mysql - 如何在准备好的语句中将此值添加为查询？

Question

Public void aMethod(String itemName, String theUsersName){  

preparedStatement = conn.prepareStatement("INSERT INTO items  (Owner_ID, Name, State)               VALUES         (?,?,?) ");
preparedStatement.setString(1, "(SELECT id FROM users WHERE username     = ' "+theUsersName+"')");
preparedStatement.setString(2, itemName); 
preparedStatement.setString(3, state);
preparedStatement.executeUpdate();

}

用户名表包含用户名和 id 在这里我给定了 theUsersName 并希望找到该用户名的 id，以设置 Item 表中新项目的 Owner_ID = theUsername 的 id。

Answer 1

如果它没有抛出异常，请尝试一下，（否则我将删除它）

String query =  "INSERT INTO items  (Owner_ID, Name, State) " +
                "SELECT ID, ? AS Name, ? AS State " +
                "FROM   users " +
                "WHERE  userName = ? " 
preparedStatement = conn.prepareStatement(query);
preparedStatement.setString(1, itemName);
preparedStatement.setString(2, state); 
preparedStatement.setString(3, theUsersName);
preparedStatement.executeUpdate();

Answer 2

通过将其插入到原始INSERT语句中：

INSERT INTO items (Owner_ID, Name, State)
VALUES (SELECT id FROM users WHERE username = ?, ?,?)

并将第一个参数替换为实际值：

preparedStatement.setString(1, theUsersName);

Answer 3

你可以这样做：使用INSERT INTO SELECT..

INSERT INTO items  (Owner_ID, Name, State)
SELECT id,?,? FROM users WHERE 
username = ?

或者像这样使用子查询：

INSERT INTO items (Owner_ID, Name, State)
VALUES ((SELECT id FROM users WHERE username = ? LIMIT 1),?,?)

在这里，您必须将限制设置为 1。

Answer 4

您首先必须从 users 表中获取 Id，然后才能传递它。看看：

preparedStatement = conn.prepareStatement("SELECT id FROM users WHERE username = ' "+theUsersName+"'");
ResultSet rs=preparedStatement.executeQuery();
 if(rs.next())
int id=rs.getInt(1);
preparedStatement.setString(1,id);