【发布时间】:2017-08-09 12:44:47
【问题描述】:
需要一点帮助。遇到一些异常问题,我在使用这个库时非常陌生。在此先感谢:)
错误:
com.mysql.jdbc.exceptions.jdbc4.MySQLSyntaxErrorException: 你的 SQL 语法有错误;查看与您的 MySQL 服务器版本相对应的手册,了解在 ''common_translations.name' 作为 nm JOIN ('common_translations.name_id' as nid) 附近使用的正确语法
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();
}
}
【问题讨论】:
-
您不能使用
?占位符系统来替换表名或列名,它只适用于值。 -
哦,好的,谢谢。你有什么建议可以明智地做到这一点吗?
-
好吧,也许您可以使用
StringBuilder根据目标表构建您的查询字符串。
标签: java mysql database jdbc prepared-statement