如何根据作为字符串数组发送的属性名从通用列表中提取特定属性？

I have a Generic List of type List<InstanceDataLog> which has huge number of properties in it. I want to pass names of a few properties to a method and want to extract a refined List from within this list.

我有一个List 类型的通用列表，其中包含大量属性。我想将一些属性的名称传递给方法，并希望从此列表中提取精炼的List。

public void Export(string col)   //a,b,c
{
    string [] selectedcol = col.Split(',');
    var grid = new GridView();
    var data = TempData["InstanceDataList"];
    List<InstanceDataLog> lst = new List<InstanceDataLog>();
    List<EToolsViewer.APIModels.InstanceDataLog> lstrefined = new List<InstanceDataLog>();
    lst=   (List<EToolsViewer.APIModels.InstanceDataLog>)TempData["InstanceDataList"];

     var r= lst.Select(e => new {e.a, e.b}).ToList();// I would like to replace these hardcoded properties with names of properties present in `selectedcol `

    grid.DataSource =r;
    grid.DataBind();
}

To clear things up further, suppose InstanceDataLog has 5 properties : a,b,c,d,e I would like pass a,b and be able to extract a new list with only properties a,b

为了进一步清理，假设InstanceDataLog有5个属性：a，b，c，d，e我想传递a，b并且能够提取仅包含属性a，b的新列表

EDIT:

编辑：

 $('#export').mousedown(function () {

 window.location = '@Url.Action("Export", "TrendsData",new { col = "a,b,c" })';

});

2 个解决方案

#1

You could use such method to get properties:

您可以使用此方法获取属性：

private object getProperty(EToolsViewer.APIModels.InstanceDataLog e, string propName)
    {
 var propInfo =typeof(EToolsViewer.APIModels.InstanceDataLog).GetProperty(propName);
    return propInfo.GetValue(e);        
}

and with another function you could get all properties you want:

并使用另一个功能，您可以获得所需的所有属性：

private dynamic getProperties(string[] props, EToolsViewer.APIModels.InstanceDataLog e )
{
    var ret = new ExpandoObject() as IDictionary<string, Object>;;
    foreach (var p in props)
    {
        ret.Add(p, getProperty(e, p));
    }       
    return ret;
}

The problem occurs if you try to assign DataSource with expando object. Solution is described hier:

如果您尝试使用expando对象分配DataSource，则会出现此问题。解决方案描述如下：

Binding a GridView to a Dynamic or ExpandoObject object

将GridView绑定到Dynamic或ExpandoObject对象

We do need one more method:

我们还需要一种方法：

public DataTable ToDataTable(IEnumerable<dynamic> items)
{
    var data = items.ToArray();
    if (data.Count() == 0) return null;

    var dt = new DataTable();
    foreach (var key in ((IDictionary<string, object>)data[0]).Keys)
    {
        dt.Columns.Add(key);
    }
    foreach (var d in data)
    {
        dt.Rows.Add(((IDictionary<string, object>)d).Values.ToArray());
    }
    return dt;
}

and use it:

并使用它：

var r = lst.Select(e => getProperties(selectedcol, e)).ToList();
grid.DataSource = ToDataTable(r);

The same thing, but ready to run for LinqPad:

同样的事情，但准备为LinqPad运行：

void Main()
{
    var samples = new[] { new Sample { A = "A", B = "B", C = "C" }, new Sample { A = "A1", B = "B2", C = "C1" } };
    var r = samples.Select(e => getProperties(new[] {"A", "C", "B"}, e)).ToList();
    r.Dump();
}

private object getProperty(Sample e, string propName)
{
    var propInfo = typeof(Sample).GetProperty(propName);
    return propInfo.GetValue(e);
}

private dynamic getProperties(string[] props, Sample e)
{
    var ret = new ExpandoObject() as IDictionary<string, Object>; ;
    foreach (var p in props)
    {
        ret.Add(p, getProperty(e, p));
    }
    return ret;
}

public class Sample
{
    public string A { get; set;}
    public string B { get; set;}
    public string C { get; set;}
}

With output:

带输出：

#2

To keep compiler type/name checking suggest to pass a Func<InstanceDataLog, TResult> instead of array of names

保持编译器类型/名称检查建议传递Func 而不是名称数组，tresult>

public void Export<TResult>(Func<InstanceDataLog, TResult> selectProperties)   
{
        var grid = new GridView();
        var data = TempData["InstanceDataList"];

        var originalList = (List<InstanceDataLog>)TempData["InstanceDataList"];

        var filteredList = originalList.Select(selectProperties).ToList();

        grid.DataSource = filteredList;
        grid.DataBind();
}

Then use it:

然后使用它：

Export(data => new { Id = data.Id, Name = data.Name });

#1