError Traceback (most recent call last) in ()

21 draw = PIL.ImageDraw.Draw(img) 22 print(os.path.join(path, filename)) ---> 23 draw.font = PIL.ImageFont.truetype((os.path.join(path, filename))+ '.ttf', 44) 24 t2 = get_display(t1) 25 w, h = draw.textsize(t2)

()

OSError: cannot open resource

我在下面遇到这个错误 PIL 还是我犯了一些错误。它只显示第一个文件路径，然后出现此错误。

unicode_text = u"\u0627"
list_of_letters = list (unicode_text) 
folder = 1 
n=1 
i=0
for i in range(0,158):
    path = r"E:\Dummy\fonts"
    dirs = os.listdir( path )
    for files in dirs:
        char = u''.join(list_of_letters) 
        t1 = arabic_reshaper.reshape(char) 
        W,H= (100, 100)
        img= PIL.Image.new('RGBA', (W, H), (255, 255, 255),)
        draw = PIL.ImageDraw.Draw(img)   
        print(os.path.join(path, filename))
        draw.font = PIL.ImageFont.truetype((os.path.join(path, filename)), 44)
        t2 = get_display(t1) 
        w, h = draw.textsize(t2)
        draw.text(((W-w)/2,(H-h)/2),t2, fill="#000000")
        path = 'E:\Dummy\sam\\'+ str(folder)
        if not os.path.exists(path):
            os.makedirs(path)
        img.save(path + '\\' + char+'.png', "PNG")
        folder+=1
            #i+=1

Answer 1

从您的回答来看，程序失败的地方不是很清楚。它提示无法打开文件。这条线在我看来很可疑:

path = 'E:\Dummy\sam\\'+ str(folder)

我建议您永远不要手动将路径与反斜杠连接起来，而是始终使用 Python 标准库来为您执行此操作。例如，您不需要关心转义某些字符。

dir_path = os.path.join('E:', 'Dummy', 'sam', str(folder))
file_name = '{}.png'.format(char)
file_path = os.path.join(dir_path, file_name)

如果您不在循环内向文件夹添加 1，而是使用内置函数 enumerate，则可以对您的代码进行另一项改进:

for folder_index, file in enumerate(dirs, start=1):
    # Do your thing here
    # The variable folder_index is incremented automatically.