Images download using Goutte - php

I have to download images using cookie in request. I could make it with file_get_contents (with stream_context_create) or curl with passing cookies. But how to make it with Goutte?
use Goutte\Client;
$client = new Client();
$response = $client->request('GET', 'https://www.google.pl/images/srpr/logo11w.png');
I make a GET request and what's the next step?
Ok, I figured that out:
$client->get('https://www.google.pl/images/srpr/logo11w.png', array('save_to' => __DIR__.'/image.jpg'));

With Goutte 2:
$client = new GuzzleHttp\Client([
'base_url' => 'https://www.google.pl',
'defaults' => [
'cookies' => true,
]
]);
$response = $client->post('/login', [
'body' => [
'login' => $login,
'password' => $password
]
]);
$response = $client->get('/images/srpr/logo11w.png');
$image = $response->getBody();
With Goutte 1 after adding "guzzle/plugin-cookie": "~3.1" to Composer:
use Guzzle\Http\Client;
use Guzzle\Plugin\Cookie\CookiePlugin;
use Guzzle\Plugin\Cookie\CookieJar\ArrayCookieJar;
$client = new Client('https://www.google.pl');
$client->addSubscriber(new CookiePlugin(new ArrayCookieJar()));
$response = $client->post('/login', '', array('login' => $login, 'password' => $password))->send();
$response = $client->get('/images/srpr/logo11w.png')->send();
$image = $response->getBody();

Related

Using guzzle to post to Facebook Conversions API [duplicate]

