Api响应和Json laravel格式

Question

我用的是Laravel 5.7。和GuzzleHttp 6.0从端点获取API响应

我将查询数据从Blade表单传递到此函数。

public static function prhmulti($multisearch, $start ,$end)

{   $city = $multisearch['city'];

  $client = new Client([
     'base_uri' =>  'https://avoindata.prh.fi/tr/',
     'query' => [
    'totalResults' => 'true',
    'maxResults' => '1000',
    'registeredOffice'=> $city,
    'companyForm'=>'OY',
    'companyRegistrationFrom'=>$start,
    'companyRegistrationTo'=>$end,
],
       'defaults'=>[
       'timeout'  => 2.0,
       'cookies' => true,
       'headers'  => [
         'content-type' => 'application/json',
         'User-Agent' =>"GuzzleHttp/Laravel-App-5.7, Copyright MikroMike"
       ]]]);

$res = $client->request('GET','v1');
$ResData = json_decode($res->getBody()->getContents());

dd ($ResData)提供来自API响应的所有数据。

但我无法将JSON返回到其他函数

return $this->multisave($ResData);



public static function multisave (data $ResData)

这将解析JSON和

{
 foreach ($data->results as $company) {
    $name = $company->name;
    $Addr = $company->addresses;
    $businessId = $company->businessId;
    $companyForm = $company->companyForm;
    $registrationDate = $company->registrationDate;

  foreach ($company->addresses as $Addr) {
      $city = $Addr->city;
      $postcode = $Addr->postCode;
      $street = $Addr->street;
   }
 }

将数据保存到Mysql。

$NewCompany = new Company();

 $NewCompany = Company::updateOrCreate($array,[
 [ 'vat_id', $businessId],
 [ 'name', $name],
 [ 'form',$companyForm],
 [ 'street', $Addr],
 [ 'postcode', $postcode],
 [ 'city', $city],
 [ 'regdate', $registrationDate],
 ]);

}

如果Parse部件和Save部件在相同的函数内，代码工作正常(只保存一个公司)，

但我需要把它们分开，因为以后更容易维护。

错误，我要返回$ResData

" Using $this when not in object context" 

信息在JSON数组中。

还有前部只存一家公司？

  foreach ($data->results as $company) {
  $name = $company->name;
  $Addr = $company->addresses;
  $businessId = $company->businessId;
  $companyForm = $company->companyForm;
  $registrationDate = $company->registrationDate;

foreach ($company->addresses as $Addr) {
   $city = $Addr->city;
   $postcode = $Addr->postCode;
   $street = $Addr->street;
 }

那么: 1)创建自己的解析JSON函数和其他将数据保存到DB的最佳方法是什么? 2)作为foreach循环只保存一个公司数据，修复它的最佳方法是什么？

谢谢MikroMike。

Answer 1

解决了我自己的问题，把公司保存到db

• 先得到数组内的总数
• 使用for-循环进行计数
• 以每个公司的前瞻循环提取信息为对象. $data = json_decode($res->getBody()->getContents())；$total =$data-> $total；for ($i = 0；$i <$total；$i++){ $NewCompany =新公司( )；foreach ($data->结果as $company) { $name = $company->name；$businessId = $company->businessId；$companyForm = $company->companyForm；$registrationDate = $company->registrationDate；$array = []；Arr:set($array，‘vat_id，$businessId)；Arr:set($array，'name’s，$businessId)；Arr::set($array，'form'，$companyForm)；Arr::set($array，'regdate'，$registrationDate)；$NewCompany =Company:updateOrCreate($array，[ 'vat_id'，$businessId，'name'，$name，'form'，$companyForm，'regdate'，$registrationDate，]；}// END OF FOREACH }// END For循环}// END OF FUCNTION } // END OF类