基于 WP 插件样板的插件返回错误 "Warning: call_user_func_array() expects parameter 1 to be a valid callback"

Question

我正在构建一个基于 Wordpress 插件样板（面向对象编程）的 Wordpress 插件，但在调用在我的自定义管理仪表板菜单中显示我的插件管理页面的函数时遇到了困难。

在我的 class 中，我调用了我的自定义菜单方法，它在菜单显示时效果很好。

private function define_admin_hooks() {

    $plugin_admin = new myPlugin_Admin( $this->get_plugin_name(), $this->get_version() );

    $this->loader->add_action( 'admin_enqueue_scripts', $plugin_admin, 'enqueue_styles' );
    $this->loader->add_action( 'admin_enqueue_scripts', $plugin_admin, 'enqueue_scripts' );

    // Load CPTs
    $this->loader->add_action( 'admin_init', $plugin_admin, 'create_cpt_person' );
    $this->loader->add_action( 'admin_menu', $plugin_admin, 'setup_admin_menus' );
    $this->loader->add_action( 'parent_file', $plugin_admin, 'set_current_menu' );

}

然后它执行我的 setup_admin_menus 方法。

public static function setup_admin_menus() {

    $page_title = 'myPlugin';
    $menu_title = 'myPlugin';
    $capability = 'manage_options';
    $menu_slug  = 'myplugin_admin';
    $function   = 'display_admin_page'; // This is what triggers the showing of my admin page once I click on the parent menu (as if you were clicking on "Pages" in your admin dashboard)
    $icon_url   = 'dashicons-admin-page';
    $position   = 3;

    add_menu_page(
        $page_title,
        $menu_title,
        $capability,
        $menu_slug,
        $function,
        $icon_url,
        $position
        );

    $submenu_pages = array(
        // Post Type :: Person
        array(
            'parent_slug'   => 'myplugin_admin',
            'page_title'    => 'People',
            'menu_title'    => 'People',
            'capability'    => 'read',
            'menu_slug'     => 'edit.php?post_type=person',
            'function'      => null
            )
        );

    // Add each submenu item to custom admin menu.
    foreach($submenu_pages as $submenu){
        add_submenu_page(
            $submenu['parent_slug'],
            $submenu['page_title'],
            $submenu['menu_title'],
            $submenu['capability'],
            $submenu['menu_slug'],
            $submenu['function']
            );
    }
}

这是调用的函数：

public static function display_admin_page() {

    echo '<div class="wrap">';
    echo '<h2>Welcome To WordPets</h2>';
    echo '<p>This is the custom admin page created from the custom admin menu.</p>';


    echo '</div><!-- end .wrap -->';
    echo '<div class="clear"></div>';
}

但是每当我进入插件的管理页面时，我都会收到 警告：call_user_func_array() 期望参数 1 是一个有效的回调，函数 'display_admin_page' 未找到或无效第 496 行 /Users/Lazhar/dev/wp-includes/plugin.php 中的函数名称 ，它只显示错误消息。

任何帮助将不胜感激:)

Answer 1

尝试将您的 display_admin_page() 方法放在 class

之外

function display_admin_page() {

    echo '<div class="wrap">';
    echo '<h2>Welcome To WordPets</h2>';
    echo '<p>This is the custom admin page created from the custom admin menu.</p>';


    echo '</div><!-- end .wrap -->';
    echo '<div class="clear"></div>';
}

Answer 2

另一种方法（而不是将函数放在 class 之外）是将 class 方法作为数组传递：

    $page_title = 'myPlugin';
    $menu_title = 'myPlugin';
    $capability = 'manage_options';
    $menu_slug  = 'myplugin_admin';
    $function   = array($this, 'display_admin_page');  //This will properly get the class method function name

这是引用函数（class 方法）的正确方法。

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