本文介绍了从不同的关系中获取最新的雄辩的laravel的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我有3张桌子
模型网址
class Url extends Model
{
public function users(){
return $this->belongsToMany(User::class);
}
public function url_status(){
return $this->hasMany(UrlStatus::class);
}
}
模型UrlStatus
class UrlStatus extends Model
{
public function url()
{
return $this->belongsTo(Url::class);
}
}
模型用户
class User extends Authenticatable
{
use Notifiable, SoftDeletes, HasRoles;
public function urls(){
return $this->belongsToMany(Url::class);
}
}
在我的控制器中,我正在查询:
In my controller I'm querying :
$url = Url::with('url_status','users')->where('list_status', true)->get();
如何获取最新的url_status?
How can I get the latest url_status?
编辑---
这是我的迁移文件中每个表的结构
This is my structure for each table in migration file
用于网址表
Schema::create('urls', function (Blueprint $table) {
$table->increments('id');
$table->string('url');
$table->string('description');
$table->boolean('list_status');
$table->timestamps();
});
URL状态
Schema::create('url_status', function (Blueprint $table) {
$table->increments('id');
$table->unsignedInteger('url_id');
$table->integer('status_code');
$table->string('status');
$table->boolean('sms');
$table->boolean('email');
$table->timestamps();
});
用户表
Schema::create('users', function (Blueprint $table) {
$table->increments('id');
$table->string('name');
$table->string('email')->unique();
$table->string('contacts');
$table->string('password');
$table->rememberToken();
$table->timestamps();
$table->softDeletes();
});
推荐答案
您也可以这样尝试
模型网址
public function latestUrlStatus(){
return $this->hasOne(UrlStatus::class)->latest();
}
以这种方式获取
$urls = Url::with('latestUrlStatus','users')->where('list_status', true)->get();
foreach($urls as $url){
echo $url->latestUrlStatus->status_code;
}
这篇关于从不同的关系中获取最新的雄辩的laravel的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!