Question

我正在使用Xerials SQLite JDBC驱动程序，并希望做你能想象到的最简单的事情：

我创建了下表：

CREATE TABLE IF NOT EXISTS households (hostname_id_pk INTEGER PRIMARY KEY, hostname TEXT UNIQUE, vm TEXT);

现在，因为我想要一个参数化的插入语句，我使用PreparedStatment如下：

public static void main(String[] args) throws ClassNotFoundException {
    Class.forName("org.sqlite.JDBC");
    Connection con = null;
    PreparedStatement updateHouseholdStmt = null;

    try {
        con = DriverManager.getConnection(DB_IDENTIFIER);           

        String getHouseholdString = "SELECT hostname_id_pk FROM households WHERE hostname = ?";
        String updateHouseholdString = "INSERT INTO households (hostname, vm) values(?, 'VM1')";

        getHouseholdStmt = con.prepareStatement(getHouseholdString);
        getHouseholdStmt.setString(1, "test1");

        updateHouseholdStmt = con.prepareStatement(updateHouseholdString);
        getHouseholdStmt.setString(1, "test1");

        ResultSet hhRS = getHouseholdStmt.executeQuery();
        int hhId = -1;

        if(hhRS.next()){
            hhId = hhRS.getInt(1);
            System.out.println(hhId);
        } else {
            System.out.println(updateHouseholdStmt.executeUpdate());
        }

    } catch (SQLException e) {
        System.err.println(e.getMessage());
    } finally {
        try {
            if (con != null) {
                con.close();
            }
            if (getHouseholdStmt != null){
                getHouseholdStmt.close();
            }
            if (updateHouseholdStmt != null) {
                updateHouseholdStmt.close();
            }
        } catch (SQLException e) {
            System.err.println(e);
        }
    }

}

在这种情况下，我在DB中唯一可以找到的是(1, , VM1)的新条目，意味着创建了条目，但由于某种原因缺少字符串参数。

当我更换“？”时在带有示例值的updateHousholdStatement()中删除.setString(...)方法行，一切正常。

我错过了什么？！我今天早上已经检查了一千次。

提前谢谢。

Answer 1

请更正以下内容：

updateHouseholdStmt = con.prepareStatement(updateHouseholdString);
getHouseholdStmt.setString(1, "test1");

到

updateHouseholdStmt = con.prepareStatement(updateHouseholdString);
updateHouseholdStmt.setString(1, "test1");

Answer 2

更新时，您正在设置并执行错误的变量：

getHouseholdStmt = con.prepareStatement(getHouseholdString);
getHouseholdStmt.setString(1, "test1");

updateHouseholdStmt = con.prepareStatement(updateHouseholdString);
getHouseholdStmt.setString(1, "test1"); //updating the select query
ResultSet hhRS = getHouseholdStmt.executeQuery(); //you're executing the select query again!

应该是：

getHouseholdStmt = con.prepareStatement(getHouseholdString);
getHouseholdStmt.setString(1, "test1");

updateHouseholdStmt = con.prepareStatement(updateHouseholdString);
updateHouseholdStmt .setString(1, "test1"); //update stmt 
ResultSet hhRS = updateHouseholdStmt .executeQuery(); //update stmt

Answer 3

首先，getHouseholdStmt语句未声明为PreparedStatement。其次，你需要executeUpdate（）你的PreparedStatement for updateHouseholdStmt（updateHouseholdStatement.executeUpdate（）;）

JDBC PreparedStatement.setString（）与SQLite无法正常工作？

3 个答案: