在php中传递变量到graphql

Question

我只是想确认一下我对如何从php中传递graphql变量到graphql查询的理解。我使用php Curl来执行我对graphql api服务器的请求.我得到的错误是 - Variable "$page" required type "Int!" not provided.

   $data='
    query($businessId : ID!, $page : Int!, $pageSize : Int!) {
       business(id: $businessId) {
          id
          isClassicInvoicing
          invoices(page: $page, pageSize: $pageSize) {'
  ...

  $variables=array(
    '$businessId'=>'"xxxx"',
    '$page'=>1,
    '$pageSize'=>5);

  $payData = urlencode($data);
  $payVariables = urlencode(json_encode($variables));
  $apiURL=$apiURL . '?' . "query=".$payData."&variables=".$payVariables ;

我得到的错误是 - Variable "$page" of required type "Int!" was not provided. 对于$businessId和$pageSize我也收到这个错误。

Answer 1

$variables的结构不正确。正确的方法是

$variables='
{
  "businessId": "xxxx",
  "page": 1,
  "pageSize": 5
}';