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我有一个函数可以减少（优化）它所拥有的代码行数。这个函数有一些重复的片段。

可以使用 lambda 函数减少此代码吗？

这是我的功能代码：

static function get_all_category_with_widgets( $status = 'all' ) {
            $all_categories = SB_Central::get_categories();
            $all_widgets    = SB_Settings::get_sb_widgets(); // Get all widgets from options variable.

            foreach ( $all_categories as $category_key => $category_value ) {
                foreach ( $all_widgets as $widget_value ) {
                    // Create one widget
                    $widget = array_merge($widget_value['widget'], array ( 'id' => $widget_value['id'], 'status' => $widget_value['status']));

                    // In this case save active and disable.
                    if ( $status == 'active_and_disable' && ( $widget_value['status'] == 'active' || $widget_value['status'] == 'disable' )) {
                        if ( $category_value[ 'category_title' ] == $widget_value[ 'category' ][ 'title' ] ) {
                            $all_categories[ $category_key ][ 'widgets' ][ ] = $widget; // Save widget
                        }
                    } elseif ( $status == 'active' && $widget_value['status'] == 'active' ) {
                        if ( $category_value[ 'category_title' ] == $widget_value[ 'category' ][ 'title' ] ) {
                            $all_categories[ $category_key ][ 'widgets' ][ ] = $widget; // Save widget
                        }
                    } elseif ( $status == 'disable' && $widget_value['status'] == 'disable' ) {
                        if ( $category_value[ 'category_title' ] == $widget_value[ 'category' ][ 'title' ] ) {
                            $all_categories[ $category_key ][ 'widgets' ][ ] = $widget; // Save widget
                        }
                    } elseif ( $status == 'deleted' && $widget_value['status'] == 'deleted' ) {
                        if ( $category_value[ 'category_title' ] == $widget_value[ 'category' ][ 'title' ] ) {
                            $all_categories[ $category_key ][ 'widgets' ][ ] = $widget; // Save widget
                        }
                    } elseif ( $status == 'all' ) {
                        if ( $category_value[ 'category_title' ] == $widget_value[ 'category' ][ 'title' ] ) {
                            $all_categories[ $category_key ][ 'widgets' ][ ] = $widget; // Save widget
                        }
                    }
                }
            }

            return $all_categories;
        }

这是我一直重复的代码：

if ( $category_value[ 'category_title' ] == $widget_value[ 'category' ][ 'title' ] ) {
                            $all_categories[ $category_key ][ 'widgets' ][ ] = $widget; // Save widget
                        }

如果您知道减少此代码的更好方法，将受到欢迎。

4

1 回答 1

1

您可以用以下更简洁的表示替换您的 if/elseif/elseif.... 块

if ( $category_value[ 'category_title' ] == $widget_value[ 'category' ][ 'title' ] ) {
    switch ($status){
        case "active_and_disable":
            // set if widget statis is active or disable
            if( $widget_value['status'] == 'active' || $widget_value['status'] == 'disable' ){
                $all_categories[ $category_key ][ 'widgets' ][ ] = $widget; // Save widget
            }
            break;
        case "all":
            // always set
            $all_categories[ $category_key ][ 'widgets' ][ ] = $widget; // Save widget
            break;
        default:
            // set when status == widget status
            if ($status == $widget_value['status']){
                $all_categories[ $category_key ][ 'widgets' ][ ] = $widget; // Save widget
            }
            break;
    }
}
于 2013-08-14T18:16:07.507 回答