接受List <t> </t>的通用方法

时间:2011-12-02 11:43:24

标签: c# generics inheritance

我有5个类,代表一行数据网格。所有这些类都继承自CoreGrid的抽象类。

我有一个导出机制,它使用反射来确定要导出的列。在这一刻,我有一种方法用于每种类型的网格(ExportOrganisations,ExportPeople,ExportEvents)但是这很糟糕,因为它们之间唯一不同的是它查找类型的部分。示例代码如下所示:

public string ExportEvents(List<EventGrid> events)
{
    DataTable report = new DataTable();

    EventGrid ev = new EventGrid();

    Type t = ev.GetType();

    PropertyInfo[] props = t.GetProperties();

    foreach (PropertyInfo prop in props)
    {
        if (!prop.Name.Contains("ID"))
        {
            report.Columns.Add(prop.Name);
        }
    }

    foreach (var item in events)
    {
        DataRow dr = report.NewRow();

        Type itemType = item.GetType();

        PropertyInfo[] itemProps = itemType.GetProperties();

        foreach (PropertyInfo prop in itemProps)
        {
            if (report.Columns.Contains(prop.Name))
            {
                if (prop.GetValue(item, null) != null)
                {
                    dr[prop.Name] = prop.GetValue(item, null).ToString().Replace(",", string.Empty);
                }
            }
        }

        report.Rows.Add(dr);
    }

    return GenerateCSVExport(report, ExportType.Events);
}

我的问题是,如何将这些方法压缩到一个方法中,该方法接受一个继承自CoreGrid的列表?

2 个答案:

答案 0 :(得分:5)

public string ExportEvents<T>(List<T> events) where T : CoreGrid
{
    DataTable report = new DataTable();

    Type t = typeof(T);

    //your magic here
}

然后使用

var result = ExportEvents<EventGrid>(eventList);

答案 1 :(得分:3)

应该是这样的。由于能够推断类型,您不必更改当前的呼叫签名,只需将所有内容指向通用方法:

//if myList is a list of CoreGrid or derived.
string export = ExportEvents(myList);


public string ExportEvents<T>(List<T> events) where T : CoreGrid
{
    DataTable report = new DataTable();

    Type t = typeof(T);

    PropertyInfo[] props = t.GetProperties();

    foreach (PropertyInfo prop in props)
    {
        if (!prop.Name.Contains("ID"))
        {
            report.Columns.Add(prop.Name);
        }
    }

    foreach (var item in events)
    {
        DataRow dr = report.NewRow();

        Type itemType = item.GetType();

        PropertyInfo[] itemProps = itemType.GetProperties();

        foreach (PropertyInfo prop in itemProps)
        {
            if (report.Columns.Contains(prop.Name))
            {
        var propValue = prop.GetValue(item, null)
                if (propValue != null)
                {
                    dr[prop.Name] = propValue.ToString().Replace(",", string.Empty);
                }
            }
        }

        report.Rows.Add(dr);
    }

    return GenerateCSVExport(report, ExportType.Events);
}