Php 如何使用laravel内部的条件
请查看我的消息表架构Php 如何使用laravel内部的条件,php,laravel,eloquent,laravel-5.4,Php,Laravel,Eloquent,Laravel 5.4,请查看我的消息表架构 Schema::create('messages', function (Blueprint $table) { $table->increments('id'); $table->integer('conversation_id')->unsigned()->index(); $table->foreign('conversation_id')->references('id')->
Schema::create('messages', function (Blueprint $table) {
$table->increments('id');
$table->integer('conversation_id')->unsigned()->index();
$table->foreign('conversation_id')->references('id')->on('conversations');
$table->integer('last_sender')->unsigned();
$table->foreign('last_sender')->references('id')->on('users');
$table->string('msg');
$table->integer('deleted')->unsigned();
$table->integer('seen')->unsigned();
$table->timestamps();
});
这是我正在使用的当前查询
Message::whereIn('conversation_id',$msgs)
->with(array('last_sender'=>function($query){
$query->select('id','userName','profilePic', 'firstName','lastName');
}))
->orderBy('conversation_id', 'desc')
->groupBy('conversation_id')
->get();
如您所见,我渴望加载最后一个\u发件人
。但我想根据条件加载它。例如,如果用户已登录且其id为10且last_sender
不=10,则我希望加载,因为我只希望在当前登录的用户id不在last_sender
列中时加载
非常感谢您的帮助。多谢各位
Message::with([
'last_sender' => function($q) use ($userId = Auth::user()->id) {
$q->where('id','!=',$userId)
}
])->get();
在条件下加载关系
Message::whereHas(
'last_sender', function($q) use ($userId = Auth::user()->id) {
$q->where('id','!=',$userId)
}
)->get();
加载与条件相关的模型实例
Message::whereHas(
'last_sender', function($q) use ($userId = Auth::user()->id) {
$q->where('id','!=',$userId)
}
)->get();
希望这就是您所要求的?是的,谢谢,让我看看它是如何工作的。再次感谢:)当心打字错误:)谢谢你,它成功了。我为其他可能陷入这种问题的人发表这篇评论。对他们来说,这就是我使用的
$userId=Auth::user()->id$leftMsg=Message::where('conversation_id',$msgs)->with(array('last_sender'=>function($query)use($userId){$query->where('id','!=',$userId);$query->select('id','userName','profilePic'firstName','lastName');}))->orderBy('conversation_id','desc->groupBy('conversation_id')->get()代码>