Laravel 在同一个 table 中返回 parents 的 children
Laravel returning children of parents in same table
使用 Laravel 5.1,我正在尝试从 MySQL 类别 table 创建菜单列表。我的服务提供商 returns 数据,但我不明白如何在 foreach 循环中创建 child 类别。当我执行循环时,只返回 child 查询的最后一行。任何指导将不胜感激。
类别Table
id | cat_name | cat_parent_id
--- | --------------| -------------
1 | Parent Cat 1 | NULL
2 | Parent Cat 2 | NULL
3 | Child Cat 1 | 2
4 | Child Cat 2 | 2
5 | Parent Cat 3 | NULL
6 | Child Cat 3 | 5
想要的结果
Parent Cat 1
Parent Cat 2
Child Cat 1
Child Cat 2
Parent Cat 3
Child Cat 3
viewComposerServiceProvider.php
public function boot()
{
$this->composeTopCategoryNavigation();
$this->composeSubCategoryNavigation();
}
private function composeTopCategoryNavigation()
{
view()->composer('partials.header', function($view)
{
$view->with('top_cats', Category::whereNull('cat_parent_id')->orderBy('cat_name', 'asc')->get());
});
}
private function composeSubCategoryNavigation()
{
view()->composer('partials.header', function($view)
{
$view->with('sub_cats', Category::whereNotNull('cat_parent_id')->orderBy('cat_name', 'asc')->get());
});
}
header 查看
<ul>
@foreach ($top_cats as $top_cat)
<?php $top_cat_slug = str_slug( $top_cat->cat_name, "-"); ?>
<li>{{ $top_cat->cat_name }}
@foreach ($sub_cats as $sub_cat)
@if ( $sub_cat->cat_parent_id === $top_cat->id )
<ul>
<li{{ $sub_cat->cat_name }}</li>
</ul>
@endif
@endforeach
</li>
@endforeach
</ul>
首先,你做的事情效率低下。您的视图遍历每个父类别的所有子类别。如果您正确定义关系并利用 Eloquent 的预先加载,您可以更轻松地获取和访问子类别。
从定义关系开始:
class Category extends Model {
//each category might have one parent
public function parent() {
return $this->belongsToOne(static::class, 'cat_parent_id');
}
//each category might have multiple children
public function children() {
return $this->hasMany(static::class, 'cat_parent_id')->orderBy('cat_name', 'asc');
}
}
一旦正确定义了关系,就可以像下面这样获取整个类别树:
view()->composer('partials.header', function($view) {
$view->with('categories', Category::with('children')->whereNull('cat_parent_id')->orderBy('cat_name', 'asc')->get());
});
不需要第二个作曲家,因为父类别已经包含子类别。
现在,您只需在视图中显示类别:
<ul>
@foreach ($categories as $parent)
<li>{{ $parent->cat_name }}
@if ($parent->children->count())
<ul>
@foreach ($parent->children as $child)
<li>{{ $child->cat_name }}</li>
@endforeach
</ul>
@endif
</li>
@endforeach
</ul>
我自己找到了解决方案。
我必须遍历结果才能访问成员变量。
@foreach($locations as $location)
<tr>
<td>
{{$location->location_id}}
</td>
<td>
@foreach($location->members as $member)
{{$member->first_name}}
@endforeach
</td>
<td>
</td>
</tr>
@endforeach
使用 Laravel 5.1,我正在尝试从 MySQL 类别 table 创建菜单列表。我的服务提供商 returns 数据,但我不明白如何在 foreach 循环中创建 child 类别。当我执行循环时,只返回 child 查询的最后一行。任何指导将不胜感激。
类别Table
id | cat_name | cat_parent_id
--- | --------------| -------------
1 | Parent Cat 1 | NULL
2 | Parent Cat 2 | NULL
3 | Child Cat 1 | 2
4 | Child Cat 2 | 2
5 | Parent Cat 3 | NULL
6 | Child Cat 3 | 5
想要的结果
Parent Cat 1
Parent Cat 2
Child Cat 1
Child Cat 2
Parent Cat 3
Child Cat 3
viewComposerServiceProvider.php
public function boot()
{
$this->composeTopCategoryNavigation();
$this->composeSubCategoryNavigation();
}
private function composeTopCategoryNavigation()
{
view()->composer('partials.header', function($view)
{
$view->with('top_cats', Category::whereNull('cat_parent_id')->orderBy('cat_name', 'asc')->get());
});
}
private function composeSubCategoryNavigation()
{
view()->composer('partials.header', function($view)
{
$view->with('sub_cats', Category::whereNotNull('cat_parent_id')->orderBy('cat_name', 'asc')->get());
});
}
header 查看
<ul>
@foreach ($top_cats as $top_cat)
<?php $top_cat_slug = str_slug( $top_cat->cat_name, "-"); ?>
<li>{{ $top_cat->cat_name }}
@foreach ($sub_cats as $sub_cat)
@if ( $sub_cat->cat_parent_id === $top_cat->id )
<ul>
<li{{ $sub_cat->cat_name }}</li>
</ul>
@endif
@endforeach
</li>
@endforeach
</ul>
首先,你做的事情效率低下。您的视图遍历每个父类别的所有子类别。如果您正确定义关系并利用 Eloquent 的预先加载,您可以更轻松地获取和访问子类别。
从定义关系开始:
class Category extends Model {
//each category might have one parent
public function parent() {
return $this->belongsToOne(static::class, 'cat_parent_id');
}
//each category might have multiple children
public function children() {
return $this->hasMany(static::class, 'cat_parent_id')->orderBy('cat_name', 'asc');
}
}
一旦正确定义了关系,就可以像下面这样获取整个类别树:
view()->composer('partials.header', function($view) {
$view->with('categories', Category::with('children')->whereNull('cat_parent_id')->orderBy('cat_name', 'asc')->get());
});
不需要第二个作曲家,因为父类别已经包含子类别。
现在,您只需在视图中显示类别:
<ul>
@foreach ($categories as $parent)
<li>{{ $parent->cat_name }}
@if ($parent->children->count())
<ul>
@foreach ($parent->children as $child)
<li>{{ $child->cat_name }}</li>
@endforeach
</ul>
@endif
</li>
@endforeach
</ul>
我自己找到了解决方案。 我必须遍历结果才能访问成员变量。
@foreach($locations as $location)
<tr>
<td>
{{$location->location_id}}
</td>
<td>
@foreach($location->members as $member)
{{$member->first_name}}
@endforeach
</td>
<td>
</td>
</tr>
@endforeach