Laravel高级搜索查询修复
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我有一个带有多个输入和选择框的搜索表单我需要帮助才能得到我的查询中的条件,以便每个部分单独工作并且一次完成。
这是我的刀片代码:
<form action="{{route('advancesearch')}}" method="post">
{{csrf_field()}}
<div class="sidebar-title">
<span>Advanced Search</span>
<i class="fa fa-caret-down show_sidebar_content" aria-hidden="true"></i>
</div>
<!-- ./sidebar-title -->
<div id="tags-filter-content" class="sidebar-content">
<div class="filter-tag-group">
@foreach($options as $option)
<div class="tag-group">
<p class="title">
<span class="filter-title show_filter_content">{{$option->title}} <span class="pull-right"><i class="fa fa-minus"></i></span></span>
</p>
<div class="filter-content">
<div class="checkbox">
@foreach($option->suboptions as $suboption)
<label for="suboptions">
<input name="suboptions[]" type="checkbox" value="{{$suboption->id}}">
{{ucfirst($suboption->title)}}
</label>
@endforeach
</div>
</div>
</div>
@endforeach
<!-- ./tag-group -->
<div class="tag-group">
<p class="title">
<span class="filter-title show_filter_content">Brand <span class="pull-right"><i class="fa fa-minus"></i></span></span>
</p>
<div class="filter-content">
<div class="checkbox">
@foreach($brands as $brand)
<label for="brands">
<input name="brands[]" type="checkbox" value="{{$brand->id}}">
{{$brand->title}}
</label>
@endforeach
</div>
</div>
</div>
<!-- ./tag-group -->
<div class="tag-group">
<p class="title">
<span class="filter-title show_filter_content">Price Range <span class="pull-right"><i class="fa fa-minus"></i></span></span>
</p>
<div class="row filter-content">
<div class="col-md-6">
<div class="form-group">
<label for="min_price" hidden>Min</label>
<input type="text" name="min_price" class="form-control" placeholder="Rp Min">
</div>
</div>
<div class="col-md-6">
<div class="form-group">
<label for="max_price" hidden>Max</label>
<input type="text" name="max_price" class="form-control" placeholder="Rp Max">
</div>
</div>
</div>
</div>
<!-- tag-group -->
<div class="text-center mt-20">
<button type="submit" class="btn btn-danger">TERPAKAN</button>
</div>
</div><!-- ./filter-tag-group -->
</div><!-- ./sidebar-content -->
</form>
这是我的路线:
Route::post('/advanced-search', 'frontendSearchController@filter')->name('advancesearch');
最后我的功能代码是:
public function advancedsearch(Request $request) {
$brands = Brand::all(); // uses for other part of the page. (not related to search function)
$options = Option::all(); // uses for other part of the page. (not related to search function)
$suboptions = DB::table('product_suboption'); // where my product_id and subopyion_id saves
//search function
$products = Product::where(function($query){
//getting inputs
$suboptions2 = Input::has('suboptions') ? Input::get('suboptions') : [];
$min_price = Input::has('min_price') ? Input::get('min_price') : null;
$max_price = Input::has('max_price') ? Input::get('max_price') : null;
$brands2 = Input::has('brands') ? Input::get('brands') : [];
//returning results
$query->where('price','>=',$min_price)
->where('price','<=',$max_price);
})->get();
return view('front.advancesearch', compact('products', 'brands', 'options'));
}
My models relations:
product
型号:
public function options(){
return $this->belongsToMany(Option::class);
}
public function suboptions(){
return $this->belongsToMany(Suboption::class, 'product_suboption', 'product_id', 'suboption_id');
}
public function brand(){
return $this->belongsTo(Brand::class);
}
Option
型号:
public function suboptions(){
return $this->hasMany(Suboption::class, 'option_id');
}
public function products(){
return $this->belongsToMany(Product::class);
}
Suboption
型号:
public function option(){
return $this->belongsTo(Option::class, 'option_id');
}
public function products(){
return $this->belongsToMany(Product::class);
}
Brand
型号:
public function products(){
return $this->hasMany(Product::class);
}
note
我的brands
搜索来自产品表,我有每个产品的列brand_id
。
但
我的suboptions
来自名为product_suboption
的第3张桌子(正如你在我的模型代码中看到的那样)我保存product_id
和suboption_id
。
这只是为了提出一个想法。你可以使用多个->where()
和渴望加载->with()
为您的查询。看看下面的查询:
$products = Product::where('price', '>=', $min_price) // you get the max and min price
->where('id', '<=', $max_price)->select('id')
->with([
"brand" => function ($query) {
$query->whereIn('id', $brand_ids); // [1, 2, 3,...]
},
"specifications" => function ($query) {
$query->where('some_column', '=', 'possible-value'); // single condition
},
"specifications.subspecifications" => function ($query) {
$query->where([
'some_column' => 'possible-value',
'another_column' => 'possible-value'
]); // you can also pass arrays of condition
}
])->get(); // This will return the products with the price set by the user
// Since we're just using ->with(), this will also return those products
// that doesn't match the other criteria specifications) so we
// still need to filter it.
最后,您可以过滤与specifications
匹配的产品, - product
与空specifications
意味着该产品与标准不符,因此我们必须将其从集合中删除。
$filtered = $products->filter(function ($product, $key) {
return count($product->brand) > 0 && count($product->specifications) > 0;
// add your other boolean conditions here
});
dd($filtered->toArray()); // your filtered products to return
您可以使用laravel orWhere
和orWhereHas
单独获得结果并且一次性完成,假设您没有选择min_price
和max_price
但是您选择了brand
然后所有带有此brnad的产品应该返回,您的查询将如下所示
$products = Product::orWhere('price','>=',$min_price)
->orWhere('price','<=',$max_price)
->orWhereHas('brand',function($query){
$query->whereIn('id', $brand_ids);
})
->orWhereHas('suboptions',function($query){
$query->whereIn('id', $suboptions_ids);
})
->orWhereHas('subspecifications',function($query){
$query->whereIn('id', $subspecifications_ids);
})->get();
$products
将有产品集合如果上述查询中陈述的任何条件匹配。
希望这可以帮助。
这是我怎么做的。注意使用when
来简化可选的where条件(不需要设置变量),以及用于约束whereHas
和with
的闭包(如果你想加载关系)。
$products = Product::query()
->when($request->min_price, function ($query, $min_price) {
return $query->where('price', '>=', $min_price);
})
->when($request->max_price, function ($query, $max_price) {
return $query->where('price', '<=', $max_price);
})
->when($request->suboptions, function ($query, $suboptions) {
$suboptionsConstraint = function ($q) use ($suboptions) {
return $q->whereIn('id', $suboptions);
};
return $query->whereHas('suboptions', $suboptionsContraint)
->with(['suboptions' => $suboptionsContraint]);
})
->when($request->brands, function ($query, $brands) {
$brandsConstraint = function ($q) use ($brands) {
return $q->whereIn('id', $brands);
};
return $query->whereHas('brands', $brandsConstraint)
->with(['brands' => $brandsConstraint]);
});
我建议使用每个分离,它帮助您轻松操作代码
作为典型条件,您的sub_option来自第三个表最后一个关系。
if(count($request['suboptions'])) {
$product->whereHas('options',function($options) use ($request) {
$options->whereHas('suboptions',function($suboption)use($request) {
$suboption->whereIn('id',$request['suboptions']);
});
});
}
对于最低价格最高价格我在产品表中假设您的价格
if(! empty($request['min_price'])) {
$product->where('price','>=',$request['min_price']);
}
if(! empty($request['max_price'])) {
$product->where('price','<=',$request['max_price']);
}
对于品牌,你说产品表中的brand_id列然后
if(count($request['brands'])) {
$product->whereIn('brand_id',$request['brands']);
}
我建议采用不同的方法。
在您的控制器上,将其更改为:
public function advancedsearch(Request $request) {
$suboptions2 = request->suboptions ? request->suboptions : null;
$min_price = request->min_price ? request->min_price : null;
$max_price = request->max_price ? request->max_price : null;
$brands2 = request->brands ? request->brands : null;
$query = Product::select('field_1', 'field_2', 'field_3')
->join('brands as b', 'b.id', '=', 'products.brand_id')
...(others joins);
// here we do the search query
if($suboptions2){
$query->where('suboptions_field', '=', $suboptions);
}
if($min_price && $max_price){
$query->where(function($q2) {
$q2->where('price', '>=', $min_price)
->where('price', '<=', $max_price)
});
}
if($brands2){
$query->where('products.brand_id', '=', $brands2);
}
// others queries
// finish it with this
$query->get();
return view('front.advancesearch', compact('products', 'brands', 'options'));
我发现这样做非常有用,因为它可以非常容易地实现其他查询。
这是我使用laravel eloquent和多个输入进行搜索的方法:
$input = Input::all(); //group all the inputs into single array
$product = Product::with('options','suboptions','brand');
//looping through your input to filter your product result
foreach ($input as $key => $value)
{
if ($value!='') {
if ($key == "max_price")
$product = $product->where('price','<=', $value);
elseif ($key == "min_price")
$product = $product->where('price','>=', $value);
elseif ($key == "brands")
$product = $product->whereIn('brand_id', $value); //assuming that your Input::get('brands') is in array format
elseif ($key == "suboptions")
$product = $product->whereIn('suboption_id', $value);
}
}
$product = $product->get();
如果没有提交输入,上面的方法将返回所有产品,并将根据输入过滤结果(如果可用),此外,在继续查询之前,使用验证清理输入也是一种很好的做法。
SOLVED
经过数周的代码播放后,我为自己找到了正确的结果(在我的情况下,它可以通过这种方式为其他人工作,也可以与其他建议的答案一起工作)
public function advancedsearch(Request $request) {
$options = Option::all();
$brands = Brand::all();
$brandss = Input::has('brands') ? Input::get('brands') : [];
$suboption = Input::has('suboptions') ? (int)Input::get('suboptions') : [];
$min_price = Input::has('min_price') ? (int)Input::get('min_price') : null;
$max_price = Input::has('max_price') ? (int)Input::get('max_price') : null;
//codes
if(count($request['suboptions'])){
$products = DB::table('products')
->join('product_suboption', function ($join) {
$suboption = Input::has('suboptions') ? Input::get('suboptions') : [];
$join->on('products.id', '=', 'product_suboption.product_id')
->where('product_suboption.suboption_id', '=', $suboption);
})
->paginate(12);
}
elseif(count($request['brands'])){
$products = DB::table('products')
->whereIn('products.brand_id', $brandss)
->paginate(12);
}
elseif(count($request['min_price']) && count($request['max_price'])){
$products = DB::table('products')
->whereBetween('price', [$min_price, $max_price])
->paginate(12);
}
return view('front.advancesearch', compact('products', 'brands', 'options')以上是关于Laravel高级搜索查询修复的主要内容,如果未能解决你的问题,请参考以下文章