链接到我的PHP结构上的内容

Question

我在Views 2中的PHP代码：

<?php
$back_image_url = db_result(db_query("select files.filepath from {content_type_team} as ctb inner join {files} as files on files.fid = ctb.field_member_photo_fid  where ctb.nid = %d",$data->nid));
$presetname = 'team_member';
$preset = imagecache_preset_by_name($presetname);
$src = $back_image_url;
$dst = imagecache_create_path($presetname, $src);
// Ensure existing derivative or try to create it on the fly
if (file_exists($dst) || imagecache_build_derivative($preset['actions'], $src, $dst)) {
  $back_image_url_preset = "<img src='". variable_get('static_url', base_path()) . "". $dst."' alt='".$data->node_title."'/>";
}

$html_code = "<li class='team_members'>";
$html_code .= "<div class='ManagementTeam_FirstInfo'>
    <div class='Image'>
        ".$back_image_url_preset."
    </div>
    <div class='Name'>".$data->node_title."</div>
    <div class='Title'>".$data->node_data_field_member_photo_field_unvan_value."</div>
    <div class='DetailLink Absolute'>
        <a>".t("detayl? bilgi")."</a>
    </div>
</div></li>";

echo $html_code;

?>

我想链接内容Ø li尝试过添加a标签，但没有工作。 像这样：

$html_code = "<a href='".$data->node."'><li class='team_members'>";

不工作 我该如何解决？

Answer 1

这样尝试

$html_code = "<li class='team_members'><a href='".$data->node."'>content</a></li>";