Получение этой ошибки со следующим запросом в SQL server 2014

---

У меня есть кнопки "Обновить"и " удалить". У них одни и те же ошибки...


Выражение небулевского типа, заданное в контексте, где ожидается условие, рядом с "ID".

Что я уже пробовал:

Private Sub btnUpdate_Click(sender As Object, e As EventArgs) Handles btnUpdate.Click
        Try
            conn = New SqlConnection(strcon)
            conn.Open()
            Dim str As String = "Update PatientInfo set Surname=@Surname,Firstname=@Firstname,MI=@MI,Gender=@Gender,Birthday=@Birthday,Address=@Address,Contact=@Contact,Email=@Email Where Patient ID=@PID"
            cmd = New SqlCommand(str, conn)
            cmd.Parameters.AddWithValue("@Surname", txtSurname.Text)
            cmd.Parameters.AddWithValue("@Firstname", txtFirstname.Text)
            cmd.Parameters.AddWithValue("@MI", txtMI.Text)
            cmd.Parameters.AddWithValue("@Gender", CmbxGender.Text)
            cmd.Parameters.AddWithValue("@Birthday", dtpBday.Text)
            cmd.Parameters.AddWithValue("@Address", txtAdd.Text)
            cmd.Parameters.AddWithValue("@Contact", txtContact.Text)
            cmd.Parameters.AddWithValue("@Email", txtEmail.Text)
            cmd.Parameters.AddWithValue("@PID", txtPID.Text)
            cmd.ExecuteNonQuery()
            MessageBox.Show("Data has been updated!")
            loadData()
            cmd.Dispose()
            conn.Close()

        Catch ex As Exception
            MessageBox.Show(ex.Message)
        End Try
    End Sub
'==========================================================================

Private Sub btnDelete_Click(sender As Object, e As EventArgs) Handles btnDelete.Click

        Dim cmd As New SqlCommand
        Try

            conn.ConnectionString = strcon
            conn.Open()
            cmd.Connection = conn
            cmd.CommandText = "Delete from PatientInfo Where Patient ID=" + txtPID.Text
            cmd.ExecuteNonQuery()
            MessageBox.Show("Data has been deleted")
            loadData()

            conn.Close()


        Catch ex As Exception
            MessageBox.Show(ex.Message)
        End Try
    End Sub

---