将所有项目传递给另一个表单

Question

我将数据网格中的 variás 选择项传递给另一个表单，但是我需要第二个表单只打开一次。我把 Form 调用放在

    foreach (DataGridViewRow row in dgvcontasreceber.SelectedRows)
    {

       SqlConnection conConexao1 = clsdb.AbreBanco();
       SqlCommand cmd1 = new SqlCommand("select dbo.empresas.id, dbo.empresas.razao_social from empresas inner join dbo.pessoa_empresa on dbo.pessoa_empresa.empresa_id = dbo.empresas.id inner join dbo.pessoa on dbo.pessoa.id = dbo.pessoa_empresa.pessoa_id where dbo.pessoa.id ='" + lblid.Text + "'", conConexao1);
       SqlDataReader dr1 = cmd1.ExecuteReader();

        if (dr1.HasRows == true)
        {
            if (dr1.Read())
            {
               var item = new clsItens
               {
                  PessoaId = txtid.Text,
                  EmpresaId = int.Parse(dr1[0].ToString()),
                  RazaoSocial = dgvcontasreceber[0, linhaAtual].Value.ToString()
               };

               items.Add(item);
             }
         }

         dr1.Close();
         cmd1.Clone();
         conConexao1.Close();

   }
   FormRecebimento abrir = new FormRecebimento(items);
   abrir.ShowDialog();

Answer 1

正如第一个答案所说，您需要找到一种方法让 FormRecebimento 接收一个集合（可以是数组、列表、IEnumerable 等），并根据您的需要在表单中使用这些值。

也许你可以从中得到一些启发：

class Item
{
    public string PessoaId { get; set };
    public int EmpresaId { get; set };
    public string RazaoSocial { get; set };
}

class FormRecebimento
{
    private List<Item> items;

    public FormRecebimento(List<Item> items)
    {
        this.items = items;
        this.initialize();
    }

    private void initialize()
    {
        /* Iterate over Item list and set the values you want
        pessoaid = int.Parse(txtid.Text);
        id_empresa = int.Parse(dgvcontasreceber[16, linhaAtual].Value.ToString());
        idrec.Text = dgvcontasreceber[0, linhaAtual].Value.ToString();
        */
    }
}

//... 

List<Item> items = new List<Item>();

foreach(DataGridViewRow row in dgvcontasreceber.SelectedRows)
{

   SqlConnection conConexao1 = clsdb.AbreBanco();
   SqlCommand cmd1 = new SqlCommand("select dbo.empresas.id, dbo.empresas.razao_social from empresas inner join dbo.pessoa_empresa on dbo.pessoa_empresa.empresa_id = dbo.empresas.id inner join dbo.pessoa on dbo.pessoa.id = dbo.pessoa_empresa.pessoa_id where dbo.pessoa.id ='" + lblid.Text + "'", conConexao1);
   SqlDataReader dr1 = cmd1.ExecuteReader();

   if (dr1.HasRows == true)
   {
      if (dr1.Read())
      {
         var item = new Item
         { 
            PessoaId = txtid.Text, 
            EmpresaId = int.Parse(dr1[0].ToString()),
            RazaoSocial = dgvcontasreceber[0, linhaAtual].Value.ToString()
         }
         items.add(item);
      }
   }

   dr1.Close();
   cmd1.Clone();
   conConexao1.Close();
}

FormRecebimento abrir = new FormRecebimento(items);
abrir.ShowDialog();

我真的不能在变量名称上使用正确的“标签”，因为我不太了解您的代码，所以我真的建议您看看互联网上的“清洁代码”资料，我确定对你有很大帮助:)

提示：http://www.macoratti.net/17/05/c_codlimp1.htm（ptBR，我相信你也会说葡萄牙语）

Answer 2

首先你没有使用SQL安全方式。

第二件事是你创建一个具有变量类型的表单实例，比如说int，然后每次在 foreach 循环中覆盖该变量。

我现在将为您编写完成的代码，因为这段代码完全是一团糟。

首先创建新的class，看起来像这样：

public YourNewClass
{
    public int Id { get; set; }
    public string razao_social { get; set; } //I am guessing this one is string in your database

    public YourNewClas()
    {
    }
}

现在在您要打开的新表单 (FormRecebimento) 中创建此变量

public List<YourNewClass> listOfMyNewClass;

现在你现在的代码重写如下：

FormRecebimento abrir = new FormRecebimento();
foreach(DataGridViewRow row in dgvcontasreceber.SelectedRows)
{
    using(SqlConnection conConexao1 = clsdb.AbreBanco()) //Use using since it will dispose of connection automatically after it finishes using it
    {
        conConexao1.Open(); // you do not have this line so maybe it is inside your function but not sure so i put it here
        using(SqlCommand cmd1 = new SqlCommand("select dbo.empresas.id, dbo.empresas.razao_social from empresas inner join dbo.pessoa_empresa on dbo.pessoa_empresa.empresa_id = dbo.empresas.id inner join dbo.pessoa on dbo.pessoa.id = dbo.pessoa_empresa.pessoa_id where dbo.pessoa.id = @ID", conConexao1))
        {
            cmd1.Parameters.AddWithValue("@ID", lblid.Text); //Use parameters when building SqlCommand since it is safer way

            SqlDataReader dr1 = cmd1.ExecuteReader();

            if (dr1.HasRows == true) // I am never using this one but just if(dr1.Read()) or while(dr1.Read()) and never had problem but i left it there  for you
            {
                if (dr1.Read())
                {
                    //We create new object type of your custom class and add it to new forms (which we created but haven't shown yet) list
                    abrir.listOfMyNewClass.Add(new YourNewClass{ Id = Convert.ToInt32(dr1[0]), razao_social = dr1[1].ToString();
                }
            }
        }
    }
    abrir.ShowDialog();
}

此代码将创建您的新表单 OUTSIDE FOREACH LOOP（因此只有一次），该表单将具有名为 listOfMyNewClass 的公共可访问变量，其类型为 List<MynewClas>

然后在 FOREACH 循环内部，我们将使用 Add() 方法在新表单中填充 listOfMyNewClass（因此我们不会每次都覆盖该变量）

然后在 OUTSIDE FOREACH LOOP 中，我们将显示新创建的表单，其中填充了 List<YourNewClass> listOfMyNewClass；

要在新表单中操作该变量，只需阅读有关 System.Collections.Generic.List here 的所有内容，如果您想将该列表用作 DataTable，请查看我的答案 here

Answer 3

目前，FormRecebimento 需要一个 pessoaid、一个 id_empresa 和一个 idrec 才能工作。如果必须将多个值传递给表单，则必须将这些属性替换为 IEnumerable<DataGridViewRow> 类型的新属性。如果我们将这个新属性称为 inputRows，那么您的代码将如下所示：

FormRecebimento abrir = new FormRecebimento();
abrir.inputRows = dgvcontasreceber.SelectedRows;
abrir.ShowDialog();

如何实现 inputRows 的显示超出了本问题的范围，但至少这应该让您朝着正确的方向开始。