【发布时间】:2016-06-18 15:13:17
【问题描述】:
我想做这样的事情:
$posts= Status::where('users_id',$user->id)->orWhere(DB::table('user_status_share.user_id', $user->id))->orderBy('created_at', 'DESC')->get();
但我得到一个错误:strtolower() 期望参数 1 是字符串,给定对象 - 如何在“orWhere”方法中更改表格?这可能吗?如果不是 - 如何在一个查询中使用 2 个表?
架构(状态):
public function up()
{
Schema::create('users_status', function (Blueprint $table) {
$table->increments('id')->unique();
$table->longText('status_text');
$table->integer('users_id')->unsigned();
$table->timestamps();
});
}
/**
* Reverse the migrations.
*
* @return void
*/
public function down()
{
Schema::drop('users_status');
}
架构(StatusShare):
public function up()
{
Schema::create('user_status_share', function (Blueprint $table) {
$table->increments('id');
$table->integer('status_id');
$table->integer('user_id');
$table->timestamps();
});
}
/**
* Reverse the migrations.
*
* @return void
*/
public function down()
{
Schema::drop('user_status_share');
}
模型状态:
namespace App\Eloquent;
use Illuminate\Database\Eloquent\Model;
class Status extends Model
{
public $timestamps = true;
protected $table = 'users_status';
protected $guarded = ['id'];
public function comments()
{
return $this->hasMany(StatusComments::class);
}
public function likes()
{
return $this->hasMany(StatusLikes::class);
}
public function shares()
{
return $this->hasMany(StatusShare::class);
}
}
模型状态共享:
<?php
namespace App\Eloquent;
use Illuminate\Database\Eloquent\Model;
class StatusShare extends Model
{
public $timestamps = true;
protected $table = 'user_status_share';
protected $guarded = ['id'];
public function status()
{
return $this->hasOne(Status::class);
}
}
【问题讨论】:
-
你想从两个表中获取数据吗?
-
请在输出中发布您的架构结构和您期望的数据格式。
-
是的,我想从 2 个表中获取数据
-
添加了架构。我期待数组