Java 如何确保HashMap的每次迭代都是事务性的
我有一个方法可以从hashmap中删除值,而不删除键 HashMap的结构如下所示,注意Kennel和Dog都是Java对象:Java 如何确保HashMap的每次迭代都是事务性的,java,spring,hashmap,crud,transactional,Java,Spring,Hashmap,Crud,Transactional,我有一个方法可以从hashmap中删除值,而不删除键 HashMap的结构如下所示,注意Kennel和Dog都是Java对象: Map<Kennel, List<Dog>> mapOfKennels; 重构您的代码,以便通过专用方法处理每个犬舍的移除。然后将@Transactional注释添加到此方法。例如在DogRepository中 @Transactional public class DogRepository{ public void deleteDo
Map<Kennel, List<Dog>> mapOfKennels;
重构您的代码,以便通过专用方法处理每个犬舍的移除。然后将@Transactional注释添加到此方法。例如在DogRepository中
@Transactional
public class DogRepository{
public void deleteDogs(List<Dog> dogs){
//delete all dogs
}
}
然后在你的客户端代码中
public void deleteDogsFromKennels(Map<Kennel, List<Dog>> mapOfKennelsAndDogs) {
//For each Kennel in the Map
for (Map.Entry<Kennel, List<Dog>> entry :mapOfKennelsAndDogs.entrySet()){
String key = entry.getKey().getId();
List<Dog> dogList = entry.getValue();
try{
dogRepository.deleteDogs(dogList);
}
catch(Exception ex){
//log failed to delete kennel with key
}
}
}
确保deleteDogs方法与deleteDogsFromKennels方法在单独的类中,否则事务将无法工作。尝试在方法级别添加@Transactional,即deleteDogsFromKennels,或者在类级别添加@Transactional,如果您不想像@ekemChitsiga在下面的回答中所说的那样重构代码。这应该行得通。
public void deleteDogsFromKennels(Map<Kennel, List<Dog>> mapOfKennelsAndDogs) {
//For each Kennel in the Map
for (Map.Entry<Kennel, List<Dog>> entry :mapOfKennelsAndDogs.entrySet()){
String key = entry.getKey().getId();
List<Dog> dogList = entry.getValue();
try{
dogRepository.deleteDogs(dogList);
}
catch(Exception ex){
//log failed to delete kennel with key
}
}
}