DataGridView动态更改数据源

基本上当我创建这个DataGridView时，我有这个代码来填补它

public void fillDataGrid(IQueryable patients) { dgvMyPatients.DataSource = patients; dgvMyPatients.Columns["Pat_Last_Name"].DisplayIndex = 0; dgvMyPatients.Columns["Pat_First_Name"].DisplayIndex = 1; dgvMyPatients.Columns["Pat_Middle_Name"].DisplayIndex = 2; dgvMyPatients.Columns["Pat_First_Name"].HeaderText = "First Name"; dgvMyPatients.Columns["Pat_Last_Name"].HeaderText = "Last Name"; dgvMyPatients.Columns["Pat_Middle_Name"].HeaderText = "Middle Name"; } public IQueryable showMyPatients() { DbClassesDataContext myDb = new DbClassesDataContext(dbPath); var patientInfo = from patients in myDb.PatientInfos where patients.Phy_ID == physcianID select patients; return patientInfo; } 

所以当我创建我的对象时，我就是这样做的

 fillDataGrid(showMyPatients()); 

但是，当我单击一个按钮时，我想将其内容更改为此查询中的内容

  private IQueryable searchPatient() { DbClassesDataContext myDb = new DbClassesDataContext(dbPath); var search = from myPatients in myDb.PatientInfos where (myPatients.Pat_ID == patient_ID && myPatients.Pat_First_Name.Contains(txtSearch.Text)) || (myPatients.Pat_ID == patient_ID && myPatients.Pat_Last_Name.Contains(txtSearch.Text)) || (myPatients.Pat_ID == patient_ID && myPatients.Pat_Middle_Name.Contains(txtSearch.Text)) select myPatients; return search; } 

然后，当我点击我的按钮它会这样做，但它不是更新数据网格为什么是这样？ fillDataGrid（searchPatient（））;

Answer 1

有了同样的问题，经过一段时间的搜索，终于找到了答案：

  DataTable dt = new DataTable(); dt.Columns.Add("Column One"); dt.Rows.Add(new object[] { "Item1" }); dt.Rows.Add(new object[] { "Item2" }); dt.Rows.Add(new object[] { "Item3.3" }); this.dataGridView1.AutoGenerateColumns = true; this.dataGridView1.Columns.Clear(); //dataGridView1.DataSource = null; dataGridView1.DataSource = dt; 

AutoGenerateColumns需要是真的，就是这样。

Answer 2

相反，如果做

 DataSource = null 

鉴于IQueryable返回CurrencyManager，它更好地刷新货币管理器：

  (dgvMyPatients.BindingContext[dataGridView1.DataSource] as CurrencyManager).Refresh(); 

CurrencyManager的

CurrencyManager.Refresh（）

Answer 3

换行：

 dgvMyPatients.DataSource = patients; 

至

 dgvMyPatients.DataSource = typeof(List<>); dgvMyPatients.DataSource = patients.ToList(); 

Answer 4

首先，您可以尝试设置：

 DataSource = null; 

在更新之前。 我个人建议使用BindingList将数据绑定到dataGridView。 这样，您不必更改dataSource – 仅包含其中包含的数据。 它的使用方式如下：

 BindingList data = new BindingList(); dgvMyPatients.DataSource = data; ... public void fillDataGrid(IQueryable patients) { data.Clear(); data.AddRange(patients); } 

此外，每次更新源时都不需要提供datagrid架构。

UPDATE

工作样本：

 public partial class Form1 : Form { private BindingList _data = new BindingList(); public Form1() { InitializeComponent(); dataGridView1.DataSource = _data; _data.Add(new SomeClass() { First = "1", Second = "1", Third = "1" }); _data.Add(new SomeClass() { First = "2", Second = "2", Third = "2" }); _data.Add(new SomeClass() { First = "3", Second = "3", Third = "3" }); _data.Add(new SomeClass() { First = "4", Second = "4", Third = "4" }); _data.Add(new SomeClass() { First = "5", Second = "5", Third = "5" }); _data.Add(new SomeClass() { First = "6", Second = "6", Third = "6" }); _data.Add(new SomeClass() { First = "7", Second = "7", Third = "7" }); _data.Add(new SomeClass() { First = "8", Second = "8", Third = "8" }); } private void button1_Click(object sender, EventArgs e) { _data.Clear(); _data.Add(new SomeClass() { First = "11", Second = "11", Third = "11" }); _data.Add(new SomeClass() { First = "21", Second = "21", Third = "21" }); _data.Add(new SomeClass() { First = "31", Second = "31", Third = "31" }); } } public class SomeClass { public string First { get; set; } public string Second { get; set; } public string Third { get; set; } }