Does anybody know the correct way to post JSON using Guzzle?
$request = $this->client->post(self::URL_REGISTER,array(
'content-type' => 'application/json'
),array(json_encode($_POST)));
I get an internal server error response from the server. It works using Chrome Postman.
For Guzzle 5, 6 and 7 you do it like this:
use GuzzleHttp\Client;
$client = new Client();
$response = $client->post('url', [
GuzzleHttp\RequestOptions::JSON => ['foo' => 'bar'] // or 'json' => [...]
]);
Docs
The simple and basic way (guzzle6):
$client = new Client([
'headers' => [ 'Content-Type' => 'application/json' ]
]);
$response = $client->post('http://api.com/CheckItOutNow',
['body' => json_encode(
[
'hello' => 'World'
]
)]
);
To get the response status code and the content of the body I did this:
echo '<pre>' . var_export($response->getStatusCode(), true) . '</pre>';
echo '<pre>' . var_export($response->getBody()->getContents(), true) . '</pre>';
For Guzzle <= 4:
It's a raw post request so putting the JSON in the body solved the problem
$request = $this->client->post(
$url,
[
'content-type' => 'application/json'
],
);
$request->setBody($data); #set body!
$response = $request->send();
This worked for me (using Guzzle 6)
$client = new Client();
$result = $client->post('http://api.example.com', [
'json' => [
'value_1' => 'number1',
'Value_group' =>
array("value_2" => "number2",
"value_3" => "number3")
]
]);
echo($result->getBody()->getContents());
$client = new \GuzzleHttp\Client();
$body['grant_type'] = "client_credentials";
$body['client_id'] = $this->client_id;
$body['client_secret'] = $this->client_secret;
$res = $client->post($url, [ 'body' => json_encode($body) ]);
$code = $res->getStatusCode();
$result = $res->json();
You can either using hardcoded json attribute as key, or you can conveniently using GuzzleHttp\RequestOptions::JSON constant.
Here is the example of using hardcoded json string.
use GuzzleHttp\Client;
$client = new Client();
$response = $client->post('url', [
'json' => ['foo' => 'bar']
]);
See Docs.
$client = new \GuzzleHttp\Client(['base_uri' => 'http://example.com/api']);
$response = $client->post('/save', [
'json' => [
'name' => 'John Doe'
]
]);
return $response->getBody();
This works for me with Guzzle 6.2 :
$gClient = new \GuzzleHttp\Client(['base_uri' => 'www.foo.bar']);
$res = $gClient->post('ws/endpoint',
array(
'headers'=>array('Content-Type'=>'application/json'),
'json'=>array('someData'=>'xxxxx','moreData'=>'zzzzzzz')
)
);
According to the documentation guzzle do the json_encode
Solution for $client->request('POST',...
For those who are using $client->request this is how you create a JSON request:
$client = new Client();
$res = $client->request('POST', "https://some-url.com/api", [
'json' => [
'paramaterName' => "parameterValue",
'paramaterName2' => "parameterValue2",
]
'headers' => [
'Content-Type' => 'application/json',
]
]);
Guzzle JSON Request Reference
Php Version: 5.6
Symfony version: 2.3
Guzzle: 5.0
I had an experience recently about sending json with Guzzle. I use Symfony 2.3 so my guzzle version can be a little older.
I will also show how to use debug mode and you can see the request before sending it,
When i made the request as shown below got the successfull response;
use GuzzleHttp\Client;
$headers = [
'Authorization' => 'Bearer ' . $token,
'Accept' => 'application/json',
"Content-Type" => "application/json"
];
$body = json_encode($requestBody);
$client = new Client();
$client->setDefaultOption('headers', $headers);
$client->setDefaultOption('verify', false);
$client->setDefaultOption('debug', true);
$response = $client->post($endPoint, array('body'=> $body));
dump($response->getBody()->getContents());
#user3379466 is correct, but here I rewrite in full:
-package that you need:
"require": {
"php" : ">=5.3.9",
"guzzlehttp/guzzle": "^3.8"
},
-php code (Digest is a type so pick different type if you need to, i have to include api server for authentication in this paragraph, some does not need to authenticate. If you use json you will need to replace any text 'xml' with 'json' and the data below should be a json string too):
$client = new Client('https://api.yourbaseapiserver.com/incidents.xml', array('version' => 'v1.3', 'request.options' => array('headers' => array('Accept' => 'application/vnd.yourbaseapiserver.v1.1+xml', 'Content-Type' => 'text/xml'), 'auth' => array('username#gmail.com', 'password', 'Digest'),)));
$url = "https://api.yourbaseapiserver.com/incidents.xml";
$data = '<incident>
<name>Incident Title2a</name>
<priority>Medium</priority>
<requester><email>dsss#mail.ca</email></requester>
<description>description2a</description>
</incident>';
$request = $client->post($url, array('content-type' => 'application/xml',));
$request->setBody($data); #set body! this is body of request object and not a body field in the header section so don't be confused.
$response = $request->send(); #you must do send() method!
echo $response->getBody(); #you should see the response body from the server on success
die;
--- Solution for * Guzzle 6 * ---
-package that you need:
"require": {
"php" : ">=5.5.0",
"guzzlehttp/guzzle": "~6.0"
},
$client = new Client([
// Base URI is used with relative requests
'base_uri' => 'https://api.compay.com/',
// You can set any number of default request options.
'timeout' => 3.0,
'auth' => array('you#gmail.ca', 'dsfddfdfpassword', 'Digest'),
'headers' => array('Accept' => 'application/vnd.comay.v1.1+xml',
'Content-Type' => 'text/xml'),
]);
$url = "https://api.compay.com/cases.xml";
$data string variable is defined same as above.
// Provide the body as a string.
$r = $client->request('POST', $url, [
'body' => $data
]);
echo $r->getBody();
die;
Simply use this it will work
$auth = base64_encode('user:'.config('mailchimp.api_key'));
//API URL
$urll = "https://".config('mailchimp.data_center').".api.mailchimp.com/3.0/batches";
//API authentication Header
$headers = array(
'Accept' => 'application/json',
'Authorization' => 'Basic '.$auth
);
$client = new Client();
$req_Memeber = new Request('POST', $urll, $headers, $userlist);
// promise
$promise = $client->sendAsync($req_Memeber)->then(function ($res){
echo "Synched";
});
$promise->wait();
I use the following code that works very reliably.
The JSON data is passed in the parameter $request, and the specific request type passed in the variable $searchType.
The code includes a trap to detect and report an unsuccessful or invalid call which will then return false.
If the call is sucessful then json_decode ($result->getBody(), $return=true) returns an array of the results.
public function callAPI($request, $searchType) {
$guzzleClient = new GuzzleHttp\Client(["base_uri" => "https://example.com"]);
try {
$result = $guzzleClient->post( $searchType, ["json" => $request]);
} catch (Exception $e) {
$error = $e->getMessage();
$error .= '<pre>'.print_r($request, $return=true).'</pre>';
$error .= 'No returnable data';
Event::logError(__LINE__, __FILE__, $error);
return false;
}
return json_decode($result->getBody(), $return=true);
}
The answer from #user3379466 can be made to work by setting $data as follows:
$data = "{'some_key' : 'some_value'}";
What our project needed was to insert a variable into an array inside the json string, which I did as follows (in case this helps anyone):
$data = "{\"collection\" : [$existing_variable]}";
So with $existing_variable being, say, 90210, you get:
echo $data;
//{"collection" : [90210]}
Also worth noting is that you might want to also set the 'Accept' => 'application/json' as well in case the endpoint you're hitting cares about that kind of thing.
Above answers did not worked for me somehow. But this works fine for me.
$client = new Client('' . $appUrl['scheme'] . '://' . $appUrl['host'] . '' . $appUrl['path']);
$request = $client->post($base_url, array('content-type' => 'application/json'), json_encode($appUrl['query']));

make the variable as string in laravel

how are you ?
I'm using Guzzle to send post method but when i tried to send the post method with variable getting error so i tried to convert it to sting but still same
this is my current code
if $ mobile value is 55454545445 , them the 'numbers' => "{$mobile}" will be different not "55454545445"
$user = Auth::user();
$mobile = $user->mobile;
$client = new Client();
$booking = Booking::where('id', '=', e($id))->first();
if($booking)
{
$booking->booking_status_id = 3;
$booking->save();
$client = new Client([
'headers' => [ 'Content-Type' => 'application/json' ]
]);
$data = array('userName' => "test",
'apiKey' => "1",
'numbers' => "{$mobile}",
'userSender' => "sender",
'msg' =>'msg',
'msgEncoding' => "UTF8",);
$dataJson = json_encode($data);
$response = $client->post('https://www.test.test',
['body' => $dataJson]
);
i used this
"0{$mobile}"
instead of
"{$mobile}"
and its work

