英文:
MySQL returns results, Laravel Eloquent returns empty
问题
这让我抓狂。我有一个非常简单的查询，在MySQL中运行正确，但在Laravel Eloquent中没有返回结果：
Laravel代码没有返回结果。
$feedback = FeaturedFeedback::where("company_id", "=", $company->id);
if ($request->category != "") {
$category = Category::where("name", "=", $request->category)->first();
}
if ($category) {
$feedback->where("category_id", "=", $category->id);
}
$feedback->get();
当我查看查询日志时，这是运行的查询：
select * from `featured_feedback` where `company_id` = 8 and `category_id` = 1
当我直接在数据库上运行该查询时，它返回了4个结果。这让我很烦恼。这只涉及一个表，SQL严格模式为false，没有连接，查询也很干净。有什么想法吗？
英文:
This is driving me nuts. I have a super simple query that runs correctly in MySQL but returns no results in Laravel Eloquent:
Laravel Code that returns no results.
$feedback = FeaturedFeedback::where("company_id", "=", $company->id);
if ($request->category != "") {
$category = Category::where("name", "=", $request->category)->first();
}
if ($category) {
$feedback->where("category_id", "=", $category->id);
}
$feedback->get();
When I look at the query logs, this is the query that runs:
select * from `featured_feedback` where `company_id` = 8 and `category_id` = 1
When I run that query directly on the DB, it returns 4 results. Driving me nuts. It's one table, SQL strict mode is false, no joins, clean query. Thoughts??
答案1
得分: 0
您正在将多个条件链接到`$feedback`查询构建器实例上，但未将最终结果捕获回`$feedback`变量中。
$feedbackQuery = FeaturedFeedback::where("company_id", "=", $company->id);
if ($request->category != "") {
$category = Category::where("name", "=", $request->category)->first();
}
if ($category) {
$feedbackQuery->where("category_id", "=", $category->id);
}
$feedback = $feedbackQuery->get();
英文:
You're chaining multiple conditions onto the $feedback query builder instance but not capturing the final result back into the $feedback variable.
$feedbackQuery = FeaturedFeedback::where("company_id", "=", $company->id);
if ($request->category != "") {
$category = Category::where("name", "=", $request->category)->first();
}
if ($category) {
$feedbackQuery->where("category_id", "=", $category->id);
}
$feedback = $feedbackQuery->get();
答案2
得分: 0
在“FeaturedFeedback”模型中添加一个关联：
public function category(){
return $this->belongsTo(Category::class);
}
使用when条件代替if条件：
$feedback = FeaturedFeedback::where("company_id", $company->id)->when(!empty($request->category), function ($query) use ($request) {
$query->whereHas('category', function ($query) use ($request) {
$query->where("name", $request->category);
});
})->get();
英文:
In the "FeaturedFeedback" model add a relationship
public function category(){
return $this->belongsTo(Category::class);
}
use when conditioning instead of if condition
$feedback = FeaturedFeedback::where("company_id",$company->id)->when(!empty($request->category),function ($query)use($request){
$query->whereHas('category',function ($query)use($request){
$query->where("name",$request->category);
});
})->get();
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