String 带Str::slug的Laravel蛞蝓
在我的前端URL生成中查看Str::slug,但只是想知道你们是如何使用路由等实现它的,例如,你们将如何更改为好的,我是这样做的:String 带Str::slug的Laravel蛞蝓,string,url,laravel-4,friendly-url,slug,String,Url,Laravel 4,Friendly Url,Slug,在我的前端URL生成中查看Str::slug,但只是想知道你们是如何使用路由等实现它的,例如,你们将如何更改为好的,我是这样做的: // I have a slug field in my courses table and a slug field in my categories table, along with a category_id field in my courses table. // Route Route::get('courses/{categorySlug}/{
// I have a slug field in my courses table and a slug field in my categories table, along with a category_id field in my courses table.
// Route
Route::get('courses/{categorySlug}/{slug?}', function($categorySlug, $slug) {
$course = Course::leftJoin('categories', 'categories.id', 'courses.category_id')
->where('categories.slug', $categorySlug)
->where('courses.slug', $slug)
->firstOrFail();
return View::make('courses.show')->with('course', $course);
});
工作起来很有魅力。它获取$categorySlug和$slug变量,然后使用它们过滤雄辩的模型课程,从数据库中获取正确的课程对象
编辑:您可以在视图中生成URL,如:
通过这样做:
<a href="{{ URL::to('courses/'.$course->category->parentCategorySlug($course->category->parent_id).'/'.$course->category->slug.'/'. $course->slug) }}" title="{{ $course->title }}">{{ $course->title }}</a>
我还实现了类似的URL映射,但我更喜欢在请求的URL中同时包含ID和slug,如下所示:
http://www.example.com/courses/1/my-laravel-course
此方法允许我从URL中给定的ID获取请求的course
对象,而不必将slug存储在我的DB表中
Route::post('courses/(:num)/(:any)', function ($courseid, $slug) {
$course = Course::where('id', '=', $courseid)->get();
return View::make('courses.show')->with('course', $course);
}
您可以使用。至于我,我创建了一个helper函数,并使用了下面的方法
您可以创建一个相关的模型Slug,并使用如下方法接近课程:
$course = Slug::where('slug', $slug) -> firstOrFail() -> course;
如何在控制器中获取此url以显示它?url应该是唯一的,在这种情况下,您将获得与
/1/my laravel course
和/1/lorem ipsum
相同的url
public static function getSlug($title, $model) {
$slug = Str::slug($title);
$slugCount = count( $model->whereRaw("url REGEXP '^{$slug}(-[0-9]*)?$'")->get() );
return ($slugCount > 0) ? "{$slug}-{$slugCount}" : $slug;
}
$course = Slug::where('slug', $slug) -> firstOrFail() -> course;