我正在尝试在我的应用程序中创建私人聊天,但是当我尝试创建表时出现此错误:org.postgresql.util.PSQLException:错误:“7”位置或附近的语法错误:14代码: public static String checkIfChatExists(String code, String friend) throws SQLException { String i = LoginManager.checkCode(code); if(i.equals("code-not-exists")) { return "invalid-code"; } ResultSet rs = conn.prepareStatement("SELECT * FROM pwchats").executeQuery(); while (rs.next()) { if(rs.getString("user1").equals(i) && rs.getString("user2").equals(friend) || rs.getString("user1").equals(friend) && rs.getString("user2").equals(i)) { return "exists"; } } PreparedStatement pre = conn.prepareStatement("INSERT INTO pwchats(user1, user2) VALUES(?, ?)"); pre.setString(1, i); pre.setString(2, friend); pre.execute(); PreparedStatement getChatId = conn.prepareStatement("SELECT * FROM pwchats WHERE user1 = ? AND user2 = ?;"); getChatId.setString(1, i); getChatId.setString(2, friend); ResultSet rss = getChatId.executeQuery(); while (rss.next()) { PreparedStatement chat = conn.prepareStatement("CREATE TABLE " + rss.getInt("id") + "chat (username text NOT NULL, created_at timestamp with time zone NOT NULL DEFAULT now(), avatar text NOT NULL, message text NOT NULL, id serial NOT NULL);"); chat.execute(); //ERROR return "done"; } return "wat"; }
1 回答
犯罪嫌疑人X
TA贡献2080条经验 获得超4个赞
您看到的错误可能来自那里:
preparedStatement chat = conn.prepareStatement("CREATE TABLE " + rss.getInt("id") + "chat (...);");
通常在 SQL 中,表名不能以数字开头。如果您要用双引号将表名引起来,Postgres 仍然允许这样做,但是每次查询时都需要双引号表名,这可能很乏味。
一种解决方案是将数字放在表名称的末尾,例如:
preparedStatement chat = conn.prepareStatement("CREATE TABLE chat" + rss.getInt("id") + " (...);");
create table 7chat(x int);
ERROR: syntax error at or near "7"
LINE 1: create table 7chat(x int);
^
create table "7chat"(x int);
-- works
create table chat7(x int);
--works
添加回答
举报
0/150
提交
取消