【发布时间】:2020-08-31 04:50:59
【问题描述】:
我编写了一个代码来过滤关系数据,但我没有得到好的结果。这是代码。父查询和同一个查询是独立工作的。
public function get(Request $request, Company $company, Survey $survey)
{
$this->authorize('update', $survey->company);
$search = $request->get('search');
$query = EsAnswer::where(['company_id' => $company->id, 'survey_id' => $survey->id])
->orderByDesc('created_at')
->with(['employee'=> function ($q) use($search) {
$q->where('first_name', 'LIKE', "%{$search}%");
$q->orwhere('last_name', 'LIKE', "%{$search}%");
$q->select('id', 'first_name', 'last_name');
}]);
if ($request->has('start_date')) {
$query->whereDate('created_at', '>=', $request->start_date);
}
if ($request->has('end_date')) {
$query->whereDate('created_at', '<=', $request->end_date);
}
$answers = $query->get()->groupBy('submission_id');
// ->paginate('100');
return $answers;
}
当我使用这个时
{
"search":""
}
我收到了这样的回复:
{
"id": 2,
"company_id": 1,
"employee_id": 1,
"survey_id": 2,
"es_question_id": 1,
"answer": "done",
"submission_id": "1",
"mark_as_read": 1,
"created_at": "2020-08-19 12:18:25",
"updated_at": "2020-08-26 06:55:21",
"employee": {
"id": 1,
"first_name": "Baseapp",
"last_name": "saw"
}
}
当我尝试按另一个表中存在的员工姓名进行搜索时,如果数据存在,则应过滤数据并给出响应,但如果员工姓名或搜索数据不存在,则数据不应给出任何响应。如何处理?
{
"search":"sure"
}
{
"id": 2,
"company_id": 1,
"employee_id": 1,
"survey_id": 2,
"es_question_id": 1,
"answer": "done",
"submission_id": "1",
"mark_as_read": 1,
"created_at": "2020-08-19 12:18:25",
"updated_at": "2020-08-26 06:55:21",
"employee": null
}
这里的数据为空,但为什么显示父数据。请帮助并给出另一个想法来过滤关系数据。
【问题讨论】:
标签: php laravel filter laravel-query-builder