查询关系 Eloquent

Query relationship Eloquent(查询关系 Eloquent)

问题描述

我有 News 模型，而且 News 有很多评论，所以我在 News 模型中做了这个:

I have News model, and News has many comments, so I did this in News model:

public function comments(){
    $this->hasMany('Comment', 'news_id');
}

但是我在 comments 表中也有字段 trashed，我只想选择没有被删除的评论.所以 废弃了 <>1.所以我想知道有没有办法做这样的事情:

But I also have field trashed in comments table, and I only want to select comments that are not trashed. So trashed <> 1. So I wonder is there a way to do something like this:

$news = News::find(123);
$news->comments->where('trashed', '<>', 1); //some sort of pseudo-code

有没有办法使用上述方法，或者我应该写这样的东西:

Is there a way to use above method or should I just write something like this:

$comments = Comment::where('trashed', '<>', 1)
    ->where('news_id', '=', $news->id)
    ->get();

推荐答案

这些都适合你，选择你最喜欢的:

Any of these should work for you, pick the one you like the most:

1. 急切加载.

1. Eager-loading.

$comments = News::find(123)->with(['comments' => function ($query) {
    $query->where('trashed', '<>', 1);
}])->get();

您可以通过 use($param) 方法将参数注入查询函数，这允许您在运行时使用动态查询值.

You can inject the parameter to query function by use($param) method, that allows you to use dynemic query value at runtime.

延迟加载

$news = News::find(123);
$comments = $news->comments()->where('trashed', '<>', 1)->get();

<小时>

不过，我忍不住注意到，您可能想做的是处理软删除，而 Laravel 有内置功能可以帮助您:http://laravel.com/docs/eloquent#soft-deleting

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