Question

private final Map<Integer, Lobby> lobbiesByID;

private boolean isPlayerInLobby(Player player) {
    return lobbiesByID.values().stream().anyMatch(l -> l.hasPlayer(player));
}

public IRemotePublisherForListener connect(Player player, int lobbyID) {
    if(!isPlayerInLobby(player)) {
        //.. insert null-check unless you trust the id
        lobbiesByID.get(lobbyID).addPlayer(player);
    }
    //.. else, log an error, or throw
    return publisher;
}

Answer 1

正确形成的SQL查询只有一个where子句（不包括CTE和子查询）。所以：

select *
from table1 a LEFT OUTER JOIN 
     (select * 
      from table99
      where col = 1
     ) b
     ON a.col1 = b.col1 AND
        a.col2 = b.col2 LEFT OUTER JOIN
     (select *
      from table99
      where col = 2
     ) c
     ON a.col1 = c.col1 AND
        a.col2 = c.col2
WHERE SIGN(a.col3) = 1;

然而，这似乎太复杂了。怎么样？

select *
from table1 a LEFT OUTER JOIN 
     table99 b
     ON a.col1 = b.col1 AND
        a.col2 = b.col2 AND
        b.col = 1 LEFT OUTER JOIN
     table99 c
     ON a.col1 = c.col1 AND
        a.col2 = c.col2 AND
        c.col = 2 
WHERE a.col3 > 0;

Answer 2

我认为你不需要在这句话ON (a.col1 = b.col1)中使用括号，我同意戈登的意见，你应该梳理你的观点和地点

HIVE错误：对于第一个LEFT OUTER JOIN的ON子句之后的WHERE子句，我得到1的EOF错误，对于hive中的以下代码

2 个答案: