我就是搞不懂《雄辩》里的人际关系。当我以为我已经得到它时，我绊倒了。
就像这里一样，我想列出每个Headquarters_Pay_Data项的country字段。
模型;
<?php
namespace Datamate;
use Illuminate'Database'Eloquent'Model;
class Headquarters_Pay_Data extends Model
{
//
protected $table = 'headquarters_pay_data';
public function postcode()
{
return $this->hasOne('Datamate'Postcode_Quarter', 'postcode', 'Vendor ZIP');
}
}
和这个;
<?php
namespace Datamate;
use Illuminate'Database'Eloquent'Model;
use Datamate'Country;
class Postcode_Quarter extends Model
{
public $table = '201502_postcode';
protected $fillable = ['country'];
}
我的控制器;
public function index()
{
//
$headquarters_pay_data = Headquarters_Pay_Data::limit(12)->get();
foreach ($headquarters_pay_data as $key => $value) {
//print $value->postcode->country; //this returns an error. Trying to get property of non-object
print "<br><br>";
print $value->getAttribute('Vendor ZIP');
print "<br><br>";
print $value->postcode; //this is JSON?! Why?
print "<br><br>";
}
示例打印出看起来像JSON的东西，即使我没有要求JSON;
RH108PJ
{"postcode":"RH108PJ","county":"E10000032","district":"","ward":"","health":"E18000008","gor":"J","parlc":"E14000652","locauth":"E07000226","wardcode":"E05007639","country":"E92000001","gor_new":"E12000008","pct":"E16000108"}
澄清……我怎样才能打印出每次付款的country ?
由于您的雄辩关系方法，postcode字段是Headquarters_Pay_Data模型上另一个模型(Postcode_Quarter)的实例。因此，$value->postcode返回该模型(作为PHP对象)。您将该模型转换为字符串(通过print它)使其尝试将自己转换为最佳格式，以用作字符串，这是一个JSON字符串。
但是，您可以访问该模型的属性，并且因为您想要国家，您可以执行以下操作:
public function index()
{
$headquarters_pay_data = Headquarters_Pay_Data::with('postcode')->limit(12)->get();
foreach ($headquarters_pay_data as $key => $value) {
print $value->postcode->country;
}
}
你会注意到，在这个例子中，我们也使用with()来"急切加载"postcode关系。这通常会使您的查询更有效，特别是在您有很多Headquarters_Pay_Data模型而不是很多Postcode_Quarter模型的情况下，但这不是必需的。