Question

我有一个简单的问题，因为我不知道应该将哪条线添加到laravel分页中。（-> paginate（5）。

public function index(Request $request)
{

    $codeSearch = $request->get('code');
    $descriptionSearch = $request->get('description');

    //$tmp = Category::all()->toArray();
    $tmp = Category::where('code','like','%' .$codeSearch. '%')->where('description','like','%' .$codeSearch. '%')->get()->toArray();

    $category = array();
    foreach ($tmp as $key => $row) {
        $policy = Category::find($row['parent_id']);

        $tmpResult = new Category();
        $tmpResult->id = $row['id'];
        $tmpResult->code = $row['code'];
        $tmpResult->description = $row['description'];
        $tmpResult->parent_id = $policy['description'];
        $tmpResult->status = $row['status'];

        array_push($category, $tmpResult);
    }

    return view('category.index', compact('category'));
}

Answer 1

无法在集合上调用分页，因此您必须在查询上运行它，只需将->get()替换为->paginate(5)即可

$tmp = Category::where('code','like','%' .$codeSearch. '%')
    ->where('description','like','%' .$codeSearch. '%')
    ->paginate(5)
    ->toArray();

Answer 2

以下->get(), ->first(), all()从数据库中获取结果，-> paginate（5）也是如此，因此，我建议您将->get()替换为paginate(5)，这样就可以摆脱toArray()，因为结果将是一个集合，您可以使用该集合使用foreach()或按索引获取值。

Answer 3

当您需要一个条件数组时，只需将最终数组转换为集合对象并使用分页概念即可。

$items = [
            'item1',
            'item2',
            'item3',
            'item4',
            'item5',
            'item6',
            'item7',
            'item8',
            'item9',
            'item10'
            ];

        // Get current page form url e.x. &page=1
        $currentPage = LengthAwarePaginator::resolveCurrentPage();

        // Create a new Laravel collection from the array data
        $itemCollection = collect($items);

        // Define how many items we want to be visible in each page
        $perPage = 1;

        // Slice the collection to get the items to display in current page
        $currentPageItems = $itemCollection->slice(($currentPage * $perPage) - $perPage, $perPage)->all();

        // Create our paginator and pass it to the view
        $paginatedItems= new LengthAwarePaginator($currentPageItems , count($itemCollection), $perPage);

        // set url path for generted links
        $paginatedItems->setPath($request->url());

        return view('items_view', ['items' => $paginatedItems]);

Answer 4

尝试

 $tmp = Category::where('code','like','%' .$codeSearch. '%')->where('description','like','%' .$codeSearch. '%')->paginate(5);

可见

 @foreach($tmp as $tm)
    //whatever operation you like to do
 @endforeach
 {{$tmp->links()}}

Answer 5

为Category的父母建立关系可能会更容易。然后，您可以在检索类别时加载父项。

class Category extends Model
{
    ...
    public function parent()
    {
        return $this->belongsTo(self::class);
    }
}

我觉得您的搜索可能正在尝试进行OR WHERE，在code字段中搜索该值或description：

$categories = Category::with('parent')
    ->where('code', 'like', '%' .$codeSearch. '%')
    ->orWhere('description', 'like', '%' .$codeSearch. '%')
    ->paginate(5);

然后在您的视图中，如果您想要父类别的描述，则可以通过以下关系获取它：

@foreach ($categories as $category)
    parent description: {{ $category->parent->description }}
@endforeach

Laravel分页有问题

5 个答案: