不允许 HSQLDB 参数标记

Question

使用 hsqldb 和 preparedStatement 给我这个错误：

java.sql.SQLSyntaxErrorException: parameter marker not allowed
at org.hsqldb.jdbc.JDBCUtil.sqlException(Unknown Source)
at org.hsqldb.jdbc.JDBCUtil.sqlException(Unknown Source)
at org.hsqldb.jdbc.JDBCStatement.fetchResult(Unknown Source)
at org.hsqldb.jdbc.JDBCStatement.executeUpdate(Unknown Source)

我试过了：

String sql = "INSERT INTO Emergency (Id, status, typed, typeb, floor, 
locationX, locationY) Values(CAST(? AS INT), CAST(? AS INT), CAST(? AS 
VARCHAR(50)), CAST(? AS VARCHAR(50)), CAST(? AS INTEGER), CAST(? AS 
INTEGER), CAST(? AS INTEGER))";

并且：

String sql = "INSERT INTO Emergency (Id, status, typed, typeb, floor, 
locationX, locationY) Values(?,?,?,?,?,?,?)";

同样的错误。

编辑：更多代码

    try {
        con = DriverManager.getConnection(
                "jdbc:hsqldb:file:hsqldb; shutdown=true", "root", "");
        Statement stmt = con.createStatement();

        // Alle Kunden ausgeben
        //double help = ((number * pow(2.0, j)) - 1);
        String sql = "INSERT INTO Emergency (Id, status, typed, typeb, floor, locationX, locationY) Values(CAST(? AS INT), CAST(? AS INT), CAST(? AS VARCHAR(50)), CAST(? AS VARCHAR(50)), CAST(? AS INTEGER), CAST(? AS INTEGER), CAST(? AS INTEGER))";
        //Id INTEGER, status INTEGER, typeD VARCHAR(50) , typeB VARCHAR(50), floor INTEGER, locationX INTEGER, locationY INTEGER

        PreparedStatement pstmt = con.prepareStatement(sql);
        pstmt.setInt(1, Counter.emergencyID);
        pstmt.setInt(2, emergency.status);
        pstmt.setString(3, emergency.typeD);
        pstmt.setString(4, emergency.typeB);
        pstmt.setInt(5, emergency.floorID);
        pstmt.setInt(6, emergency.locationXY[0]);
        pstmt.setInt(7, emergency.locationXY[1]);
        Statement st = null;

        st = con.createStatement();    // statements

        int i = st.executeUpdate(sql);    // run the query

        Counter.emergencyID++;

        if (i == -1) {
            System.out.println("db error : " + sql);
        }

        st.close();
        System.out.println("Eintrag in DB");

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

        if (con != null) {
            try {
                con.close();
            } catch (SQLException e) {
                e.printStackTrace();
            }
        }
    }

我正在添加所有值然后执行它。

有人可以向我解释这里出了什么问题吗？没有找到任何有用的搜索。

Answer 1

你不需要转换你的类型，你只需要知道 PreparedStatement 是如何工作的。
你正在使用Statement和Prepared Statement，但你不需要Statement，你只需要执行Prepared Statement，所以改用这个:

    String sql = "INSERT INTO Emergency (Id, status, typed, typeb, floor, 
                  locationX, locationY) Values(?,?,?,?,?,?,?)";

    PreparedStatement pstmt = connection.prepareStatement(sql);

        pstmt.setInt(1, Counter.emergencyID);
        pstmt.setInt(2, emergency.status);
        pstmt.setString(3, emergency.typeD);
        pstmt.setString(4, emergency.typeB);
        pstmt.setInt(5, emergency.floorID);
        pstmt.setInt(6, emergency.locationXY[0]);
        pstmt.setInt(7, emergency.locationXY[1]);

        int i = pstmt.executeUpdate();//execute pstmt no need to st

不允许 HSQLDB 参数标记

HSQLDB parameter marker not allowed

java

jdbc

prepared-statement

hsqldb