Laravel。 如何从控制器返回link（）分页

Question

I'm trying to return a pagination from the controller using the link function, but I return an empty value. What could be the problem?

$models= Model::with('table2', 'table3')
->simplePaginate(5);
$str= "";
foreach($models as model){
       $str .= model['name'];
}
return response()->json([
        'rows' => $str,
        'links' => $models->links()
    ], 200);

laravel 5.5

If I return

return $models->appends(['rows' => $str])->links();

I get what I want. But when you try to return an array or a json, the link is empty

Answer 1

Use either the toArray or toJson methods, depending on what you need. For example::

$models = Model::with('table2', 'table3')->paginate(5)->toArray();
return [
    'models' => $models
];