Laravel Sync方法仅发送第二个数据 [英] Laravel Sync method only sending the 2nd data

问题描述

这段代码应该从数据库中删除所有旧数据，并在添加新数据时(使用sync())

This piece of code should be deleting all old data from the database and when add the new ones (using sync())

现在我有一个包含用户的项目，并且可以通过复选框将用户链接到项目.

Now I have a project with users, and a user can be linked to a project with a checkbox.

因此，选中此复选框将触发此功能，但是例如，当我说 user 1 和 user 2 正在通过此功能添加到数据透视表，它将仅发送用户2 ，而用户1 无法通过，这是怎么回事?

So on checkbox checked this function will trigger, but for example when I say that user 1 and user 2 are going through this fuction to get added to the pivot table it will only send user 2, and user 1 will not get through, what is going wrong?

当我添加3个用户用户1 ，用户2 ，用户3 时，只有 user 2 被添加.

And when I add 3 users user 1, user 2, user 3, only user 2 will get added.

控制器

public function update(CreateProjectRequest $request)
{
    if($request->get('contribute'))
    {
        foreach($request->get('contribute') as $k => $contribute)
        {
            if($contribute == 1)
            {
                $project = $this->project->find($request->project_id);
                $project->users()->sync(array($k));

            }
        }
    }

    $project = $this->project->find($request->project_id);
    $project->fill($request->input())->save();

    return redirect('project');
}

刀片

@foreach($users as $user)
            <tr>
                <td>
                    {{$user->firstname}} {{$user->middlename}} {{$user->lastname}}
                </td>
                <td>
                    {!! Form::checkbox('contribute['.$user->id.']', '1', $user->projects->contains('id', $project->id)) !!}
                </td>
            </tr>
@endforeach

在 dd($ request-> input()); 上，在我的更新方法开始时(选择至少3个用户)，这将得到回报:

On a dd($request->input()); at the start of my update method (with selecting atleast 3 users) this will get in return:

  array:9 [▼
  "_method" => "PATCH"
  "_token" => "0uIZNn6zwZjVKfgE0ckhDULeYda0OaLzKVdUgoM8"
  "name" => "Dire Straits"
  "completion_date" => "2015-05-18"
  "DataTables_Table_0_length" => "10"
  "contribute" => array:3 [▼
    1 => "1"
    3 => "1"
    2 => "1"
  ]
  "completed" => "1"
  "active" => "0"
  "project_id" => "11"
]

因此， 1/3/2 将是 user_id 和 =>1 应该是值.

So 1 / 3 / 2 would be the user_id and => 1 should be the value.

推荐答案

问题是 sync 在 loop 中被调用3次，因此每次同步一个值时.您必须在 sync ex:

The problem is that sync gets called in loop 3 times so each time it's syncing one value. You have to pass an array of the ids in sync ex:

$project->users()->sync([1,3,2]);

或者，如果您愿意，可以在 contribute == 1 时使用 attach ，在 contribute == 0 时使用 detach >

Or if you want you can use attach when contribute==1 and detach when contribute==0

或者如果 contribute 在取消选择用户时不返回输入，而仅在选择用户时返回，则可以尝试:

Or if contribute doesnt return input when a user is deselected and it only returns when it is selected then you can try:

$this->project->users()->sync(array_keys($request->get('contribute'));

我刚刚注意到您还有另一个错误，除非您通过一次调用更新许多项目，否则应将下面的行放在函数的第一行.

I just noticed that you have another bug unless you are updating many projects with one call you should put the line below on the first line of your function.

$project = $this->project->find($request->project_id);

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