0

我有这个java代码：

ArrayList<MessageMap> ids = getNewMail(UserID, type, maxIdMessage);
Message[] message = new Message[ids.size()];
if(ids.size()>0){
   ResultSet rs;
   int j =0;
   for(MappaMessaggi i : ids){
      pstmt = conn.prepareStatement("SELECT * FROM Messaggi WHERE MessageID = ?");
      pstmt.setInt(1, i.getMessageID());
      rs = pstmt.executeQuery();
      rs.next();
      .
      .
      .
   }

哪里getNewMail是：

public synchronized ArrayList<MessageMap> getNewMail(int UserID,int type,int max){
   ArrayList<MessageMap> map = new ArrayList<MessageMap>();
   try {
      pstmt = conn.prepareStatement("SELECT * FROM MessageMap WHERE UserID = ? AND TipoID = ? AND MessageID > ?");
      .
      .//fill arraylist with resultSet 
      .
}

我知道只需一个查询就可以做到这一点，但我不知道如何......有人可以睁开我的眼睛吗？：） 谢谢！！！

编辑：我尝试：

SELECT i.*
FROM Messaggi AS i
INNER JOIN MessageMap AS p i.MessageID = p.MessageID
WHERE p.MessageID = 1 AND p.UserID = 1 AND p.TipoID = 2

但我检索到：

#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'i.MessageID = p.MessageID WHERE p.MessageID = 1 AND p.UserID = 1 AND p.TipoID = ' at line 3
4

2 回答 2

1

根据您的问题和查询，我找到的简单解决方案可能是这个查询（也许还有其他列与您的表相关）

SELECT A.*
FROM MessageMap A
    INNER JOIN Message B on A.MessageId = B.MessageId
WHERE
    A.UserId = ?
    AND A.TipoID = ?
    AND A.MessageId = ?

另外，请说明您是否需要这 3 个参数，否则您无法发送其中任何一个。

于 2012-06-28T15:08:05.337 回答
1

试试这个：

SELECT i.*, p.*
FROM Messaggi i
INNER JOIN MessageMap p ON i.MessageID = p.MessageID
WHERE i.MessageID = ? AND p.UserID = ? AND p.TipoID = ?
于 2012-06-28T15:09:08.000 回答