英文:
Laravel Eloquent How to apply condition on "with" datas on the output?
问题
以下是您的代码的翻译部分：
$prestations = Prestation::with(['service' => function($service) use($searchService) {$service->select(['id','name'])->where('name', 'regexp', "/$searchService/i");},'facility' => function($facility) use($searchPartenaire) {$facility->select(['id','name'])->where('name', 'regexp', "/$searchPartenaire/i");}])->where('name', 'regexp', "/$search/i")->orderBy($orderBy, $orderDirection)->simplePaginate(50);$res = ['results' => $prestations,'total' => Prestation::all()->count(),];
问题在于输出中，当“service”和“facility”名称与$searchService和$searchPartenaire不相等时，这些值会被替换为"null"。因此，我不希望在输出中有与搜索变量不相等的值。感谢您。
英文:
here my code :
$prestations = Prestation::with(['service' => function($service) use($searchService) {$service->select(['id','name'])->where('name', 'regexp', "/$searchService/i");},'facility' => function($facility) use($searchPartenaire) {$facility->select(['id','name'])->where('name', 'regexp', "/$searchPartenaire/i");}])->where('name', 'regexp', "/$search/i")->orderBy($orderBy, $orderDirection)->simplePaginate(50);$res = ['results' => $prestations,'total' => Prestation::all()->count(),];
The problem is that in the output of all datas where "service" and "facility" names are not equal on the $searchService and $searchPartenaire the values are replaced by "null".
So i don't want to have values in the output where the search variables are not equals.
Thank you.
答案1
得分: 1
1st create instances of Prestation $prestations = Prestation::with('service','facility')
then apply the condition; this is a good approach in search.
英文:
you can try like this
$prestations = Prestation::with('service','facility');$prestations->whereHas('service', function ($query) use ($searchPartenaire) {$query->Where('name', 'like', '%' . $searchPartenaire . '%');});$prestations->whereHas('facility', function ($query) use ($searchPartenaire) {$query->Where('name', 'like', '%' . $searchPartenaire . '%');});$prestations->where('name', 'like', '%'.$search.'%')->orderBy($orderBy, $orderDirection)->simplePaginate(50);return $res = ['results' => $prestations,'total' => Prestation::all()->count(),];
1st create instances of Prestation $prestations = Prestation::with('service','facility')
then apply the condtion this is good approach in seach
答案2
得分: 0
这是Kamlesh Paul建议后的我的代码：
$prestations = Prestation::with('service','facility');$prestations->whereHas('service', function ($query) use ($searchService) {$query->where('name', 'regexp', "/$searchService/i");});$prestations->whereHas('facility', function ($query) use ($searchPartenaire) {$query->where('name', 'regexp', "/$searchPartenaire/i");});$prestations->where('name', 'regexp', "/$search/i")->orderBy($orderBy, $orderDirection)->simplePaginate(50);$res = ['results' => $prestations,'total' => Prestation::all()->count(),];return $res;
但是存在无限的HTTP请求调用问题，我认为问题可能出在当where条件找不到相等的名称时。有人有建议吗？谢谢。
英文:
Here my code after Kamlesh Paul suggestion :
$prestations = Prestation::with('service','facility');$prestations->whereHas('service', function ($query) use ($searchService) {$query->where('name', 'regexp', "/$searchService/i");});$prestations->whereHas('facility', function ($query) use ($searchPartenaire) {$query->where('name', 'regexp', "/$searchPartenaire/i");});$prestations->where('name', 'regexp', "/$search/i")->orderBy($orderBy, $orderDirection)->simplePaginate(50);$res = ['results' => $prestations,'total' => Prestation::all()->count(),];return $res;
But there is an infinite calls of http request, i think that the problem is when the where don't find an equal name, anyone have a suggest ?
Thank's.
答案3
得分: 0
我终于找到了一个非常相似的解决方案：
$prestations = Prestation::with('service', 'facility')->whereHas('service', function ($query) use ($searchService) {$query->where('name', 'regexp', "/$searchService/i");})->whereHas('facility', function ($query) use ($searchPartenaire) {$query->where('name', 'regexp', "/$searchPartenaire/i");})->where('name', 'regexp', "/$search/i")->orderBy($orderBy, $orderDirection)->simplePaginate(50);$res = ['results' => $prestations,'total' => Prestation::all()->count(),];感谢您的帮助。
英文:
I finnaly found a solution very similar :
$prestations = Prestation::with('service','facility')->whereHas('service', function ($query) use ($searchService) {$query->where('name', 'regexp', "/$searchService/i");})->whereHas('facility', function ($query) use ($searchPartenaire) {$query->where('name', 'regexp', "/$searchPartenaire/i");})->where('name', 'regexp', "/$search/i")->orderBy($orderBy, $orderDirection)->simplePaginate(50);$res = ['results' => $prestations,'total' => Prestation::all()->count(),];
Thank's for your help.
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论