如何仅显示特定用户ID的类别
问题描述:
我有以下代码行,其中包含编辑附件的链接。我只希望某些用户ID能够看到这个编辑链接。我怎样才能使这个链接,这是类=编辑附件,显示一个特定的用户ID?如何仅显示特定用户ID的类别
<div class="details">
<div class="filename">29982792291_9e1fc5b238_k-1.jpg</div>
<div class="uploaded">November 9, 2016</div>
<div class="file-size">1 MB</div>
<div class="dimensions">2048 × 1365</div>
<a class="edit-attachment" href="https://www.citynarrative.com/wp-admin/post.php?post=308&action=edit&image-editor" target="_blank">Edit Image</a>
<button type="button" class="button-link delete-attachment">Delete Permanently</button>
<div class="compat-meta">
</div>
</div>
答
取决于您在谈论用户标识时的含义。
您也许可以使用函数edit_post_link($link, $before, $after, $id, $class);
其中$ id是后(或附件)ID
这只会根据用户capatibilities显示链接。
更多细节edit_post_link
$user = wp_get_current_user();
$allowed_roles = array('editor', 'administrator', 'author'); // can be authorized ids
// but you'll need to replace $user->roles by $user->ID in array_intersect
if(array_intersect($allowed_roles, $user->roles)) { ?>
<div class="details">
<div class="filename">29982792291_9e1fc5b238_k-1.jpg</div>
<div class="uploaded">November 9, 2016</div>
<div class="file-size">1 MB</div>
<div class="dimensions">2048 × 1365</div>
<?php echo edit_post_link(__('Edit Image', text-domain), '', '', $id);?>
<button type="button" class="button-link delete-attachment">Delete Permanently</button>
<div class="compat-meta">
</div>
</div>
<?php } ?>
您还可以检查用户登录其作用,以显示你的链接与current_user_can()
答
试试这个代码
$user_array = array(1, 2 , 3); // list of user id
$user_id = get_current_user_id(); // login user id
if (in_array($user_id, $user_array))
{
?>
<div class="details">
<div class="filename">29982792291_9e1fc5b238_k-1.jpg</div>
<div class="uploaded">November 9, 2016</div>
<div class="file-size">1 MB</div>
<div class="dimensions">2048 × 1365</div>
<a class="edit-attachment" href="https://www.citynarrative.com/wp-admin/post.php?post=308&action=edit&image-editor" target="_blank">Edit Image</a>
<button type="button" class="button-link delete-attachment">Delete Permanently</button>
<div class="compat-meta">
</div>
</div>
<?php
}
您需要可以使用PHP或JavaScript来确定UserID是否与所需的ID匹配。代码将如下所示: if(user.id === 235){/ *代码显示相关按钮* /}; – user3492940