我正在使用 Xerials SQLite JDBC 驱动程序,并且想要做您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,所以我使用PreparedStatment,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,我在数据库中唯一能找到的是一个带有 (1, , VM1)
的新条目,这意味着该条目已创建,但由于某种原因缺少字符串参数.
当我替换“?”时在 updateHousholdStatement()
中使用示例值并删除 .setString(...)
方法行,一切正常。
我到底错过了什么?!今天早上我已经检查了一千遍了。
提前谢谢。
最佳答案
请更正以下行:
updateHouseholdStmt = con.prepareStatement(updateHouseholdString);
getHouseholdStmt.setString(1, "test1");
至
updateHouseholdStmt = con.prepareStatement(updateHouseholdString);
updateHouseholdStmt.setString(1, "test1");
关于java - JDBCPreparedStatement.setString() 无法与 SQLite 一起正常工作?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/21989816/