拉拉维尔更新连接表

Question

所以我有 2 个具有 OneToMany 关系的表(项目和project_types(。

public function project_types()
{
return $this->belongsTo('AppModelsProjectType', 'project_type_id');
}
public function projects()
{
return $this->hasMany('AppModelsProject', 'project_type_id');
}

我的项目控制器：

public function update(Request $request, $id)
{
if ($request->isMethod('get')){
$types = ProjectType::all()
->pluck('name');
// dd($types);
return view('projects.form', compact('types'), ['project' => Project::find($id)]);
}
$rules = [
'name' => 'required',
'description' => 'required',
'project_code' => 'required',
];
$validator = Validator::make($request->all(), $rules);
if ($validator->fails())
return response()->json([
'fail' => true,
'errors' => $validator->errors()
]);
$project = Project::find($id);
$project->name = $request->name;
$project->description = $request->description;
$project->project_code = $request->project_code;
$project->project_type_id = $request->project_type_id;
$project->start_p = $request->start_p;
$project->start_r = $request->start_r;
$project->end_p = $request->end_p;
$project->end_r = $request->end_r;
$project->days_p = $request->days_p;
$project->days_r = $request->days_r;
$project->date_requested = $request->date_requested;
$project->save();
return response()->json([
'fail' => false,
'redirect_url' => url('projects')
]);
}

我的表格：

<div class="form-group row">
{!! Form::label("project_types", trans('project.Type'),["class"=>"col-form-label col-md-3 col-lg-2"]) !!}
<div class="col-md-8">
{!! Form::select("project->project_type_id", $types, null, ["class"=>"form-control".($errors->has('project->project_type_id')?" is-invalid":""),'placeholder'=>'Project Type']) !!}
<span id="error-name" class="invalid-feedback"></span>
</div>
</div>

在我的表单视图中，我有一个包含所有项目类型名称的下拉列表(所以我想我的关系很好(，但是当我尝试更新项目信息时，表单会发送 NULL 以获取project_type_id。我在控制台中没有收到任何错误。

谁能帮忙！谢谢！

Answer 1

尝试这样的事情：

public function projects()
{
return $this->hasMany(Project::class, 'project_type_id', 'id');
}
public function project_types()
{
return $this->belongsTo(ProjectType::class, 'project_type_id', 'id');
}

在您的控制器中更改以下内容：

if ($request->isMethod('get')){
$types = ProjectType::all()
->pluck('name');
// dd($types);
return view('projects.form', compact('types'), ['project' => Project::find($id)]);
}

对此：

if ($request->isMethod('get')){
$types = ProjectType::all()
->pluck('name');
$project = Project::find($id);
// dd($types);
return view('projects.form', compact('types', 'project'));
}

根据您的刀片模板，我猜您在Form::select()中有一些错误

Laravel文档表单和HTML

Form::select('select_name', $types->toArray());

Answer 2

好的发现了我的错误。我正在选择项目类型名称，我也需要 id。

$types = ProjectType::all()
->pluck('name', 'id');
return view('projects.form', compact('types', 'project'));

并以以下形式：

{!! Form::label("project_types", trans('project.Type'),["class"=>"col-form-label col-md-3 col-lg-2"]) !!}
<div class="col-md-8">
{!! Form::select("project_type_id", $types, null, ["class"=>"form-control".($errors->has('project_type_id')?" is-invalid":""),'placeholder'=>'Project Type']) !!}
<span id="error-name" class="invalid-feedback"></span>
</div>