尝试在 gridview 中添加 link
Trying to add a link within a gridview
我已经创建了一个 sql 数据库,其中一个名为 "Link" 的列有网站 url,用于我网站内和网站外的网站。例如,我有一份 link 的 Geico 保险。问题是当我将此列添加到我的 gridview 时,它只显示网站的 url。我需要它成为我可以点击并在新页面中显示的网站的 link,我只想要 "website" 这个词作为 link。
<asp:GridView id="GridView1" runat="server" AutoGenerateColumns="False" BackColor="White" BorderColor="#999999" BorderStyle="None" BorderWidth="1px" CellPadding="3" DataSourceID="apps2" GridLines="Vertical" AllowSorting="True">
<AlternatingRowStyle BackColor="#DCDCDC" />
<Columns>
<asp:HyperLinkField HeaderText="Link" Text="Website" DataNavigateUrlFields="Link" DataNavigateUrlFormatString="{0}" target=_blank/>
</asp:HyperLinkField>
<asp:BoundField DataField="Source" HeaderText="Source" SortExpression="Source">
</asp:BoundField>
<asp:BoundField DataField="Name" HeaderText="Name" SortExpression="Name">
</asp:BoundField>
<asp:BoundField DataField="Address" HeaderText="Address" SortExpression="Address">
</asp:BoundField>
<asp:BoundField DataField="City" HeaderText="City" SortExpression="City">
</asp:BoundField>
<asp:BoundField DataField="Primary_Number" HeaderText="Phone Number" SortExpression="Primary_Number">
</asp:BoundField>
<asp:BoundField DataField="Backup_Number" HeaderText="Backup Number" SortExpression="Backup_Number">
</asp:BoundField>
<asp:BoundField DataField="Fax_Number" HeaderText="Fax Number" SortExpression="Fax_Number">
</asp:BoundField>
<asp:BoundField DataField="Alarm_Reset_Code" HeaderText="Alarm Reset Code" SortExpression="Alarm_Reset_Code">
</asp:BoundField>
<asp:BoundField DataField="First_Contact_Name" HeaderText="1st Contact Name" SortExpression="First_Contact_Name">
</asp:BoundField>
<asp:BoundField DataField="First_Contact_Number" HeaderText="1st Contact #" SortExpression="First_Contact_Number">
</asp:BoundField>
<asp:BoundField DataField="Second_Contact_Name" HeaderText="2nd Contact Name" SortExpression="Second_Contact_Name">
</asp:BoundField>
<asp:BoundField DataField="Second_Contact_Number" HeaderText="2nd Contact #" SortExpression="Second_Contact_Number">
</asp:BoundField>
<asp:BoundField DataField="Third_Contact_Name" HeaderText="3rd Contact Name" SortExpression="Third_Contact_Name">
</asp:BoundField>
<asp:BoundField DataField="Third_Contact_Number" HeaderText="3rd Contact #" SortExpression="Third_Contact_Number">
</asp:BoundField>
<asp:BoundField DataField="Notes" HeaderText="Notes" SortExpression="Notes">
</asp:BoundField>
<asp:BoundField DataField="Modified" HeaderText="Date Modified" SortExpression="Modified">
</asp:BoundField>
</Columns>
当我使用这个脚本时,gridview 没有在我的网站上启动。我收到服务器错误,错误中突出显示的是 hyperlink 字段的行。
Gridview 最好使用 <ItemTemplate>
来显示网站链接。这样您就可以完全自定义超链接列的 UI/Design。
示例:
<ItemTemplate>
<a href="<%# Bind("Link") %>" target="_blank">Website</a>
</ItemTemplate>
我已经创建了一个 sql 数据库,其中一个名为 "Link" 的列有网站 url,用于我网站内和网站外的网站。例如,我有一份 link 的 Geico 保险。问题是当我将此列添加到我的 gridview 时,它只显示网站的 url。我需要它成为我可以点击并在新页面中显示的网站的 link,我只想要 "website" 这个词作为 link。
<asp:GridView id="GridView1" runat="server" AutoGenerateColumns="False" BackColor="White" BorderColor="#999999" BorderStyle="None" BorderWidth="1px" CellPadding="3" DataSourceID="apps2" GridLines="Vertical" AllowSorting="True">
<AlternatingRowStyle BackColor="#DCDCDC" />
<Columns>
<asp:HyperLinkField HeaderText="Link" Text="Website" DataNavigateUrlFields="Link" DataNavigateUrlFormatString="{0}" target=_blank/>
</asp:HyperLinkField>
<asp:BoundField DataField="Source" HeaderText="Source" SortExpression="Source">
</asp:BoundField>
<asp:BoundField DataField="Name" HeaderText="Name" SortExpression="Name">
</asp:BoundField>
<asp:BoundField DataField="Address" HeaderText="Address" SortExpression="Address">
</asp:BoundField>
<asp:BoundField DataField="City" HeaderText="City" SortExpression="City">
</asp:BoundField>
<asp:BoundField DataField="Primary_Number" HeaderText="Phone Number" SortExpression="Primary_Number">
</asp:BoundField>
<asp:BoundField DataField="Backup_Number" HeaderText="Backup Number" SortExpression="Backup_Number">
</asp:BoundField>
<asp:BoundField DataField="Fax_Number" HeaderText="Fax Number" SortExpression="Fax_Number">
</asp:BoundField>
<asp:BoundField DataField="Alarm_Reset_Code" HeaderText="Alarm Reset Code" SortExpression="Alarm_Reset_Code">
</asp:BoundField>
<asp:BoundField DataField="First_Contact_Name" HeaderText="1st Contact Name" SortExpression="First_Contact_Name">
</asp:BoundField>
<asp:BoundField DataField="First_Contact_Number" HeaderText="1st Contact #" SortExpression="First_Contact_Number">
</asp:BoundField>
<asp:BoundField DataField="Second_Contact_Name" HeaderText="2nd Contact Name" SortExpression="Second_Contact_Name">
</asp:BoundField>
<asp:BoundField DataField="Second_Contact_Number" HeaderText="2nd Contact #" SortExpression="Second_Contact_Number">
</asp:BoundField>
<asp:BoundField DataField="Third_Contact_Name" HeaderText="3rd Contact Name" SortExpression="Third_Contact_Name">
</asp:BoundField>
<asp:BoundField DataField="Third_Contact_Number" HeaderText="3rd Contact #" SortExpression="Third_Contact_Number">
</asp:BoundField>
<asp:BoundField DataField="Notes" HeaderText="Notes" SortExpression="Notes">
</asp:BoundField>
<asp:BoundField DataField="Modified" HeaderText="Date Modified" SortExpression="Modified">
</asp:BoundField>
</Columns>
当我使用这个脚本时,gridview 没有在我的网站上启动。我收到服务器错误,错误中突出显示的是 hyperlink 字段的行。
Gridview 最好使用 <ItemTemplate>
来显示网站链接。这样您就可以完全自定义超链接列的 UI/Design。
示例:
<ItemTemplate>
<a href="<%# Bind("Link") %>" target="_blank">Website</a>
</ItemTemplate>