Question

以下SQL语句正在创建一条错误消息： “消息：命令执行期间遇到致命错误。” “内部异常：必须定义参数'@LastUserID'。”

如果我直接使用LAST_INSERT_ID（）而不是LastUserID，则在执行这种操作时，它始终返回零（因此在第二次插入时失败）。

我没有看到我的语法与mySQL文档中的语法不同。

有人可以帮我吗？

string Query = @"INSERT INTO login (" +
                                    "LOGIN_EMAIL," +
                                    "LOGIN_PASSWORD," +
                                    "LOGIN_SALT," +
                                    "LOGIN_LAST_LOGIN_DATE," +
                                    //         "LOGIN_LAST_LOGIN_LOCATION," +
                                    "LOGIN_ACCOUNT_STATUS," +
                                    "LOGIN_LOGIN_ATTEMPTS," +
                                    "LOGIN_CREATED_DATE) " +
                                "VALUES (" +
                                    "@Parameter2," +
                                    "@Parameter3," +
                                    "@Parameter4," +
                                    "@Parameter5," +
                                    //                                        "@Parameter6," +
                                    "@Parameter6," +
                                    "@Parameter7," +
                                    "@Parameter8); " +
                                "SET @LastUserID = LAST_INSERT_ID(); " + 
                                "INSERT INTO user_role (" +
                                    "USER_ROLE_USER_ID," +
                                    "USER_ROLE_ROLE," +
                                    "USER_ROLE_STATUS," +
                                    "USER_ROLE_CREATED_DATE) " +
                                "SELECT " +
                                    "@LastUserID," +
                                    "@Parameter9," +
                                    "@Parameter10," +
                                    "@Parameter11 " +
                                "FROM dual WHERE NOT EXISTS (SELECT USER_ROLE_USER_ID FROM user_role " +
                                "WHERE USER_ROLE_USER_ID = @LastUserID AND USER_ROLE_ROLE = @Parameter9)";

                    MySqlCommand oCommand = new MySqlCommand(Query, oMySQLConnecion);

                oCommand.Transaction = tr;

Answer 1

简单修复：将“ $ LastUserID”替换为“ $'LastUserID'”。衣食住行与众不同。

Answer 2

创建一个过程，在该过程中，您首先执行插入操作，缓存最后插入的ID，然后执行另一个插入操作，如果最后一次插入操作无效，则使用bool打印输出参数。这样您就可以正确调试它。

通常，应避免使用字符串来生成sql命令，否则可能会遇到包含意外字符的参数或被injection击中的麻烦。

为什么设置SQL变量会创建错误？

2 个答案: