我在视图刀片中的条件导致我通常使用本机php编码:
$ jlhkriteria = Kriteria :: count();
kriterias表:id_kriteria,kriteria_name
@for ($i = 1; $i <= $jlhkriteria; $i ++)
<?php $queri1 = mysqli_query($mysqli, "SELECT * FROM kriterias WHERE id_kriteria='$i' ORDER BY id_kriteria ASC");
$kriteria1 = mysqli_fetch_array($queri1, MYSQLI_ASSOC); ?>
@for ($j = $i + 1; $j <= $jlhkriteria ; $j ++)
<?php $queri2 = mysqli_query($mysqli, "SELECT * FROM kriterias WHERE id_kriteria='$j' ORDER BY id_kriteria ASC");
$kriteria2 = mysqli_fetch_array($queri2, MYSQLI_ASSOC); ?>
<?php echo $i . $j; ?>
<?php echo $kriteria1['kriteria_name']; ?>
<?php echo $kriteria2['kriteria_name']; ?>
@endfor
@endfor
在Laravel 5中我使用了这段代码,因为我认为View中的查询是不好的想法,任何想法都可以在MVC中进行,而不在Blade视图中查询。
答案 0 :(得分:1)
您需要创建包含业务逻辑的模型和控制器类,然后将数据传递给视图。类似的东西:
型号:
class Kriteria extends Model
{
protected $table = 'kriterias';
}
控制器:
class KriteriaController extends Controller
{
public function index()
{
$kriteria = Kriteria::orderBy('id_kriteria')->get();
return view('index', compact('kriteria'));
}
}
查看:
@for($i=0; $i < count($kriteria); $i ++)
@for($j = $i + 1; $j < count($kriteria) ; $j ++)
{{ $i . $j }}
{{$kriteria[$i]->kriteria_name.' '.$kriteria[$j]->kriteria_name}}
@endfor
@endfor