雄辩的 each() 方法返回零值

Question

这个问题与Eloquent save() method not working inside for loop有关。

过程图：上传文件 > 使用默认值 0 创建表行 > 继续迭代 0 部分（例如，如果最后一个 3-> 4,5,6）

我想在第一次上传图片后继续向特定项目添加图片。

$input['id']：持有项目id值。

我确实像这样应用 each() 方法：

*** code part executed after multiple file upload ***

TABLE STATUS AT FIRST STAGE :
+---------+-------------+-----------+
| id      | project_id  | order_id  |  
+---------+-------------+-----------+
| 1       | 15          | 1         |
+---------+-------------+-----------+
| 2       | 15          | 2         |
+---------+-------------+-----------+
| 3       | 16          | 1         |  --> Added in the past
+---------+-------------+-----------+
| 4       | 16          | 0         |  --> Created on first stage** : default 0.
+---------+-------------+-----------+ 
| 5       | 16          | 0         |  --> Created on first stage** : default 0.
+---------+-------------+-----------+


*** code continues ***

$start = Images::select('order_id')->where('project_id', '=', $input['id'])
->orderBy('order_id', 'DESC')->first();  

Images::where('project_id', '=', $input['id'])->where('order_id', '=', 0)->get()
       ->each(function($ord) use (&$start) {
              $ord->order_id = $start++;
              $ord->save();
});

上面给出的代码执行后的表格结果（仍在处理中）

+---------+-------------+-----------+
| id      | project_id  | order_id  |  
+---------+-------------+-----------+
| 1       | 15          | 1         |
+---------+-------------+-----------+
| 2       | 15          | 2         |
+---------+-------------+-----------+
| 3       | 16          | 1         |  --> Added in the past
+---------+-------------+-----------+
| 4       | 16          | 0         |  --> New / Present Action : 1st file 
+---------+-------------+-----------+ 
| 5       | 16          | 0         |  --> New / Present Action : 2nd one
+---------+-------------+-----------+

期望：

+---------+-------------+-----------+
| id      | project_id  | order_id  |  
+---------+-------------+-----------+
| 1       | 15          | 1         |
+---------+-------------+-----------+
| 2       | 15          | 2         |
+---------+-------------+-----------+
| 3       | 16          | 1         |  --> Added in the past
+---------+-------------+-----------+
| 4       | 16          | 2         |  --> New / Present Action : 1st file 
+---------+-------------+-----------+ 
| 5       | 16          | 3         |  --> New / Present Action : 2nd one
+---------+-------------+-----------+

有什么问题？错误的逻辑或 each() ？我不知道，但我尝试使用 foreach 并得到相同的错误结果。任何帮助将不胜感激。谢谢！

Answer 1

$start = Images::select('order_id')->where('project_id', '=', $input['id'])
->orderBy('order_id', 'DESC')->first();  

返回Images的对象

$start->order_id 应该包含您的 id-number

Answer 2

尝试在递增时访问 id 属性。

Images::where('project_id', '=', $input['id'])->where('order_id', '=', 0)->get()
       ->each(function($ord) use (&$start) {
              $ord->order_id = $start->id++;
              $ord->save();
});