Question

我正在尝试将一个超链接字段列添加到GridView，该文件从服务器上的文件目录中提取文件名。我正在使用下面的代码，但它并不完全正常。它显示路径和文件名，但它不是可点击的超链接字段。我不确定我做错了什么，有人可以帮帮我吗？

public void GetFilesAndFolders()
    {
        using (IREPEntities dbContext = new IREPEntities())
        {

            String vcharTempFileLocation = (from a in dbContext.tbl_ApplicationSetting
                                            where a.vcharKey == "vcharTempFileLocation"
                                            select a).Single().vcharValue;

            DataTable gridviewSource = DisplayFilesInGridView();
            DataRow gridviewRow;

            //Get All Folders Or Directories and add in table  
            DirectoryInfo directory = new DirectoryInfo(vcharTempFileLocation);
            DirectoryInfo[] subDirectories = directory.GetDirectories();

            foreach (DirectoryInfo dirInfo in subDirectories)
            {
                gridviewRow = gridviewSource.NewRow();
                gridviewRow["Claim"] = ddlClaimNumber.SelectedItem;
                gridviewRow["Name"] = dirInfo.Name;
                dynamic newlink = new HyperLinkField();

                newlink = vcharTempFileLocation + dirInfo.Name;

                gridviewRow["link"] = newlink;
                gridviewRow["Application"] = chkApplicationType.SelectedItem;

                gridviewSource.Rows.Add(gridviewRow);
            }




            //Get files in all directories  

            FileInfo[] files = directory.GetFiles("*.*", SearchOption.AllDirectories);
            foreach (FileInfo fileInfo in files)
            {
                gridviewRow = gridviewSource.NewRow();
                gridviewRow["Claim"] = ddlClaimNumber.SelectedItem;
                gridviewRow["Name"] = fileInfo.Name;
                dynamic newlink = new HyperLinkField();

                newlink = vcharTempFileLocation + fileInfo.Name;

                gridviewRow["link"] = newlink;
                gridviewRow["Application"] = chkApplicationType.SelectedItem;                                   

                gridviewSource.Rows.Add(gridviewRow);
            }                        


            gvBatchDetails.DataSource = gridviewSource;
            gvBatchDetails.DataBind();

        }

    }
    private DataTable DisplayFilesInGridView()
    {
        DataTable dtgridviewSource = new DataTable();
        dtgridviewSource.Columns.Add(new DataColumn("Claim", typeof(System.String)));
        dtgridviewSource.Columns.Add(new DataColumn("Name", typeof(System.String)));
        dtgridviewSource.Columns.Add(new DataColumn("Application", typeof(System.String)));
        dtgridviewSource.Columns.Add(new DataColumn("link", typeof(System.String)));

        return dtgridviewSource;
    }


}

Answer 1

不是手动构建GridView，而是创建可绑定的数据源。

首先创建一个自定义类来保存详细信息：

public class FileOrFolderDetail
{
    public string Claim{get;set;}
    public string Name{get;set;}
    public string Link{get;set'}
    public string Application{get;set;}
}

接下来，创建新班级的List：

var fileOrFolderDetails = new List<FileOrFolderDetail>();

此时，像你一样经历你的两个循环，但是创建新的FileOrFolderDetail个对象并将它们添加到List。

最后，将您的GridView绑定到fileOrFolderDetails，并在ASPX中设置HyperLinkField以使用Link作为DataNavigateUrl。

让我们从那里开始，我们可以根据需要进行调整。

如何将HyperLinkField添加到GridView数据绑定到文件目录？

1 个答案: