我在gridview表中显示文件创建日期时遇到了问题。
这是page_load
代码
if (!IsPostBack)
{
string[] filePaths = Directory.GetFiles(Server.MapPath("~/XMLFiles/"));
List<Thing> lst = new List<Thing>();
foreach (string filePath in filePaths)
{
string filename = Path.GetFileName(filePath); // you might add NULL check and _ check, if that has 2 elements after split.
lst.Add(new Thing() { FileDate = filename.Split('_')[0], FileName = filename.Split('_')[1], FilePath = filePath });
}
GridView1.DataSource = lst;
GridView1.DataBind();
}
错误在这一行
lst.Add(new Thing() { FileDate = filename.Split('_')[0], FileName = filename.Split('_')[1], FilePath = filePath });
这是课程Thing
public class Thing
{
public string FilePath { get; set; }
public string FileName { get; set; }
public string FileDate { get; set; }
}
答案 0 :(得分:0)
好的,做一件事......将你的代码分成两行......如下所示:
String FileDate = filename.Split('_')[0];
String FileName = filename.Split('_')[1];
String FilePath = filePath;
然后检查导致问题的原因。然后检查数据源以获取文件名。