Question

需要一点帮助。遇到一些异常问题，我在使用这个库时非常新。在此先感谢：）

错误：

com.mysql.jdbc.exceptions.jdbc4.MySQLSyntaxErrorException：您的SQL语法中有错误;查看与您的MySQL服务器版本对应的手册，以便在＃＆quot;＆＃39; common_translations.name＆＃39;附近使用正确的语法。 as nm JOIN（＆＃39; common_translations.name_id＆＃39; as nid）

我的代码：

private DatabaseConnection db;
private final HashMap <String, String> statements;

public DatabaseReader(DatabaseConnection db) {
    statements = new HashMap<String, String>() {
        private static final long serialVersionUID = -1827340576955092045L;
    {
        put("odds","vfl::%");
        put("common_translations", "vhc::%");
        put("common_translations","vdr::%");
        put("common_translations", "vto::%");
        put("common_translations","vbl::%");
        put("common_translations", "vf::%");
        put("odds","vsm::%");
        put("odds", "rgs::%");
        put("odds", "srrgs::%");
    }};

    this.db = db;
}

public void read() {
    try {
        Connection connection = db.connect(db.getUrl_common_translation());
        PreparedStatement ps = (PreparedStatement) connection.prepareStatement("SELECT nm.id, nid.key, nm.name FROM ? as nm JOIN (? as nid)\r\n" + 
                "                 ON (nm.id = nid.id) where nid.key like ? and nm.typeId=8 and nm.sourceId=-1 and nm.languageCode='en'");           
        for(Entry <String,String> e : statements.entrySet()) {
            ps.setString(1, e.getKey() + ".name");
            ps.setString(2, e.getKey() + ".name_id");
            ps.setString(3, e.getValue());
            ResultSet rs = ps.executeQuery();
            while(rs.next()) {
                int id = rs.getInt("id");
                //String tag = rs.getString("tag");
                //String translation = rs.getString("translation");     
                System.out.println(id);
            }
        }



    } catch (SQLException e) {
        e.printStackTrace();
    }   
}

Answer 1

您正在使用错误的方式设置列名称：

ps.setString(1, e.getKey() + ".name");
ps.setString(2, e.getKey() + ".name_id");

将在引号之间输入：

FROM "something.name" as

这是一种错误的语法。

相反，您必须直接设置名称而不使用这样的预处理语句：

Connection connection = db.connect(db.getUrl_common_translation());
PreparedStatement ps = (PreparedStatement) connection.prepareStatement();
for (Entry<String, String> e : statements.entrySet()) {
    String query = "SELECT nm.id, nid.key, nm.name FROM " + e.getKey() + ".name" +" as nm "
            //----------------------------------------------^__________________^
            + "JOIN (" + e.getKey() + ".name_id" + " as nid) ON (nm.id = nid.id) "
            //-----------^_____________________^
            + "where nid.key like ? and nm.typeId=8 "
            + "and nm.sourceId=-1 and nm.languageCode='en'";
    ps.setString(1, e.getValue());
    ResultSet rs = ps.executeQuery(query);
    //------------------------------^^

但在此之前你必须检查名称不应该包含感染查询的东西（以避免语法错误和sql注入），所以你需要先制作一些控制器

Answer 2

从DB的角度来看，您将尝试执行查询：

SELECT nm.id, nid.key, nm.name 
  FROM :param1 as nm 
       JOIN (:param2 as nid) ON (nm.id = nid.id) 
 where nid.key like :param3 and nm.typeId=8 and nm.sourceId=-1 and nm.languageCode='en'

有一些参数。但是您不能将表或视图作为参数传递。这毫无意义。

要解决您的问题，您可以使用此代码段：

String sql = "SELECT nm.id, nid.key, nm.name FROM %s as nm JOIN (%s as nid)\r\n" + 
            "                 ON (nm.id = nid.id) where nid.key like ? and nm.typeId=8 and nm.sourceId=-1 and nm.languageCode='en'"
for(Entry <String,String> e : statements.entrySet()) {
    PreparedStatement ps = (PreparedStatement) connection.prepareStatement(
          String.format(sql, e.getKey() + ".name", e.getKey() + ".name_id")
    );           
    ps.setString(1, e.getValue());
        .... your code here ....
}

使用HashMap准备语句，例外

2 个答案: