自定义输出文件名

Question

with open('media_cdn/account/pdf01.pdf', 'wb') as out_file:
        output.write(out_file)

我该如何格式化它以使每个输出都有一个自定义名称？例如，我尝试过：

author_id=str(instance.author.id)

with open('media_cdn/account/{author_id}-pdf01.pdf', 'wb') as out_file:
    output.write(out_file)

但是，文件输出为"{author_id}-pdf01.pdf"

Answer 1

用途：

with open("media_cdn/account/{0}-pdf01.pdf".format(author_id), 'wb') as out_file:

代替

with open('media_cdn/account/{author_id}-pdf01.pdf', 'wb') as out_file:

'media_cdn/account/{author_id}-pdf01.pdf'是字符串。此字符串的{author_id}部分将由author_id变量本身取代。

Answer 2

难道你不能做到这一点

author_id=str(instance.author.id)

with open(f'media_cdn/account/{author_id}-pdf01.pdf', 'wb') as out_file:
    output.write(out_file)