Laravel 如果不在where()中,则雄辩地创建列
我有这样一个问题:Laravel 如果不在where()中,则雄辩地创建列,laravel,laravel-4,Laravel,Laravel 4,我有这样一个问题: $valid_statuses = array('1', '2', '3'); $results = User::with(['posts' => function($query) use ($valid_statuses) { $query->whereIn('status', $valid_statuses); } ]) ->get(); +----+---------+--------+ |
$valid_statuses = array('1', '2', '3');
$results = User::with(['posts' => function($query) use ($valid_statuses) {
$query->whereIn('status', $valid_statuses);
}
])
->get();
+----+---------+--------+
| id | post_id | status |
+----+---------+--------+
| 1 | 1 | 1 |
| 2 | 1 | 2 |
| 3 | 2 | 1 |
| 4 | 2 | 3 |
+----+---------+--------+
该表的一个示例如下所示:
$valid_statuses = array('1', '2', '3');
$results = User::with(['posts' => function($query) use ($valid_statuses) {
$query->whereIn('status', $valid_statuses);
}
])
->get();
+----+---------+--------+
| id | post_id | status |
+----+---------+--------+
| 1 | 1 | 1 |
| 2 | 1 | 2 |
| 3 | 2 | 1 |
| 4 | 2 | 3 |
+----+---------+--------+
我需要查询做的是,如果post\u id
缺少valid\u status
,它需要将缺少的状态添加到查询结果中。输出结果查询应如下所示:
+----+---------+--------+
| id | post_id | status |
+----+---------+--------+
| 1 | 1 | 1 |
| 2 | 1 | 2 |
| 3 | 1 | 3 |
| 4 | 2 | 1 |
| 5 | 2 | 2 |
| 6 | 2 | 3 |
+----+---------+--------+
我该怎么做呢?如果帖子和状态模型之间存在多对多关系,您可以这样做 维护原始数据库记录:
foreach($valid_statuses as $status_id) {
$have_status = Post::whereHas('status_id', $status_id)->lists('id');
$need_status = Post::whereNotIn('id', $valid_posts)->lists('id');
Status::find($status_id)->posts()->sync($need_status, false);
}
或者如果你不在乎原始记录:
foreach($valid_statuses as $status_id) {
Status::find($status_id)->posts()->sync(Post::all()->lists('id'));
}
是否要更新特定用户的所有帖子,使其具有所有3种状态,而您不知道哪些状态已创建?否,我想选择所有帖子,但如果该特定
帖子id
不存在该状态,请填写缺少的状态。i、 e.如果post\u id
的状态为1和2,而不是3,则让查询将3添加到结果中。具有post\u id和状态的表的名称是什么?它有自己的型号吗?是否在Post模型中将其设置为归属关系?