所以我有这个代码:def validate_username(username): find_username = cursor.execute("SELECT username FROM user WHERE username = ?", (username, )) if find_username.fetchone() is not None: username_is_valid = False invalid_reason = 'Username already exists' elif len(username) > 255: username_is_valid = False invalid_reason = 'Invalid format' elif not match(r"[^@]+@[^@]+\.[^@]+", username): username_is_valid = False invalid_reason = 'Username must be an email address' else: username_is_valid = True invalid_reason = None return username_is_valid, invalid_reason因此,显然,它将返回其中一个:(False, 'Username already exists')(False, 'Invalid format')(False, 'Username must be an email address')(True, None)现在我想做的是将这两个参数传递到变量中,以便它们可以重用,打印等,但由于某种原因,我无法真正找到如何在线进行;我假设是因为我没有正确地表达我的问题。也许这里有人可以提供一些指导?基本上,这里的目标是使用用户名运行函数,然后如果用户名无效,请打印为什么它无效,如果它有效,则继续。
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慕田峪7331174
TA贡献1828条经验 获得超13个赞
只需将输出分配给两个变量,如下所示:
username_is_valid, invalid_reason = validate_username(username)
或者,如果将输出分配给单个变量,则可以使用切片访问其元素:
username_validation = validate_username(username)
username_is_valid = username_validation[0]
invalid_readon = username_validation[1]
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