MySQL存储过程，处理多个游标和查询结果

本文介绍了MySQL存储过程，处理多个游标和查询结果的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

如何在同一例程中使用两个游标?如果我删除了第二个游标声明并获取了循环，一切都将正常工作.该例程用于在我的webapp中添加朋友.它使用当前用户的ID和我们要添加的朋友的电子邮件作为朋友，然后检查电子邮件是否具有相应的用户ID，如果不存在朋友关系，它将创建一个.除此以外的任何其他常规解决方案也都很好.

How can I use two cursors in the same routine? If I remove the second cursor declaration and fetch loop everthing works fine. The routine is used for adding a friend in my webapp. It takes the id of the current user and the email of the friend we want to add as a friend, then it checks if the email has a corresponding user id and if no friend relation exists it will create one. Any other routine solution than this one would be great as well.

DROP PROCEDURE IF EXISTS addNewFriend; DELIMITER // CREATE PROCEDURE addNewFriend(IN inUserId INT UNSIGNED, IN inFriendEmail VARCHAR(80)) BEGIN DECLARE tempFriendId INT UNSIGNED DEFAULT 0; DECLARE tempId INT UNSIGNED DEFAULT 0; DECLARE done INT DEFAULT 0; DECLARE cur CURSOR FOR SELECT id FROM users WHERE email = inFriendEmail; DECLARE CONTINUE HANDLER FOR NOT FOUND SET done = 1; OPEN cur; REPEAT FETCH cur INTO tempFriendId; UNTIL done = 1 END REPEAT; CLOSE cur; DECLARE cur CURSOR FOR SELECT user_id FROM users_friends WHERE user_id = tempFriendId OR friend_id = tempFriendId; DECLARE CONTINUE HANDLER FOR NOT FOUND SET done = 1; OPEN cur; REPEAT FETCH cur INTO tempId; UNTIL done = 1 END REPEAT; CLOSE cur; IF tempFriendId != 0 AND tempId != 0 THEN INSERT INTO users_friends (user_id, friend_id) VALUES(inUserId, tempFriendId); END IF; SELECT tempFriendId as friendId; END // DELIMITER ;

推荐答案

我终于编写了一个执行相同功能的不同函数:

I have finally written a different function that does the same thing:

DROP PROCEDURE IF EXISTS addNewFriend; DELIMITER // CREATE PROCEDURE addNewFriend(IN inUserId INT UNSIGNED, IN inFriendEmail VARCHAR(80)) BEGIN SET @tempFriendId = (SELECT id FROM users WHERE email = inFriendEmail); SET @tempUsersFriendsUserId = (SELECT user_id FROM users_friends WHERE user_id = inUserId AND friend_id = @tempFriendId); IF @tempFriendId IS NOT NULL AND @tempUsersFriendsUserId IS NULL THEN INSERT INTO users_friends (user_id, friend_id) VALUES(inUserId, @tempFriendId); END IF; SELECT @tempFriendId as friendId; END // DELIMITER ;

我希望这是一个更好的解决方案，无论如何它都能正常工作.感谢您告诉我不要在不需要时使用游标.

I hope this is a better solution, it works fine anyway. Thanks for telling me not to use cursors when not necessary.