Question

我需要根据用户的状态重定向用户页面。我在Laravel 5.2中的My Controller中编写了一个方法，其中状态1正在工作但是当我添加else时，如果函数它在errror之后生成。

FatalErrorException in ProjectCollaboratorsController.php line 219: syntax error, unexpected 'return' (T_RETURN)

我的控制器方法如下

public function show($id){
 if (Permission::where('status', 1)->where('project_id', $id)->exists())
 {

    $project = Project::find($id);
    $tasks = $this->getTasks($id);
    $files = $this->getFiles($id);
    $comments = $this->getComments($id);
    $collaborators = $this->getCollaborators($id);
    $permissions = $this->getPermissions($id);



    return view('collaborators.show');
    }
 else
 {


    (Permission::where('status', 2)->where('project_id', $id)->exists()) 
         return view('collaborators.manager');

  }

  else if 
  {

     (Permission::where('status', 3)->where('project_id', $id)->exists()) 
         return view('collaborators.user');
  }
}

Answer 1

您的语法完全错误，请尝试使用此结构实现..

public function yourFunctionName()
{
    if(condition1)
    {
            code
    }
    elseif(condition2)
    {
        code
    }
    else
    {
      code
    }
}

Answer 2

试试这段代码，你会很高兴

public function show($id){

    if (Permission::where('status', 1)->where('project_id', $id)->exists())
    {
        $project = Project::find($id);
        $tasks = $this->getTasks($id);
        $files = $this->getFiles($id);
        $comments = $this->getComments($id);
        $collaborators = $this->getCollaborators($id);
        $permissions = $this->getPermissions($id);
        return view('collaborators.show');
    }
     else if(Permission::where('status', 2)->where('project_id', $id)->exists())
    {     
         return view('collaborators.manager');
    }
    else if (Permission::where('status', 3)->where('project_id', $id)->exists()) 
    {     
         return view('collaborators.user');
    }
}

如何在Laravel 5.2中重定向不同的页面

2 个答案: