Dictionary LazyMap的模型
如何转换/反序列化这些模型Dictionary LazyMap的模型,dictionary,groovy,deserialization,Dictionary,Groovy,Deserialization,如何转换/反序列化这些模型 public class AccessCredentials { String userName = '' String password = '' LoginOptions loginOptions = new LoginOptions() } public class LoginOptions { String partnerId = '' String applicationId = '' } 进入LazyMap,如:
public class AccessCredentials {
String userName = ''
String password = ''
LoginOptions loginOptions = new LoginOptions()
}
public class LoginOptions {
String partnerId = ''
String applicationId = ''
}
进入LazyMap,如:
[
userName : userName,
password : password,
loginOptions : [
partnerId : partnerId,
applicationId : applicationId
]
]
你可以用。例如
如果你想快速破解的话,你可以试试这样
def objectMapper(o) {
o.class.declaredFields.findAll { !it.synthetic }.collectEntries {
switch(o."$it.name".class.name) {
case ~/^java\..*/:
case ~/^javax\..*/:
case ~/^com\.sun\..*/:
case ~/^sun\..*/:
return [(it.name):o."$it.name"]
default:
return [(it.name):objectMapper(o."$it.name")]
}
}
}
objectMapper(new AccessCredentials())
尽管快速破解的问题在于,你忘记了它们在稍后破解时很快,只剩下一个破解;-) 你试了什么?我看不出这种简单的转换有什么问题很好!没有外部库有什么简单的方法吗?@Opal我从不介意。在许多面向web的项目中,jackson很可能已经是一个过渡的dep了。
def objectMapper(o) {
o.class.declaredFields.findAll { !it.synthetic }.collectEntries {
switch(o."$it.name".class.name) {
case ~/^java\..*/:
case ~/^javax\..*/:
case ~/^com\.sun\..*/:
case ~/^sun\..*/:
return [(it.name):o."$it.name"]
default:
return [(it.name):objectMapper(o."$it.name")]
}
}
}
objectMapper(new AccessCredentials())