Question

我有一个具有以下关系的WorkPackage模型。

public function work_items()
{
    return $this->hasMany(WorkItem::class);
}

WorkPackage具有“标题”，“ project_id”列，而work_items具有“标题”，“ work_package_id”列。现在我要搜索用户键入的关键字是否与两个表中的标题都匹配。

这是我尝试过的功能

public function getWorkPackageProposals($projectId, $title)
{

    $result['data'] = $this->workPackage->with(['work_items' => function ($query) {
        $query->where('title', 'LIKE', "%{$title}%");
    }])
        ->where('project_id', $projectId)
        ->where('title', 'LIKE', "%{$title}%")
        ->get();
    return $result;
}

，但不起作用。请找到以下我用于创建WorkPackage对象的代码。

public function __construct(WorkPackage $workPackage)
{
    $this->workPackage = $workPackage;
    $this->setModel(WorkPackage::class);
    $this->workItemRepository = app(WorkItemRepository::class);
}

Answer 1

whereHas()用于为相关模型指定其他过滤器以进行检查：

$result['data'] = $this->workPackage->whereHas('work_items', function ($query) use ($title) {
        $query->where('title', 'LIKE', "%{$title}%");
    })
    ->where('project_id', $projectId)
    ->where('title', 'LIKE', "%{$title}%")
    ->get();
return $result;

Answer 2

以下代码对我有用。感谢大家的帮助，以得到正确的答案。

   $result['data'] = $this->workPackage->with(['work_items' => function ($q) use ($title) {
        $q->where('title', 'LIKE', "%{$title}%");
    }])
        ->where('project_id', $projectId)
        ->where('title', 'LIKE', "%{$title}%")
        ->get();
    return $result;

Answer 3

我认为您的问题将通过join方法解决：

$result = $this->workPackage
->join('work_items', 'work_packages.id', '=', 'work_items.package_id')
->where('work_items.title', 'LIKE', '%{$term}%')
->orWhere('work_packages.title', 'LIKE', '%{$term}%')
->get();

Laravel与关系中的条件相似

3 个答案: