Question

我要提出请假申请，老板可以在其中接受或拒绝请假请求

我在数据库中的表是这样的

id |    name     |      email           | person_in_charge |
1      Michael        michael.com                2
2      Johan          johan.com                  4
3      Lorem          lorem.com                  2
4      Ipsum          ipsum.com                  5
5      Dolor          dolor.com

我想，如果用户ID为5，则他可以查看所有请求。如果用户ID 2登录，则他可以查看用户ID 1和3的请假请求。如果用户ID 4登录，则可以查看用户ID 2，用户ID 1和用户ID 3的请假请求。

我正在使用while循环，但是出现错误消息“试图获取非对象的属性”

这是代码

    $id='5';

    $query= Users::where('id',$id)->first();

    if($query->pic_for==null)
    {
        $query = Users::where('pic_for',$id)->first();
        $pic_for = $query->id;

        $pic=array($query->id);
        while($pic_for!=null)
        {
            $query2 = Users::where('pic_for',$pic_for)->first();
            //dd($query2);
            if($query2->pic_for==null)
            {
                break;
            }else
            {
                $pic[]=$query2->id;
                $pic_for = $query2->id;
            }

        }
        dd($pic);
    }

你知道我想念哪里吗？

谢谢

Answer 1

我会尝试的：

// prepare an empty $result collection that will store the ids of the users
$result = collect();

// probably a number, not a string, in your database: we should keep the right type
$id = 5;
addRelatedUsersToResult($id, &$result);

dd($result);

// add all the related users ids to the result 
// and call itself for each realted user (to check their relations)
function addRelatedUsersToResult($id, &$result) {
    $relatedUsers = Users::where('pic_for', $id)->get();

    // add the related users and checks their children
    foreach ($relatedUsers as $relatedUser) {
        $result->push($relatedUser->id);
        addRelatedUsersToResult($relatedUser->id, $result);
    }    
}

Answer 2

尝试一下：

$user = Users::find(5); //get the inital user
$picFor = $user->pic_for; //get its pointer
$pictures = []; //initialize the result array
while (!is_null($user) && !is_null($picFor)) { //if I have a valid user and pointer
  $user = Users::find($picFor); //get the user for the pointer
  if ($user !== null) { //if I have a valid user
    $picFor = $user->pic_for; //get the next pointer
    $pictures[]= $user->id; //add the user id to the result array
  }   
}

您遇到的主要问题是缺少一些空检查。

在Laravel中尝试获取ID时尝试获取非对象的属性

2 个答案: