所以我使用此代码将帖子更改为'Mentor Q& A'作为标签,但该函数返回一个奇怪的错误
Parse error: syntax error, unexpected '$menu' (T_VARIABLE) ...
这是我正在尝试运行的功能:
function change_post_label() {
global $menu;
global $submenu;
$menu[5][0] = 'Mentor Q&A';
$submenu['edit.php'][5][0] = 'Mentor Q&A';
$submenu['edit.php'][10][0] = 'Add Mentor Q&A';
$submenu['edit.php'][16][0] = 'Mentor Q&A Tags';
echo '';
function change_post_object() {
global $wp_post_types;
$labels = &$wp_post_types['post']->labels;
$labels->name = 'Mentor Q&A';
$labels->singular_name = 'Mentor Q&A';
$labels->add_new = 'Add Mentor Q&A';
$labels->add_new_item = 'Add Mentor Q&A';
$labels->edit_item = 'Edit Mentor Q&A';
$labels->new_item = 'Mentor Q&A';
$labels->view_item = 'View Mentor Q&A';
$labels->search_items = 'Search Mentor Q&A';
$labels->not_found = 'No Mentor Q&A found';
$labels->not_found_in_trash = 'No Mentor Q&A found in Trash';
$labels->all_items = 'All Mentor Q&A';
$labels->menu_name = 'Mentor Q&A';
$labels->name_admin_bar = 'Mentor Q&A';
}
add_action( 'admin_menu', 'change_post_label' );
add_action( 'init', 'change_post_object');
这个函数也在子主题的functions.php中调用,我不确定它是否与它有任何关系。
答案 0 :(得分:0)
我认为功能应该正确关闭
function change_post_label() {
global $menu;
global $submenu;
$menu[5][0] = 'Mentor Q&A';
$submenu['edit.php'][5][0] = 'Mentor Q&A';
$submenu['edit.php'][10][0] = 'Add Mentor Q&A';
$submenu['edit.php'][16][0] = 'Mentor Q&A Tags';
echo '';
}
答案 1 :(得分:0)
解决。问题是我复制的代码中的缩进(如果是)是使代码无效的实际空格,doh!
经验教训。