如何为 mysql 5.5.44 创建触发器？答案

【问题标题】：How to create a trigger for mysql 5.5.44?如何为 mysql 5.5.44 创建触发器？
【发布时间】：2016-10-20 03:12:11
【问题描述】：

s = "CREATE TABLE " + tableName +"\n" +
    "(\n" +
    "    " + tablelower + "_currentid INT PRIMARY KEY AUTO_INCREMENT,\n" +
    "    " + tablelower + "_id VARCHAR(8) NOT NULL,\n" +
    "    " + tablelower + "_name VARCHAR(45) NOT NULL,\n" +
    "    " + tablelower + "_type VARCHAR(45) NOT NULL,\n" +
    "    " + tablelower + "_topic  VARCHAR(255) NOT NULL,\n" +
    "    " + tablelower + "_pin VARCHAR(6) NOT NULL,\n" +
    "    " + tablelower + "_device VARCHAR(100) NOT NULL,\n" +
    "    " + tablelower + "_device_id INT NOT NULL,\n" +
    "    FOREIGN KEY(" + tablelower + "_device_id) REFERENCES Devices(device_currentid)\n" +
    ");\n" +
    "\n" +
    " delimiter | \n" +
    " CREATE TRIGGER " + tablelower + "_trigger BEFORE INSERT ON " + tableName + 
    " FOR EACH ROW\n" +
    " BEGIN\n" +
    "   SET new." + tablelower + "_id = CONCAT('" + topic + "',LPAD((SELECT AUTO_INCREMENT FROM information_schema.TABLES WHERE TABLE_SCHEMA = DATABASE() AND TABLE_NAME = '" + tableName + "'),4,'0'));\n" +
    "   SET new." + tablelower + "_topic = CONCAT((SELECT device_topic FROM Devices WHERE device_name LIKE new." + tablelower + "_device),'/',(new." + tablelower + "_id));\n" +
    " END;\n" +
    " | \n" +
    " delimiter ;";

    mysqlconn.createStatement().execute(s);

上面是代码，它将使用给定的名称为 tableName 和 tablelower 创建一个表和一个触发器，它们是字符串变量。这是我写的第一个版本，我收到以下错误：

You have an error in your SQL syntax; check the manual that corresponds 
to your MySQL server version for the right syntax to use near 'delimiter | 
CREATE TRIGGER tablename_trigger BEFORE INSERT ON tablename FO' at line 14

在 google 帮助后，我找到了这个线程 Error while creating trigger through JDBC on mysql5.5 和文档 http://dev.mysql.com/doc/refman/5.5/en/trigger-syntax.html，我已经将我的代码更改为 StringBuilder：

            tableCreation.append("CREATE TABLE " + tableName);
            tableCreation.append("(");
            tableCreation.append(tablelower + "_currentid INT PRIMARY KEY AUTO_INCREMENT,");
            tableCreation.append(tablelower + "_id VARCHAR(10) NOT NULL,");
            tableCreation.append(tablelower + "_name VARCHAR(45) NOT NULL,");
            tableCreation.append(tablelower + "_type VARCHAR(45) NOT NULL,");
            tableCreation.append(tablelower + "_topic  VARCHAR(255) NOT NULL,");
            tableCreation.append(tablelower + "_pin VARCHAR(6) NOT NULL,");
            tableCreation.append(tablelower + "_device VARCHAR(100) NOT NULL,");
            tableCreation.append(tablelower + "_device_id INT NOT NULL,");
            tableCreation.append("FOREIGN KEY(" + tablelower + "_device_id) REFERENCES Devices(device_currentid)");
            tableCreation.append(" ); ");

            tableCreation.append("DELIMITER // ");
            tableCreation.append(" CREATE");
            tableCreation.append(" TRIGGER " + tablelower + "id_trigger ");
            tableCreation.append(" BEFORE INSERT");
            tableCreation.append(" ON " + tableName + " FOR EACH ROW");
            tableCreation.append(" BEGIN");
            tableCreation.append(" SET new." + tablelower + "_id = CONCAT('" + topic + "',LPAD((SELECT AUTO_INCREMENT FROM information_schema.TABLES WHERE TABLE_SCHEMA = DATABASE() AND TABLE_NAME = '" + tableName + "'),4,'0'));");
            tableCreation.append(" SET new." + tablelower + "_topic = CONCAT((SELECT device_topic FROM Devices WHERE device_name LIKE new." + tablelower + "_device),'/',(new." + tablelower + "_id));");
            tableCreation.append(" END;//");
            tableCreation.append("DELIMITER ; ");

            mysqlconn.createStatement().execute(tableCreation.toString());

但在这些更改之后，我仍然收到此错误：

您的 SQL 语法有错误；检查手册对应于您的 MySQL 服务器版本，以便使用正确的语法 'DELIMITER // 在插入之前创建触发器 reedswitchid_trigger 在第 1 行打开 ReedSwitches。

我正在通过 java 执行此操作，并且 MySQL 服务器位于 Raspberry pi 2 上。如需更多信息评论和注意我是 SQL 的初学者。谢谢

编辑：

您的 SQL 语法有错误；检查与>您的 MySQL 服务器版本相对应的手册，以在第 1 行的“SET >new.lightsensor_topic = CONCAT((SELECT device_topic FROM Devices WHERE >devic”附近使用正确的语法。退出

             tabl = "CREATE TABLE " + tableName  +
                "(" + " " + tablelower + "_currentid INT PRIMARY KEY AUTO_INCREMENT, " +
                tablelower + "_id VARCHAR(8) NOT NULL, " +
                tablelower + "_name VARCHAR(45) NOT NULL, " +
                tablelower + "_type VARCHAR(45) NOT NULL, " +
                tablelower + "_topic  VARCHAR(255) NOT NULL, " +
                tablelower + "_pin  VARCHAR(6) NOT NULL, " +
                tablelower + "_device VARCHAR(100) NOT NULL, " +
                tablelower + "_device_id INT NOT NULL, " +
                "FOREIGN KEY(" + tablelower + "_device_id) REFERENCES Devices(device_currentid)" +
                ")";
             trigg=
                " CREATE TRIGGER " + tablelower + "_trigger BEFORE INSERT ON " + tableName + 
                " FOR EACH ROW" +
                "   SET new." + tablelower + "_id = CONCAT('" + topic + "',LPAD((SELECT AUTO_INCREMENT FROM information_schema.TABLES WHERE TABLE_SCHEMA = DATABASE() AND TABLE_NAME = '" + tableName + "'),4,'0'))" +
                "   SET new." + tablelower + "_topic = CONCAT((SELECT device_topic FROM Devices WHERE device_name LIKE new." + tablelower + "_device),'/',(new." + tablelower + "_id))" +
                " END";

        mysqlconn = data.getConnection();
        mysqlconn.createStatement().execute(tabl);
        mysqlconn.createStatement().execute(trigg);

【问题讨论】：

标签： java mysql syntax-error mysql-5.5

【解决方案1】：

JDBC Stateent#executeXXX 方法旨在执行单个 SQL 语句，而不是一次执行多个语句。
即使某些数据库/驱动程序实现允许这样做，这也是一种非标准方式，并且不可移植，因此请不要使用它。

将你的代码分成两个语句：

String s1 = "CREATE TABLE ( ....... )";
String s2 = "CREATE TRIGGER .......END";
 mysqlconn.createStatement().execute( s1 );
 mysqlconn.createStatement().execute( s2 );

两个重要说明：

不要在 JDBC 中的 SQL 命令末尾使用分号 ;
不要使用DELIMITER x命令——不是SQL命令，而是mysql-client命令，数据库无法识别

【讨论】：