c# - System.Data.SQLite 参数问题

Question

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");
}

它执行得很好，但是当我查看我的“brugere”表时，它插入了以下值:'@uNameParam'、'@bNavnParam'、'@passwdParam'、'@pc_idParam'、'@noterParam'、'@ licens_idParam'字面意思。而不是更换它们。

我已经尝试制作断点并检查参数，它们确实具有正确的分配值。所以这也不是问题。

我现在一直在修补这个问题，但没有运气，有人能帮忙吗？

哦，作为引用，这里是 DBConnection 类中的 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;
    }
}

Answer 1

您应该删除 INSERT 语句中参数名称周围的引号。

所以代替

VALUES ('@uNameParam', '@bNavnParam', '@passwdParam', '@pc_idParam',
        '@noterParam', '@licens_idParam')

使用

VALUES (@uNameParam, @bNavnParam, @passwdParam, @pc_idParam,
        @noterParam, @licens_idParam)