How to get information when 500 error - php

Question

<?php
$this->client = new \GuzzleHttp\Client;
try {
$response = $this->client->request('get', $url);
$result = json_decode($response->getBody());
Log::save($result);
} catch (RequestException $e) {
Log::save($e->getMessage());
}
If the target page has internal server error 500 situation, I will get the error message Client error: 404, actually the target page has show what the bug is if I use the browser to open it, I want to save those PHP error message to log, but I don't know how to got when using Guzzle.

Answer 1

There are a couple of different ways you can proceed:
If you are using Guzzle 6+ and you are using the default
HandlerStack (creating the client with no handler new \GuzzleHttp\Client(); the default behaviour will be for the request
to throw a GuzzleHttp\Exception\ServerException exception. In this case, all you have to do is wrap your request in a try / catch block (as shown above).
If you are manually creating your
HandlerStack $stack = new \GuzzleHttp\HandlerStack(); $client = new \GuzzleHttp\Client(['handler' => > $stack,]); then the \GuzzleHttp\Middleware::httpErrors middleware has not been loaded onto the stack, and you will have to trap it within your own middleware or by passing a callable to the 'on_headers' request option.
References:
Guzzle Request Options - http_errors
Guzzle Handlers and Middleware
Guzzle Request Options - on_headers