无法向表中插入记录

发布于 2024-07-07 04:27:36 字数 905 浏览 8 评论 0原文

我编写了一个程序，其中包括从数据库写入和读取。当我运行应用程序并尝试执行写入时，我调用以下方法：

public static void AddMessage(string callID, string content)
    {
        string select =
            "INSERT INTO Sporocilo (oznaka_klica, smer, vsebina, prebrano, cas_zapisa) VALUES (@callId, 0, @content, 0, @insertTime)";
        SqlCommand cmd = new SqlCommand(select, conn);
        cmd.Parameters.AddWithValue("callId", callID.ToString());
        cmd.Parameters.AddWithValue("content", content);
        cmd.Parameters.AddWithValue("insertTime", "10.10.2008");
        try
        {
            conn.Open();
            cmd.ExecuteScalar();
        }
        catch(Exception ex)
        {
            string sDummy = ex.ToString();
        }
        finally
        {
            conn.Close();
        }
    }

方法调用后，我从表中读取所有记录并将它们显示在表单中。可以看到刷新之前插入的记录，但是当我退出应用程序并查看表格时，我看不到该记录。

有谁知道什么会导致这种行为？

原文

I wrote a program which includes writing and reading from database. When I run the app and try to perform writing I call the following method:

public static void AddMessage(string callID, string content)
    {
        string select =
            "INSERT INTO Sporocilo (oznaka_klica, smer, vsebina, prebrano, cas_zapisa) VALUES (@callId, 0, @content, 0, @insertTime)";
        SqlCommand cmd = new SqlCommand(select, conn);
        cmd.Parameters.AddWithValue("callId", callID.ToString());
        cmd.Parameters.AddWithValue("content", content);
        cmd.Parameters.AddWithValue("insertTime", "10.10.2008");
        try
        {
            conn.Open();
            cmd.ExecuteScalar();
        }
        catch(Exception ex)
        {
            string sDummy = ex.ToString();
        }
        finally
        {
            conn.Close();
        }
    }

After the method call I read all the records from the table and display them in the form. The record inserted before refresh could be seen but then when I exit the app and look at the table I don't see the record.

Does anyone know what could cause such behavior?

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牵强ㄟ 2024-07-14 04:27:36

您在此之后执行提交吗？它可能正在运行您的语句，但随后不提交更改并执行隐式回滚。

我认为异常处理看起来很狡猾。除非你能以某种方式真正处理它，否则抓住它是没有意义的。框架的顶层是捕获和报告意外异常的地方。

回复收藏 0 原文

花开浅夏 2024-07-14 04:27:36

您是否尝试过将 ExecuteScalar 方法的返回值设置为 int，然后根据表检查值？

ExecuteScalar -

执行查询，并返回查询返回的结果集中第一行的第一列。额外的列或行将被忽略

public static int AddMessage(string callID, string content)
    {
        Int32 newProdID = 0
        string select =
            "INSERT INTO Sporocilo (oznaka_klica, smer, vsebina, prebrano, cas_zapisa) VALUES (@callId, 0, @content, 0, @insertTime); SELECT CAST(scope_identity() AS int);";
        SqlCommand cmd = new SqlCommand(select, conn);
        cmd.Parameters.AddWithValue("callId", callID.ToString());
        cmd.Parameters.AddWithValue("content", content);
        cmd.Parameters.AddWithValue("insertTime", "10.10.2008");
        try
        {
            conn.Open();
            newProdID = (Int32)cmd.ExecuteScalar();
        }
        catch(Exception ex)
        {
            string sDummy = ex.ToString();
        }
        finally
        {
            conn.Close();
        }
        return (int)newProdID
    }

Have you tried setting the return value of the ExecuteScalar method to an int and then checking value against the table?

ExecuteScalar -

Executes the query, and returns the first column of the first row in the result set returned by the query. Additional columns or rows are ignored

public static int AddMessage(string callID, string content)
    {
        Int32 newProdID = 0
        string select =
            "INSERT INTO Sporocilo (oznaka_klica, smer, vsebina, prebrano, cas_zapisa) VALUES (@callId, 0, @content, 0, @insertTime); SELECT CAST(scope_identity() AS int);";
        SqlCommand cmd = new SqlCommand(select, conn);
        cmd.Parameters.AddWithValue("callId", callID.ToString());
        cmd.Parameters.AddWithValue("content", content);
        cmd.Parameters.AddWithValue("insertTime", "10.10.2008");
        try
        {
            conn.Open();
            newProdID = (Int32)cmd.ExecuteScalar();
        }
        catch(Exception ex)
        {
            string sDummy = ex.ToString();
        }
        finally
        {
            conn.Close();
        }
        return (int)newProdID
    }

回复收藏 0 原文

我不吻晚风 2024-07-14 04:27:36

我发现了问题。我修改了自动生成的连接字符串

connectionString="数据
源=.\SQLEXPRESS;AttachDbFilename=|DataDirectory|\URSZRDB.mdf;集成
安全性=True；用户实例=True"

”

connectionString="数据
源=.\SQLEXPRESS;AttachDbFilename=C:\Users\Niko\Documents\Visual
Studio 2008\Projects\URSZRWAPChat\URSZRWAPChat\URSZRDB.mdf;集成
安全性=True；用户实例=True"

现在它可以工作了。

这不是我第一次编写这种程序，到目前为止一切都进展顺利......