Laravel Slugs与Str :: slug

Question

看看我的前端URL生成的Str :: slug,但只是想知道你们如何用路由等来实现它,例如,你们如何将http://www.example.com/courses/1更改为http ://www.example.com/courses/this-course

Answer 1

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}/{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);
});

奇迹般有效.它获取$ categorySlug和$ slug变量然后使用它们来过滤Eloquent模型课程以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一个URL,如:

http://www.example.com/courses/it-training/mcse

通过做类似的事情:

<a href="{{ URL::to('courses/'.$course->category->parentCategorySlug($course->category->parent_id).'/'.$course->category->slug.'/'. $course->slug) }}" title="{{ $course->title }}">{{ $course->title }}</a>

A在我的类别中有一个方法,如下所示,检索父类别slug.虽然使用某种类型的演示者类可以让你简单地使用$ course-> url,但是我还没有完成这项工作,这可以更好地实现.我会更新答案.

public function parentCategorySlug($parentId)
{
    if ($parentId === '0')
    {
        return $this->slug;
    }

    return $this->where('id', $parentId)->first()->slug;
}

Answer 2

您可以使用cvierbrock的Eloquent-Sluggable包.