Reactjs 狂饮:罐头';无法捕获异常详细信息
我用PayPal API模拟负面响应,以便在客户通过Reactjs 狂饮:罐头';无法捕获异常详细信息,reactjs,laravel,guzzle,paypal-rest-sdk,Reactjs,Laravel,Guzzle,Paypal Rest Sdk,我用PayPal API模拟负面响应,以便在客户通过onApprove方法批准付款时,获得响应并正确处理付款的关键部分 我正在使用GuzzleHttp+Laravel来获取客户的批准。我在complete对象中获得COMPLETED状态。因此,请求工作正常 $client = new \GuzzleHttp\Client(); return $response = $client->request( 'POST', 'https://api-m.sandbox.paypa
onApprove
方法批准付款时,获得响应并正确处理付款的关键部分
我正在使用GuzzleHttp
+Laravel
来获取客户的批准。我在complete对象中获得COMPLETED
状态。因此,请求工作正常
$client = new \GuzzleHttp\Client();
return
$response = $client->request(
'POST',
'https://api-m.sandbox.paypal.com/v2/checkout/orders/' . $paypalOrderId . '/capture',
[
'headers' => [
'Content-Type' => 'application/json',
'Authorization' => 'Bearer ' . $access_token,
// 'PayPal-Mock-Response' => json_encode(["mock_application_codes" => "INTERNAL_SERVER_ERROR"]),
'PayPal-Mock-Response' => json_encode(["mock_application_codes" => "INSTRUMENT_DECLINED"]),
],
],
);
我通过在标题中添加一个=>'PayPal-Mock-response'=>json\u encode([“Mock\u application\u codes”=>“INSTRUMENT\u densed”])
来模拟响应错误,当然,如果不在try catch
块中,它将中断代码
这是请求没有try-catch时Laravel的输出:
"message": "Client error: `POST https://api-m.sandbox.paypal.com/v2/checkout/orders/6JE880117B631364H/capture` resulted in a `422 Unprocessable Entity`
response:\n{\n \"name\": \"UNPROCESSABLE_ENTITY\",\n \"details\": [\n {\n \"issue\": \"INSTRUMENT_DECLINED\",\n \"description\": \"The (truncated...)\n",
"exception": "GuzzleHttp\\Exception\\ClientException",
"file": "C:\\xampp\\htdocs\\react\\React-Laravel\\vinos-gdl\\vendor\\guzzlehttp\\guzzle\\src\\Exception\\RequestException.php",
"line": 113,
这是预期的错误。因为我用模拟响应
标题强制它
当我在try catch
块中插入代码时,“问题”就出现了。我获得了预期的异常
,但无法从响应中获得任何详细信息
:
catch (ServerException $e) {
if ($e->hasResponse()) {
return response()->json(['msg' => 'Server Error', 'error' => $e->getResponse()], 500);
}
return response()->json([
'msg' => 'Server Error',
'request' => $e->getRequest(),
$e->hasResponse() ? $e->getResponse() : ""
]);
// return response()->json(['msg' => 'Client Error', 'error' => $e->getRequest()]);
} catch (ClientException $e) {
if ($e->hasResponse()) {
return response()->json(['msg' => 'Client Error', 'error' => $e->getResponse()], 400);
}
return response()->json([
'msg' => 'Client Error',
'request' => $e->getRequest(),
$e->hasResponse() ? $e->getResponse() : ""
]);
// return response()->json(['msg' => 'Server Error', 'error' => report($e)]);
}
catch (BadResponseException $e){
return response()->json(['error' => $e]);
}
这是客户端异常的输出:
{
"msg": "Server Error",
"error": {} // also empty
}
错误:请求失败,状态代码为400
如果我强制执行服务器异常
:
{
"msg": "Server Error",
"error": {} // also empty
}
将抛出异常
,但是,这肯定不足以正确处理错误。我需要从响应中获取详细信息
前端,以防有人想看到它:
const onApprove = (data, actions) => {
console.log("payment approved by user", data);
const orderID = data.orderID;
return axios
.post("/paypal/rest-api/capture-order", {
orderID: orderID,
})
.then((res) => {
console.log("success creating order", res.data);
})
.catch((err) => {
console.error(err);
});
};
异常应该是具有getResponse方法的BadResponseException的实例。然后可以将响应主体强制转换为字符串
$response = json_decode($ex->getResponse()->getBody()->getContents(), true);