SELECT IDENTITY未获取最新创建的ID

Question

在另一个问题中，我问过如何配置ExecuteScalar并使用SCOPE_INDENTITY来获取我新创建的PK PersonID 。 将人输入到表中的第一组插入语句正在工作， PersonID在新创建的记录中位于＃39。 当我在AddNewCustomer方法上设置一个断点并按代码进行操作时， newPersonID设置为“ 1”并一直停留在该位置。 有谁知道为什么我为newPersonID新建的记录未显示“ 39或40”。

protected void AddNewCustomer(object sender, EventArgs e)
{

string nFirstName = ((TextBox)GridView1.FooterRow.FindControl("txtFirstName")).Text;
string nLastName = ((TextBox)GridView1.FooterRow.FindControl("txtLastName")).Text;
string nEmergency = ((TextBox)GridView1.FooterRow.FindControl("txtEmergency")).Text;
string nCell = ((TextBox)GridView1.FooterRow.FindControl("txtCell")).Text;
string nAge = ((TextBox)GridView1.FooterRow.FindControl("txtAge")).Text;
string nActivityCard = ((TextBox)GridView1.FooterRow.FindControl("txtActivityCard")).Text;
string nInitials = ((TextBox)GridView1.FooterRow.FindControl("txtInitials")).Text;
string nBoat = ((TextBox)GridView1.FooterRow.FindControl("txtBoat")).Text;
string nGroup = ((TextBox)GridView1.FooterRow.FindControl("txtGroup")).Text;


    SqlConnection con = new SqlConnection(strConnString);
    SqlCommand cmd = new SqlCommand("INSERT INTO Person(FirstName, LastName, Emergency#, Cell#, Age, ActivityCard, CraftType, Initials, Group#) " +
    "values(@FirstName, @LastName, @Emergency, @Cell, @Age, @ActivityCard, @Boat, @Initials, @Group); " +
    "SELECT SCOPE_IDENTITY();");

    cmd.Parameters.Add("@FirstName", SqlDbType.VarChar).Value = nFirstName;
    cmd.Parameters.Add("@LastName", SqlDbType.VarChar).Value = nLastName;
    cmd.Parameters.Add("@Emergency", SqlDbType.NChar).Value = nEmergency;
    cmd.Parameters.Add("@Cell", SqlDbType.NChar).Value = nCell;
    cmd.Parameters.Add("@Age", SqlDbType.NChar).Value = nAge;
    cmd.Parameters.Add("@ActivityCard", SqlDbType.NChar).Value = nActivityCard;
    cmd.Parameters.Add("@Initials", SqlDbType.NChar).Value = nInitials;
    cmd.Parameters.Add("@Boat", SqlDbType.VarChar).Value = nBoat;
    cmd.Parameters.Add("@Group", SqlDbType.VarChar).Value = nGroup;
    cmd.Parameters.AddWithValue("@Date", TextBox1.Text);
    cmd.Parameters.AddWithValue("@Time", ddlTripTime.SelectedItem.ToString());
    cmd.Parameters.AddWithValue("@Type", ddlTripType.SelectedItem.ToString());


    cmd.Connection = con;
    con.Open();
    int newPersonID = System.Convert.ToInt32(cmd.ExecuteScalar());
    con.Close();

    SqlCommand cmd1 = new SqlCommand();
    cmd1.CommandType = CommandType.Text;
    cmd1.CommandText ="insert into TripSchedule(TripType, PersonID, Time, Date) values (@Type, @newPerson, @Time, @Date);" +
    "SELECT Person.PersonID, Person.FirstName AS FirstName, Person.LastName AS LastName, Person.Emergency# AS Emergency#, Person.Cell# AS Cell#, Person.Age AS Age, " +
    "Person.ActivityCard AS ActivityCard, Person.CraftType AS CraftType, Person.Initials AS Initials, Person.Group# AS Group# " +
    "FROM Person INNER JOIN " +
    "TripSchedule ON Person.PersonID = TripSchedule.PersonID where TripSchedule.Date = @Date and " +
    "TripSchedule.Time = @Time and TripSchedule.TripType = @Type;";

    cmd1.Parameters.AddWithValue("@Date", TextBox1.Text);
    cmd1.Parameters.AddWithValue("@Time", ddlTripTime.SelectedItem);
    cmd1.Parameters.AddWithValue("@Type", ddlTripType.SelectedItem);
    cmd1.Parameters.AddWithValue("@newPerson", newPersonID);
    GridView1.DataSource = GetData(cmd);
    GridView1.DataBind();

}

编辑每个OP对另一个答案的评论，这是实际的错误消息：

DataBinding: 'System.Data.DataRowView' does not contain a property with the 
name 'FirstName'. 

这是我的GridView的第一个ItemTemplate。 当我在gridview的文本框中输入新数据，然后单击链接到AddNewCustomer方法的提交按钮时，出现此错误消息

Answer 1

您的查询包括2个选择

SqlCommand cmd = new SqlCommand("SELECT Person.PersonID, 
...
"SELECT SCOPE_IDENTITY();");

所以当你执行

int newPersonID = System.Convert.ToInt32(cmd.ExecuteScalar());

您总是得到Person.PersonID为1。

更新：

1）

您将获得DataBinding：'System.Data.DataRowView'不包含名称为'FirstName'的属性。 因为您将数据绑定到

GridView1.DataSource = GetData(cmd);
GridView1.DataBind();

因此，如果需要绑定任何数据，则删除的SELECT应该位于查询的末尾。 例：

INSERT ...;
INSERT ...;
SELECT ...

这将更具逻辑性，因为您先插入数据，然后选择一些要显示在页面上的结果。

2）您可以将2个INSERT查询“合并”为1个调用。 您只需要做

INSERT ...;
INSERT ... VALUES(..., SCOPE_IDENTITY());
SELECT ...

即：

SqlCommand cmd = new SqlCommand("INSERT INTO Person(FirstName, LastName, Emergency#, Cell#, Age, ActivityCard, CraftType, Initials, Group#) " +
    "values(@FirstName, @LastName, @Emergency, @Cell, @Age, @ActivityCard, @Boat, @Initials, @Group);"
+ "insert into TripSchedule(TripType, PersonID, Time, Date) values (@Type, SCOPE_IDENTITY(), @Time, @Date);" +
    "SELECT Person.PersonID, Person.FirstName AS FirstName, Person.LastName AS LastName, Person.Emergency# AS Emergency#, Person.Cell# AS Cell#, Person.Age AS Age, " +
    "Person.ActivityCard AS ActivityCard, Person.CraftType AS CraftType, Person.Initials AS Initials, Person.Group# AS Group# " +
    "FROM Person INNER JOIN " +
    "TripSchedule ON Person.PersonID = TripSchedule.PersonID where TripSchedule.Date = @Date and " +
    "TripSchedule.Time = @Time and TripSchedule.TripType = @Type;";

底线：

将所有sql代码移到存储过程中。