16

我有一个包含一系列事务对象的列表。我要做的是在加载表单时在 Datagridview 控件中显示这些事务对象,基本上 Datagridview 应该代表事务寄存器的某些内容,以显示列表中每个事务对象的数据。

我必须承认在使用 Datagridviews 时缺乏经验,而且我在理解我需要在这里做什么时遇到了一些困难。

我的问题是,如何获取列表中每个对象的详细信息以显示在 Datagridview 中?

这是我的代码。

首先是事务类:

public class Transaction
{
    // Class properties
    private decimal amount;
    private string type;
    private decimal balance;
    private string date;
    private string transNum;
    private string description;

    // Constructor to create transaction object with values set.
    public Transaction(decimal amount, string type, decimal currBal, string date, string num, string descrip)
    {
        this.amount = amount;
        this.type = type;
        this.balance = currBal;
        this.date = date;
        this.transNum = num;
        this.description = descrip;
    }

    // Get and Set accessors to allow manipulation of values.
    public decimal Amount
    {
        get
        {
            return amount;
        }
        set
        {
            amount = value;
        }
    }
    public string Type
    {
        get
        {
            return type;
        }
        set
        {
            type = value;
        }
    }
    public decimal Balance
    {
        get
        {
            return balance;
        }
        set
        {
            balance = value;
        }
    }
    public string Date
    {
        get
        {
            return date;
        }
        set
        {
            date = value;
        }
    }
    public string TransNum
    {
        get
        {
            return transNum;
        }
        set
        {
            transNum = value;
        }
    }
    public string Description
    {
        get
        {
            return description;
        }
        set
        {
            description = value;
        }
    }

    public decimal addCredit(decimal balance, decimal credit)
    {
        decimal newBalance;
        newBalance = balance + credit;
        return newBalance;
    }

    public decimal subtractDebit(decimal balance, decimal debit)
    {
        decimal newBalance;
        newBalance = balance - debit;
        return newBalance;
    }
    }
}

现在“注册”表单的代码:

    public partial class Register : Form
{
    List<Transaction> tranList = new List<Transaction>();

    public Register(List<Transaction> List)
    {
        InitializeComponent();
        this.tranList = List;
    }

    private void Register_Load(object sender, System.EventArgs e)
    {
        //regView represents the Datagridview that I'm trying to work with
        regView.AutoSize = true;
        regView.DataSource = tranList;
        regView.Rows.Add(tranList[0]);
    }
}

这是我得到的输出。 寄存器输出

4

3 回答 3

11

实际上有两种高级方法。

1) 将手动创建的行直接添加到DataGridView. 在这种情况下,您必须随着事情的变化手动更新/删除它们。如果您不打算在初始化后更改/更改显示的内容,这种方法是“可以的”。如果你这样做,它就会变得站不住脚。

要直接添加它,您需要创建一个DataGridViewRow,并用各个值填充它,然后将 添加DataGridViewRowDataGridView.Rows.

2)数据绑定DGV。有很多关于数据绑定到DataGridView. 在某些情况下,将数据添加到 aDataTable中,然后从中提取 aDataView并将 绑定DataGridViewDataView. 其他人发现直接绑定到集合更容易。

CodeProject 有一篇不错的文章可以帮助您开始这条道路,但快速的 Google 搜索会产生许多其他文章。

http://www.codeproject.com/Articles/24656/A-Detailed-Data-Binding-Tutorial

于 2013-04-28T01:14:38.810 回答
5

用作 DGV:

DataGridView groupListDataGridView;

柱子:

DataGridViewTextBoxColumn groupListNameColumn;

列设置应该是这样的:

groupListNameColumn.DataPropertyName = "name";

使用此属性,否则将添加所有列。

groupListDataGridView.AutoGenerateColumns = false;

像这样填充:

private void populateGroupList() {
    groupListDataGridView.DataSource = null;
    formattedGroupList = new SortableBindingList<DataGridGroupObject>();
    foreach (GroupObject go in StartUp.GroupList) {
        DataGridGroupObject dggo = new DataGridGroupObject();
        dggo.id = go.Id;
        dggo.name = go.Name;
        formattedGroupList.Add(dggo);
    }
    groupListDataGridView.DataSource = formattedGroupList;
    groupListDataGridView.Invalidate();
}

和型号:

public class DataGridGroupObject
{
    public int id { get; set; }      //this will be match id column
    public string name { get; set; } // this will be match name column
}
于 2013-07-17T08:19:13.117 回答
4

只需using System.Linq;在顶部添加。然后你可以这样做:

//This will create a custom datasource for the DataGridView.
var transactionsDataSource = tranList.Select(x => new
{
        Amount = x.amount,
        Type = x.type,
        Balance = x.balance,
        Date = x.date,
        TransNum = x.transNum
        Description = x.description
}).ToList();

//This will assign the datasource. All the columns you listed will show up, and every row
//of data in the list will populate into the DataGridView.
regView.DataSource = transactionsDataSource;
于 2021-02-05T21:28:15.247 回答