guzzle

How to read the response effective URL in Guzzle ~6.0

£可爱£侵袭症+ 提交于 2019-12-03 04:07:00
问题 I've been searching for about 2 hours and I can't figure it out how to read the final response uri. In previous versions of PHP Guzzle you just call $response->getEffectiveUrl() and you get it. I expected to have something similar in the new version so the final code looks like this: $response = $httpClient->post('http://service.com/login', [ 'form_params' => [ 'user' => $user, 'padss' => $pass, ] ]); $url = $response->getEffectiveUrl(); But in the latest version $response is now a GuzzleHttp

How do i do HTTP basic authentication using Guzzle?

拜拜、爱过 提交于 2019-12-03 04:04:16
问题 I want to do basic access authentication using Guzzle and i am very new to programming . i have no clue what to do. I tried to do this using curl but my environment requires using guzzle. 回答1: If you're using Guzzle 5.0 or newer , the docs say that basic auth is specified using the auth parameter: $client = new GuzzleHttp\Client(); $response = $client->get('http://www.server.com/endpoint', [ 'auth' => [ 'username', 'password' ] ]); Please note that the syntax is different if you're using

Guzzle ~6.0 multipart and form_params

匿名 (未验证) 提交于 2019-12-03 01:23:02
可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效,请关闭广告屏蔽插件后再试): 问题: I am trying to upload file and send post parameters at the same time like this: $response = $client->post('http://example.com/api', [ 'form_params' => [ 'name' => 'Example name', ], 'multipart' => [ [ 'name' => 'image', 'contents' => fopen('/path/to/image', 'r') ] ] ]); However my form_params fields are ignored and only the multipart fields are present in my post body. Can I send both at all with guzzle 6.0 ? 回答1: I ran into the same problem. You need to add your form_params to the multipart array. Where 'name' is the form element name and

Composer autoloader not loading GuzzleHttp\\ClientInterface

匿名 (未验证) 提交于 2019-12-03 00:46:02
可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效,请关闭广告屏蔽插件后再试): 问题: I'm trying to use Guzzle , but I'm getting the following fatal error: Fatal error: Class 'GuzzleHttp\ClientInterface' not found in /var/www/myapp/vendor/guzzlehttp/guzzle/src/functions.php on line 13 I'm autoloading with composer autoloader: require 'vendor/autoload.php'; use Guzzle\Http\Client; $client = new Client(); $requests = Array( $client->createRequest('GET', 'ams1.myapp.com:8080/api/ffmpeg_make_snapshots.php'), $client->createRequest('GET', 'ams2.myapp.com:8080/api/ffmpeg_make_snapshots.php'), $client->createRequest('GET', 'ams3

Copy remote file using Guzzle

纵饮孤独 提交于 2019-12-02 20:38:48
I'm trying to copy a remote file (image PNG, GIF, JPG ...) to my server. I use Guzzle since I sometimes get 404 with copy() even if the file exists and I also need to do a basic auth. This script is within a long script launched in command triggered by a cron job. I'm pretty new to Guzzle and I successfully copy the image but my files have wrong mime type. I must be doing something wrong here. Please suggest me a good way to do this (including checking success/failure of copy and mime type check). If file has no mime type I would pop an error with details informations. Here is the code:

How to read the response effective URL in Guzzle ~6.0

不想你离开。 提交于 2019-12-02 19:04:23
I've been searching for about 2 hours and I can't figure it out how to read the final response uri. In previous versions of PHP Guzzle you just call $response->getEffectiveUrl() and you get it. I expected to have something similar in the new version so the final code looks like this: $response = $httpClient->post('http://service.com/login', [ 'form_params' => [ 'user' => $user, 'padss' => $pass, ] ]); $url = $response->getEffectiveUrl(); But in the latest version $response is now a GuzzleHttp\Psr7\Response and there is no method which allow me to retrieve the uri. I read about the redirects

How do i do HTTP basic authentication using Guzzle?

痴心易碎 提交于 2019-12-02 17:23:49
I want to do basic access authentication using Guzzle and i am very new to programming . i have no clue what to do. I tried to do this using curl but my environment requires using guzzle. If you're using Guzzle 5.0 or newer , the docs say that basic auth is specified using the auth parameter: $client = new GuzzleHttp\Client(); $response = $client->get('http://www.server.com/endpoint', [ 'auth' => [ 'username', 'password' ] ]); Please note that the syntax is different if you're using Guzzle 3.0 or earlier . The constructor is different, and you also need to explicitly use the send method on a

3rd party dependency conflict in developing Wordpress Plugin

痞子三分冷 提交于 2019-12-02 13:10:27
问题 I am developing a plugin that used composer.. meaning it has a vendor folder inside the plugin folder which includes Guzzle HTTP dependency On wordpress site we installed this plugin, there is an existing plugin that has Guzzle HTTP Now when we activate this plugin i am getting an error something like this: Fatal error: Cannot redeclare GuzzleHttp\uri_template() (previously declared in /nas/content/staging/project/wp-content/plugins/my-plugin/vendor/guzzlehttp/guzzle/src/functions.php:17) in

3rd party dependency conflict in developing Wordpress Plugin

本小妞迷上赌 提交于 2019-12-02 07:02:53
I am developing a plugin that used composer.. meaning it has a vendor folder inside the plugin folder which includes Guzzle HTTP dependency On wordpress site we installed this plugin, there is an existing plugin that has Guzzle HTTP Now when we activate this plugin i am getting an error something like this: Fatal error: Cannot redeclare GuzzleHttp\uri_template() (previously declared in /nas/content/staging/project/wp-content/plugins/my-plugin/vendor/guzzlehttp/guzzle/src/functions.php:17) in /nas/content/staging/project/wp-content/plugins/other-plugin/includes/lib/aws-sdk/GuzzleHttp/functions

Guzzle6 error Invalid resource type: array when send a GuzzleHttp\Psr7\Request

六眼飞鱼酱① 提交于 2019-12-02 04:05:01
问题 I am trying to send a code by using GuzzleHttp\Psr7\Request,somehow I get error Invalid resource type: array, the following is my codes $params = ["name"=>"myName","id"=>"myId"]; $client = new Client(); $request = new Request('PUT','https://api.hitbox.tv/media/live/myName?authToken=myToken',["content-type"=>'application/json'],["json"=>$params]); $response = $client->send($request); I'm following this guide. 回答1: If you want to use JSON in the request, just create it with json_encode() :