在我的网站ebaysecrets.co.il中,我的用户可以选择查看内容而不启用评论,除非他们登录。
看起来像是:
http://i.stack.imgur.com/e2Jsw.png
现在它显示为正常文本,而不是链接,我想将其转换为链接。
守则就是这样:
在 skin_forum 下,我的 forumIndexTemplate 包含以下代码:
<if test="usercanpost:|:$forum_data['_user_can_post']">
<li>
<a
href='{parse url="module=post&section=post&do=new_post&f={$forum_data['id']}" base="publicWithApp"}'
title='{$this->lang->words['topic_start']}'
accesskey='s'>{$this->lang->words['topic_start']}
</a>
</li>
<else />
<li class='disabled'>
<span>
<if test="isGuestPostTopicTop:|: ! $this->memberData['member_id']">
{$this->lang->words['forum_no_start_topic_guest']}
<else />
{$this->lang->words['forum_no_start_topic']}
</if>
</span>
</li>
</if>
<if test="moderationDropdownLink:|:$this->memberData['is_mod'] == 1">
<li class='non_button'>
<a href='#' id='forum_mod_options' class='ipbmenu'>{$this->lang->words['forum_management']}</a>
</li>
</if>
我尝试了一些这样的改变:
<li class='disabled'>
<span>
<if test="isGuestPostTopic:|: ! $this->memberData['member_id']">
<a href="index.php?app=core&module=global§ion=login">
{$this->lang->words['forum_no_start_topic_guest']}
<else />
{$this->lang->words['forum_no_start_topic']}
</a>
</if>
</span>
</li>
但没有成功!
我需要插入的正确语法是什么,因此它将作为指向此的链接:
ebaysecrets.co.il/index.php?app=core&module=global§ion=login