EJBQL condition query for help in EJB, LIST array

flypigzj

public List findExamScoreinterface(List userid){


      String connselect="select o from TCJExamScore o where o.userID IN(:userID)";
      Query query=em.createQuery(connselect);
      query.setParameter("userID",userid);
       return query.getResultList();
}
What's wrong in my writing, why can't I make it?

syde8868

/**
* DB 検索
*/
public void findUsers() {
    Query query;
    String namePattern = null;
    //Into the force されたユーザネーム means
    if (!user.getUserName().equals("")) {
        namePattern ='%' + user.getUserName() +'%';
    }
    //Look at it.
    if (user.getUserCode() == ""&&user.getUserName() == "") {query = em.createQuery("from User");
    }else if(user.getUserCode() != ""&&user.getUserName() == "") {
query = em.createQuery("from User where user_code = :code");
query.setParameter("code", user.getUserCode());
To
log.info(">>>ユーザコード検索");
}
else if (user.getUserCode() == ""&&user.getUserName() != "") {
To
query = em.createQuery("from User where user_name like :userName");
query.setParameter("userName", namePattern);
To
log.info(">>>ユーザ名検索");
To
}
else {

query = em.createQuery("from User where user_code = :code and user_name like :userName");
To
query.setParameter("code", user.getUserCode());
query.setParameter("userName", namePattern);
To
log.info(">>>ユーザコード・ユーザ名検索");

}
To
userList = query.getResultList();
log.info(">>>show userlist-#0", userList);
To
if (userList.isEmpty()) {
log.info(">>>ユーザリストが空");
return;
}
for(User u: userList) {
log.info(">>>show userList-#0 #1", u.getUserCode(), u.getUserName(), u.getPassword());
}
To
}

syde8868

Ha, every time it is sorted out, it will be sent out accidentally. Many of the previous ones are for processing the input query conditions, so there is no practical significance. You can refer to it yourself. You can delete log.info directly. , I used the JBOSS-SEAM framework to do it, ha ha

syde8868

query.setParameter("userID",userid);
In this sentence, userID corresponds to an attribute of an entity BEAN,
In fact, the corresponding should be a field in the database, it looks like we set
The database field type is not a list, right? There is no error when compiling this.
Errors can occur when actually running.

syde8868

I wrote the full search step in my program to death
   ArrayList<String> userCode = new ArrayList<String>();
   userCode.add("H0001");
   userCode.add("H0002");
   query = em.createQuery("from User where user_code IN(:userCode)");
   query.setParameter("userCode", userCode);
   userList = query.getResultList();
It's easy to use. I don’t know where the original owner’s error is. You can pass the userid to and
Print the result after execution to see if it is an error elsewhere.

flypigzj

ArrayList<String> userCode = new ArrayList<String>();
       userCode.add("wanglijuan");
      userCode.add("H0002");
      Query query;
      try{
   query = em.createQuery("select o from Subject o where o.subjectid IN(:subjectid)");
   query.setParameter("subjectid", userCode);
      }
      catch(Exception e){
               throw new EJBException(e);
      }
   return query.getResultList()

According to your method, I can't make it to death if I write it directly, and I make a mistake.