首页 > 解决方案 > Laravel 6未定义变量:工作

问题描述

我认为没有错误,但 Laravel 显示以下错误:

未定义变量:工作

public function jobCreate(Request $request){

    $data['jobs'] = DB::table('jobs')->get();
    $job_category = JobCategory::all();
    // $job_locations = JobLocation::all();
    if(Auth::user()->user_type == 'admin'){
        return view('backend.job_circuler.create',compact('job_category',$data));
    }
    elseif(Auth::user()->user_type == 'customer'){
        return view('frontend.user.job_circuler.create', compact('job','job_category'));
    }
    else {
        abort(404);
    }

}

我正确导入了所有这些东西,但没有工作。

                <div class="form-group row" id="location">
                    <label class="col-md-3 col-from-label">
                        {{translate('Location')}}
                        <span class="text-danger">*</span>
                    </label>
                    <div class="col-md-9">
                        <select class="form-control aiz-selectpicker" name="location_id" id="location_id" data-live-search="true" >
                            <option >select location</option>
                            @foreach($jobs as $job)
                            <option value="{{ $job->location_id }}">
                                {{ $job->joblocation->location  }}
                            </option>
                            @endforeach
                        </select>
                        @error('location')
                 <div class="alert alert-danger">{{ $message }}</div>
                            @enderror
                    </div>
                </div>
             

仍然显示此错误

标签: laraveleloquentlaravel-6laravel-controllerundefined-variable

解决方案


根据错误消息,您发布的视图是backend.job_circuler.create. 我对吗?

错误说,变量$jobs丢失并且是正确的:您$job_category作为字符串传递,但$data作为数组传递。您确实可以将数组传递给,compact()但前提是它们包含变量名而不是实际变量。你为什么还要把$jobs里面包起来$data

$jobs = DB::table('jobs')->get();
$job_category = JobCategory::all();
if(Auth::user()->user_type == 'admin'){
    return view('backend.job_circuler.create', compact('job_category', 'jobs'));
}

当然,反之亦然:

$data['jobs'] = DB::table('jobs')->get();
$data['job_category'] = JobCategory::all();
if(Auth::user()->user_type == 'admin'){
    return view('backend.job_circuler.create', $data);
}

或者,如果您出于某种原因不想更改 if 上方的代码,您仍然可以这样做:

$data['jobs'] = DB::table('jobs')->get();
$job_category = JobCategory::all();
if(Auth::user()->user_type == 'admin'){
    return view('backend.job_circuler.create', [ ...$data, 'job_category' => $job_category ]);
}

或这个:

$data['jobs'] = DB::table('jobs')->get();
$job_category = JobCategory::all();
if(Auth::user()->user_type == 'admin'){
    return view('backend.job_circuler.create', [ ...$data, ...compact('job_category') ]);
}

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