Question

我目前正在使用大量if / elseif语句过滤数据库中的用户。希望有一种更简单的方法。现在，我正在为每个可能的查询编写具有以下5个值的if语句：

位置
性别
年龄
出生月
出生日期

请注意，我正在使用Laravel 5.7

HTML：按预先填充的选择下拉列表进行过滤

<!-- LOCATION -->
<div>
    <select id="location" name="location">
        <option value="">No Preference</option>
        @foreach ($country as $c)
        <option value="{{ $c->country_name }}">{{ $c->country_name }}</option>
        @endforeach
    </select>
</div>
<!-- GENDER -->
<div>
    <select id="gender" name="gender">
        <option value="">No Preference</option>
        <option value="Male">Male</option>
        <option value="Female">Female</option>
    </select>
</div>
<!-- AGE -->
<div>
    <select id="age" name="age">
        <option value="">No Preference</option>
        <option value="1">1</option>
        <option value="2">2</option>
        ETC...
    </select>
</div>
<!-- BIRTH MONTH -->
<div>
    <select id="month" name="month">
        <option value="">No Preference</option>
        <option value="01">January</option>
        <option value="02">February</option>
        ETC...
    </select>
</div>
<!-- BIRTH DATE -->
<select id="day" name="day">
    <option value="">No Preference</option>
    <option value="1">1</option>
    <option value="2">2</option>
    ETC...
</select>
</div>
<!-- FILTER RESULTS BTN -->
<div>
    <button id="ajaxSubmit">
        Filter Results
    </button>
</div>

JS过滤结果

$('#ajaxSubmit').click(function() {
    console.log('clicked');
    var location = $('#location').val();
    var gender = $('#gender').val();
    var age = $('#age').val()
    var month = $('#month').val()
    var day = $('#day').val()

     $.get('/child-sponsorship/filter?location=' + location + '&gender=' + gender + '&age=' + age + '&month=' + month + '&day=' + day, function(data){
        console.log(data);
        $('#kids_results').empty();
        $('#kids_results').html(data);
    });
});

控制器：此方法有效，但这是if语句的噩梦

细致的滤网可能会发生各种变化...

        public function filter(Request $request){ 

        $location = $request->location;
        $gender = $request->gender;
        $age = $request->age;
        $month = $request->month;
        $day = $request->day;

        $country = DB::table('countries')->orderBy('country_name')->get();

        //IF ALL FILTERS ARE EMPTY
        if(empty($location) && empty($gender) && empty($age) && empty($month) && empty($day)){
            $sponsorKid = Kid::orderBy('id')->get();
        }//IF ALL FILTER ARE TRUE
        elseif(!empty($location) && !empty($gender) && !empty($age) && !empty($month) && !empty($day)){
           $sponsorKid = DB::table('kids')
           ->where('current_country', $location)
           ->where('gender', $gender)
           ->where('age', $age)
           ->where('month', $month)
           ->where('day', $day)
           ->orderby('id')
           ->get(); 
        }//IF LOCATION ONLY IS TRUE
        elseif(empty($gender) && empty($age) && empty($month) && empty($day)){
            $sponsorKid = DB::table('kids')
            ->where('current_country', $location)
            ->orderby('id')->get();
        }//IF GENDER ONLY IS TRUE
        elseif(empty($location) && empty($age) && empty($month) && empty($day))

//etc..... 

        //return Response::json($sponsorKid);
        $view = View::make('child-sponsorship.filter_results', compact('sponsorKid'));
        $view = $view->render();
        return $view;
}

有更简便的方法吗？

我尝试使用LIKE过滤控制器，但没有运气：

   $sponsorKid = Kid::query()
   ->where('countries', 'LIKE', "%{$location}%") 
   ->orWhere('gender', 'LIKE', "%{$gender}%")
   ->orWhere('age', 'LIKE', "%{$age}%")
   ->orWhere('month', 'LIKE', "%{$month}%")
   ->orWhere('day', 'LIKE', "%{$day}%")
   ->orderby('id')
   ->get();

        $view = View::make('child-sponsorship.filter_results', compact('sponsorKid'));
        $view = $view->render();
        return $view;

任何帮助将不胜感激。

Answer 1

尝试使用conditional-clauses

何时查询构建器方法

 $sponsorKid = Kid::when($location,function($q,$location) {
     return  $q->where('countries', 'LIKE', "%."$location."%") 
   })->when($gender,function($q,$gender) {
      return  $q->orWhere('gender', 'LIKE', "%".$gender."%")
   })->when($age,function($q,$age) {
      return  $q->orWhere('age', 'LIKE', "%".$age."%")
   })->when($month,function($q,$month) { 
      return $q->orWhere('month', 'LIKE', "%".$month."%")
   })->when($day,function($q,$day) {
      return $q->orWhere('day', 'LIKE', "%".$day."%")
   })->orderby('id')
   ->get();

Answer 2

我相信您有一个名为Kid.php的儿童表模型，您可以使用该模型启动查询并执行以下操作：

//create a query of Kid model
$sponsorKid = Kid::query();

// add a where clause for location only if location is not empty, 
//otherwise you have the default query
if(!empty($location)) {
    $sponsorKid = $sponsorKid->where('location', $location);
}
// add a where clause for gender only if gender is not empty, 
//otherwise you have the default query, or if this condition and 
//above is true then you will have a query which has 2 wheres 
if(!empty($gender)) {
    $sponsorKid = $sponsorKid->where('gender', $gender);
}

if(!empty($age)) {
    $sponsorKid = $sponsorKid->where('age', $age);
}

if(!empty($month)) {
    $sponsorKid = $sponsorKid->where('age', $age);
}

if(!empty($day)) {
    $sponsorKid = $sponsorKid->where('day', $day);
}

//now that you have a final query pull the records 
$sponsorKid = $sponsorKid->orderBy('id')->get();

$view = View::make('child-sponsorship.filter_results', compact('sponsorKid'));
    $view = $view->render();
    return $view;

在Laravel中过滤没有if语句的数据库

2 个答案: