Question

我在这个话题中停留了一个小时，我没有看到任何我想念的东西。我正试图做一个懒惰的急切负荷：

$p = Project::all();
$p = $p->load('skills', 'user', 'comments', 'suggests');
var_dump($p->skills());

但是这会引发错误：

Call to undefined method Illuminate\Database\Eloquent\Collection::skills()

Answer 1

不要将$ p-＆gt; load（）分配给$ p。

$p = Project::all();
$p->load('skills', 'user', 'comments', 'suggests');
var_dump($p->skills());

Answer 2

您在这里尝试load关系而不是Model，因为$p包含Models的集合，因此此代码无效

$p = Project::all();
$p = $p->load('skills', 'user', 'comments', 'suggests');
var_dump($p->skills()); <-- $p is a collection not a Model

您可以使用循环或使用集合中的索引来指定要加载关系的Model，如：

$p = Project::all();
$p = $p->load('skills', 'user', 'comments', 'suggests');

// Get the first Model from collection and load it's first
// skill from collection of skills objects
dd($p->first()->skills()->first());

或使用索引指定集合中的项目，例如：

$skills = $p->get(2)->skills()->get(); // Get skills from second Project, 1st = 1
dd($skills->get(2)); // Get second skill item from collection of skills

您也可以运行循环

$p = Project::all();
foreach($p as $project) {
    $skills = $project->skills()->get();
    dd($skills->first());
    dd($skills->get(2)) // 2nd item/skill, 1st = 1 (not 0)

    // Or another loop
    foreach($skills as $skill) {
       // Use $skill->field_name
    }
}

Laravel懒得急切的装载

2 个答案: