我有以下代码:
try
{
//Create connection
SQLiteConnection conn = DBConnection.OpenDB();
//Verify user input, normally you give dbType a size, but Text is an exception
var uNavnParam = new SQLiteParameter("@uNavnParam", SqlDbType.Text) { Value = uNavn };
var bNavnParam = new SQLiteParameter("@bNavnParam", SqlDbType.Text) { Value = bNavn };
var passwdParam = new SQLiteParameter("@passwdParam", SqlDbType.Text) {Value = passwd};
var pc_idParam = new SQLiteParameter("@pc_idParam", SqlDbType.TinyInt) { Value = pc_id };
var noterParam = new SQLiteParameter("@noterParam", SqlDbType.Text) { Value = noter };
var licens_idParam = new SQLiteParameter("@licens_idParam", SqlDbType.TinyInt) { Value = licens_id };
var insertSQL = new SQLiteCommand("INSERT INTO Brugere (navn, brugernavn, password, pc_id, noter, licens_id)" +
"VALUES ('@uNameParam', '@bNavnParam', '@passwdParam', '@pc_idParam', '@noterParam', '@licens_idParam')", conn);
insertSQL.Parameters.Add(uNavnParam); //replace paramenter with verified userinput
insertSQL.Parameters.Add(bNavnParam);
insertSQL.Parameters.Add(passwdParam);
insertSQL.Parameters.Add(pc_idParam);
insertSQL.Parameters.Add(noterParam);
insertSQL.Parameters.Add(licens_idParam);
insertSQL.ExecuteNonQuery(); //Execute query
//Close connection
DBConnection.CloseDB(conn);
//Let the user know that it was changed succesfully
this.Text = "Succes! Changed!";
}
catch(SQLiteException e)
{
//Catch error
MessageBox.Show(e.ToString(), "ALARM");
}它可以完美地执行,但当我查看我的"brugere“表时,它插入了以下值:'@uNameParam','@bNavnParam','@passwdParam','@pc_idParam','@noterParam','@licens_idParam‘。而不是替换它们。
我已经尝试创建断点,并检查了参数,它们确实具有正确的赋值。所以这也不是问题所在。
我现在已经修补了很多次了,没有运气,有人能帮上忙吗?
哦,作为参考,下面是来自DBConnection类的OpenDB方法:
public static SQLiteConnection OpenDB()
{
try
{
//Gets connectionstring from app.config
const string myConnectString = "data source=data;";
var conn = new SQLiteConnection(myConnectString);
conn.Open();
return conn;
}
catch (SQLiteException e)
{
MessageBox.Show(e.ToString(), "ALARM");
return null;
}
}发布于 2009-04-15 12:00:30
感谢rwwilden和Jorge Villuendas,答案是:
var insertSQL = new SQLiteCommand("INSERT INTO Brugere (navn, brugernavn, password, pc_id, noter, licens_id)" +
" VALUES (@uNavnParam, @bNavnParam, @passwdParam, @pc_idParam, @noterParam, @licens_idParam)", conn);
insertSQL.Parameters.AddWithValue("@uNavnParam", uNavn);
insertSQL.Parameters.AddWithValue("@bNavnParam", bNavn);
insertSQL.Parameters.AddWithValue("@passwdParam", passwd);
insertSQL.Parameters.AddWithValue("@pc_idParam", pc_id);
insertSQL.Parameters.AddWithValue("@noterParam", noter);
insertSQL.Parameters.AddWithValue("@licens_idParam", licens_id);
insertSQL.ExecuteNonQuery(); //Execute queryhttps://stackoverflow.com/questions/751172
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