Php Insert语句在存储过程中不起作用

我最近在一个AWS EC2 Ubuntu实例上上传了我的MySQL数据库，现在我遇到了一个问题，我开发的存储过程在该实例上不能正常工作。它在Windows上的XAMPP实例上运行得非常好，所以我不完全确定问题出在哪里，因为我不熟悉Linux和AWS

以下是存储过程：

DELIMITER $$
CREATE PROCEDURE sp_PopulateAbstract
(conID INT, eID INT, abs TEXT)
BEGIN
    DECLARE existID INT;
    DECLARE abID INT;

    SELECT a.contributorID INTO existID
    FROM abstract a
    WHERE a.contributorID = conID and a.eventID = eID;

    IF conID = existID
    THEN
       --UPDATE statement works fine on Ubuntu instance
       UPDATE abstract
       SET abstract = abs
       WHERE eventID = eID and contributorID = conID; 
    ELSE
       --This statement doesn't work 
       INSERT INTO abstract
       (eventID, contributorID, abstract)
       VALUES (eID, conID, abs);
    END IF;

    SELECT a.abstractID INTO abID
    FROM abstract a
    WHERE a.eventID = eID and a.contributorID = conID;

    SELECT abID as abstractID;
END$$

DELIMITER $$
BEGIN
    DECLARE existID INT;
    DECLARE abID INT;

    SELECT a.contributorID INTO existID
    FROM eventcontributor a 
    WHERE a.contributorID = conID and a.eventID = eID;

    IF conID = existID
    THEN
        --cause of problem
        SELECT a.contributorID INTO existID
        FROM abstract a
        WHERE a.contributorID = conID and a.eventID = eID;

        IF conID = existID
        THEN 
            UPDATE abstract
            SET abstract = abs
            WHERE eventID = eID and contributorID = conID; 
        ELSE 
            INSERT INTO abstract
            (eventID, contributorID, abstract)
            VALUES (eID, conID, abs);
        END IF;

        SELECT a.abstractID INTO abID
        FROM abstract a
        WHERE a.eventID = eID and a.contributorID = conID;

        DELETE
        FROM abstractattachments
        WHERE abstractID = abID;

        SELECT abID as abstractID;
    ELSE
        SELECT -1 as abstractID;
    END IF;
END$$

这很奇怪，因为

UPDATE

语句在Ubuntu实例上运行良好，而

INSERT

语句则不行。

INSERT

语句不插入任何内容，最后一个

SELECT

语句返回

NULL

。如果执行了

UPDATE

语句，则最后一个

SELECT

语句确实会返回一些内容，并且数据会相应地更新

---

我已经将

INSERT

语句作为独立语句运行，它工作正常，但在存储过程中不起作用。XAMPP实例和Ubuntu实例之间有什么区别？我已经找到了导致INSERT语句不起作用的原因，这与我在原始问题中遗漏的存储过程部分有关，我认为这不是问题的原因。以下是完整的程序：

DELIMITER $$
CREATE PROCEDURE sp_PopulateAbstract
(conID INT, eID INT, abs TEXT)
BEGIN
    DECLARE existID INT;
    DECLARE abID INT;

    SELECT a.contributorID INTO existID
    FROM abstract a
    WHERE a.contributorID = conID and a.eventID = eID;

    IF conID = existID
    THEN
       --UPDATE statement works fine on Ubuntu instance
       UPDATE abstract
       SET abstract = abs
       WHERE eventID = eID and contributorID = conID; 
    ELSE
       --This statement doesn't work 
       INSERT INTO abstract
       (eventID, contributorID, abstract)
       VALUES (eID, conID, abs);
    END IF;

    SELECT a.abstractID INTO abID
    FROM abstract a
    WHERE a.eventID = eID and a.contributorID = conID;

    SELECT abID as abstractID;
END$$

DELIMITER $$
BEGIN
    DECLARE existID INT;
    DECLARE abID INT;

    SELECT a.contributorID INTO existID
    FROM eventcontributor a 
    WHERE a.contributorID = conID and a.eventID = eID;

    IF conID = existID
    THEN
        --cause of problem
        SELECT a.contributorID INTO existID
        FROM abstract a
        WHERE a.contributorID = conID and a.eventID = eID;

        IF conID = existID
        THEN 
            UPDATE abstract
            SET abstract = abs
            WHERE eventID = eID and contributorID = conID; 
        ELSE 
            INSERT INTO abstract
            (eventID, contributorID, abstract)
            VALUES (eID, conID, abs);
        END IF;

        SELECT a.abstractID INTO abID
        FROM abstract a
        WHERE a.eventID = eID and a.contributorID = conID;

        DELETE
        FROM abstractattachments
        WHERE abstractID = abID;

        SELECT abID as abstractID;
    ELSE
        SELECT -1 as abstractID;
    END IF;
END$$

在这个完整的过程中，问题实际上是在变量

existID

中重新选择一个值时引起的。出于某种原因，现在真正的问题是，如果MySQL的Ubuntu实例已经包含一个变量值，那么它不会覆盖该变量值。也许这与MySQL的版本不同于Ubuntu和我的Windows XAMPP上的版本有关

因此，变量

existID

保留其初始值，嵌套的

IF

语句测试为

true

，这意味着实际执行的是

UPDATE

语句，而不是

INSERT

语句

解决此问题的方法是

声明另一个变量，恰当地命名为
existID2
，然后使用此变量，而不是重复使用
existID

解决方案：

DELIMITER $$
BEGIN
    DECLARE existID INT;
    DECLARE existID2 INT;
    DECLARE abID INT;

    SELECT a.contributorID INTO existID
    FROM eventcontributor a 
    WHERE a.contributorID = conID and a.eventID = eID;

    IF conID = existID
    THEN
        --using a new, empty variable
        SELECT a.contributorID INTO existID2
        FROM abstract a
        WHERE a.contributorID = conID and a.eventID = eID;

        IF conID = existID2
        THEN 
            UPDATE abstract
            SET abstract = abs
            WHERE eventID = eID and contributorID = conID; 
        ELSE 
            INSERT INTO abstract
            (eventID, contributorID, abstract)
            VALUES (eID, conID, abs);
        END IF;

        SELECT a.abstractID INTO abID
        FROM abstract a
        WHERE a.eventID = eID and a.contributorID = conID;

        DELETE
        FROM abstractattachments
        WHERE abstractID = abID;

        SELECT abID as abstractID;
    ELSE
        SELECT -1 as abstractID;
    END IF;
END$$

为什么我的MySQL本地XAMPP实例可以重用变量，而我的AWS Ubuntu MySQL实例不能重用变量，这真的让我很困惑。这对我来说真的没有意义。
一次执行一条语句。如果执行“if”语句，那么为什么执行“else”语句。在时间，如果或否则将执行