Java JDBC PreparedStatement.setString（）不能与SQLite一起正常工作？

我使用的是Xerials SQLite JDBC驱动程序，我想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，由于我需要一个参数化的插入语句，我使用PreparedStation，如下所示：

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，我能在DB中找到的唯一东西是一个带有

（1，VM1）

的新条目，这意味着该条目已创建，但字符串参数由于某种原因丢失

当我用示例值替换

updateHousholdStatement（）

中的“？”并删除

.setString（…）

方法行时，一切正常

我到底错过了什么？！今天早上我已经查了一千遍了

---

Thx提前

请更正以下行：

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

到

---

请更正以下行：

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

到

---

更新时，您设置并执行了错误的变量：

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

---

更新时，您设置并执行了错误的变量：

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

---

首先，gethouseholdsmt语句没有声明为PreparedStatement。

---

第二，您需要执行updateUpdate（）您的updateHouseholdStmt的PreparedStatement（updateHouseholdStatement.executeUpdate（）；）

首先，getHouseholdStmt语句没有声明为PreparedStatement。

---

第二，您需要执行updateUpdate（）您为UpdateHouseholdsmt准备的语句（updateHouseholdStatement.executeUpdate（）；）

dddaaarn。。。好的，谢谢，伙计们：）这就是当你站到床的另一边时会发生的事情。。。好的，谢谢，伙计们：）这就是当你站在床的另一边时会发生的事情。