Laravel - 在非对象上调用成员函数 paginate()
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Laravel - Call to a member function paginate() on a non-object
提问by Lovelock
I am making a search page in Laravel and cant understand how to use the paginate method for this usage.
我正在 Laravel 中制作一个搜索页面，但无法理解如何将分页方法用于此用途。
I have used it else where, but not like this.
我在其他地方使用过它，但不是这样。
So my route is:
所以我的路线是：
Route::get('search', function()
{
$q = Input::get('srch-term');
$searchTerms = explode(' ', $q);
$query = DB::table('blogs');
foreach($searchTerms as $term)
{
$query->where('blogtitle', 'LIKE', '%'. $term .'%')
->where('frontpage', '1')
->orderBy('id', 'desc');
}
$results = $query->get()->paginate(15);
$countBuilds = count($results);
return View::make('search', compact('results'));
});
That gives me the error in the title. If i dont use the paginate method then it works fine.
这给了我标题中的错误。如果我不使用 paginate 方法，那么它工作正常。
Where am I going wrong?
我哪里错了？
P.S. As a side note, I know I shouldn't be doing my DB things in the route, still getting my head around how to use functions from controllers etc in routes but thats not for here.
PS 作为旁注，我知道我不应该在路线中做我的数据库事情，仍然在思考如何在路线中使用控制器等的功能，但这不是在这里。
回答by Arryangga Aliev Pratamaputra
Route::get('search', function()
{
$q = Input::get('srch-term');
$searchTerms = explode(' ', $q);
$query = DB::table('blogs');
foreach($searchTerms as $term)
{
$query = $query->where('blogtitle', 'LIKE', '%'. $term .'%')
->where('frontpage', '1')
->orderBy('id', 'desc');
}
//remove ->get()
$results = $query->paginate(15);
$countBuilds = count($results);
return View::make('search', compact('results'));
});
