我正在尝试使用 Kotlin 实现干净的架构。该过程的流程将是：
usecase --> get rowresult from DB --> map rowresult to entity --> entity used by the usecase to check business rules
代码示例：
UserTable
------------------
id (varchar)
email (varchar)
password (varchar)
gender (varchar)
phone (varchar)
anotherAttribute1
anotherAttribute2
.
anotherAttributeN
class UserEntity {
val id: String,
val email: String,
val password: String,
//Business rules
fun isUserAllowedToLogin(): Boolean {
//validate password
}
}
interface UserDataStore {
fun getUser(email: String): User
}
class UserDataStoreImplementation {
fun getUser(email: String): User {
//query …如果这些特殊字符粘在一个单词上，如何在Python中创建RE以在这些特殊字符前面添加空格？,?!
这是输入字符串：
myString= 'I like him, but is he good? Maybe he is good , smart, and strong.'
所需的输出（如果特殊字符不粘在单词上，则不会被修改）：
modifiedString= 'I like him , but is he good ? Maybe he is good , smart , and strong.'
我已经尝试过这个回复：
modifiedString= re.sub('\w,' , ' ,' ,myString)
但它给出了错误的结果。它删除逗号之前的最后一个字符，这是结果示例：
modifiedString = 'I like hi , but is he good? Maybe he is goo , smar , and strong.'
有什么建议可以解决这个问题吗？