C# System.Data.SQLite 参数问题

System.Data.SQLite parameter issue

提问by CasperT

I have the following code:

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");
}

It executes perfectly, but when I view my "brugere" table, it has inserted the values: '@uNameParam', '@bNavnParam', '@passwdParam', '@pc_idParam', '@noterParam', '@licens_idParam' literally. Instead of replacing them.

它完美地执行，但是当我查看我的“brugere”表时，它插入了值：'@uNameParam'、'@bNavnParam'、'@passwdParam'、'@pc_idParam'、'@noterParam'、'@licens_idParam' . 而不是取代它们。

I have tried making a breakpoint and checked the parameters, they do have the correct assigned values. So that is not the issue either.

我尝试创建断点并检查参数，它们确实具有正确的分配值。所以这也不是问题。

I have been tinkering with this a lot now, with no luck, can anyone help?

我现在一直在修补这个，没有运气，有人可以帮忙吗？

Oh and for reference, here is the OpenDB method from the DBConnection class:

哦，作为参考，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;
    }
}

采纳答案by Ronald Wildenberg

You should remove the quotes around your parameter names in the INSERT statement.

您应该删除 INSERT 语句中参数名称周围的引号。

So instead of

所以代替

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

use

用

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

回答by CasperT

Thanks to rwwilden and Jorge Villuendas, the answer is:

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 query

回答by Stark

replace

代替

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

值（'@uNameParam'、'@bNavnParam'、'@passwdParam'、'@pc_idParam'、'@noterParam'、'@licens_idParam'）

with

和

VALUES (?, ?, ?, ?, ?, ?)

值 (?, ?, ?, ?, ?, ?)

回答by bro

When you use System.Data.SqlClientthen you provide parameter types from System.Data.SqlDbTypeenumeration.

当您使用System.Data.SqlClientthen 时，您可以从System.Data.SqlDbType枚举中提供参数类型。

Butif you use System.Data.SQLitethen you have to use **System.Data.DbType**enumeration.

但是如果你使用System.Data.SQLite那么你必须使用**System.Data.DbType**枚举。