Laravel Gazzel Making multiple requests to same URL without login multiple times

So I have this class setup in laravel. It is using following header to initialize requests.
$this->xml = "<?xml version=\"1.0\"?>
<query xmlns=\"http://www.someurl.com/queryLanguage/v1.0\">
<logon>
<userName>".config('some.username')."</userName>
<password>".config('some.password')."</password>
<deviceName>".config('some.device')."</deviceName>
</logon>";
Then I have like 10 requests to make to the same url with above auth details.
so im doing it like.
$xml1 = $this->xml;
$xml1 .= "some xml";
$options = [
'headers' => [
'Content-Type' => 'text/xml; charset=UTF8',
],
'body' => $xml1,
];
$client = new Client();
$response = $client->request('POST', config('some.apiurl'), $options);
$xml2 = $this->xml;
$xml2 .= "some xml";
$options = [
'headers' => [
'Content-Type' => 'text/xml; charset=UTF8',
],
'body' => $xml2,
];
$client = new Client();
$response = $client->request('POST', config('some.apiurl'), $options);
$xml3 = $this->xml;
$xml3 .= "some xml";
$options = [
'headers' => [
'Content-Type' => 'text/xml; charset=UTF8',
],
'body' => $xml3,
];
$client = new Client();
$response = $client->request('POST', config('some.apiurl'), $options);
as you can see with each requests it makes a new login and eventually ended up getting too many concurrent logons error from remote server. so my question is how we use this api login information and just login once with guzzel and then use it for multiple requests later.
Thanks in advance.
You can use Guzzle CookieJar to save your session and send it to the next request.
http://docs.guzzlephp.org/en/stable/quickstart.html#cookies
#Request-1
$login = $client->request("POST" , $url , [
"headers" => $header,
"body" => $xmlLogin
]);
// Here you should get all cookies from $login request and save it to $tmpCookieJar variable for example.
// For request-2 it will depends on the requirement of the endpoint or api.
#Request-2
$action2 = $client->request("POST" , $url , [
"headers" => $header,
"body" => $xmlLogin, // If necessary
"cookies" => $tmpCookieJar
]);

Using Guzzle 6 in Laravel to communicate with an external API

When I do this on the commandline:
curl -X POST -d '{"userDetails":{"username":"myself","password":"Myself123"}}' https://sub.domain.com/energy/api/login
it returns a sessionid like expected.
When I use Guzzle 6 on Laravel 5.5 and do this:
$client = new GuzzleHttp\Client();
$login = $client->post('https://sub.domain.com/energy/api/login', [
'userDetails' => [
'username' => 'myself',
'password' => 'Myself123'
]
]);
I get this error:
Server error: `POST https://sub.domain.com/energy/api/login` resulted in a `503 Service Unavailable` response: org.json.JSONException: JSONObject["userDetails"] not found.
What am I doing wrong?
Use this to debug and send request, sample here.
include these in class
//FOR GUZZLE
use GuzzleHttp\Client;
use GuzzleHttp\Exception\GuzzleException;
use GuzzleHttp\Exception\ServerException;
use GuzzleHttp\Exception\BadResponseException;
use GuzzleHttp\Exception\ClientException;
try{
$endPoint = "http://server_ip/security/subject";
$options = [ "auth" => [$username, $password], "headers" => ["header1" => value, "header2" => value] ];
$client = new Client();
$response = $client->get($endPoint, $options);
$responseArray = json_decode($response->getBody()->getContents(), true);
$status = $response->getStatusCode();
}catch(BadResponseException $e){
$response = $e->getResponse();
$reason = $response->getReasonPhrase();
$exceptionMessage = (!empty($reason)) ? $reason. ' email or password.' : 'Unauthorized email or password.';
}
You are not passing as a json. Try this:
$req = [
'userDetails' => [
'username' => 'myself',
'password' => 'Myself123'
]
];
$client = new GuzzleHttp\Client();
$login = $client->post('https://sub.domain.com/energy/api/login',
[
'json' => $req
]
);
return $login;

How to get past login screen on Guzzle call

I have to send information to an external website using cURL. I set up Guzzle on my Laravel application. I have the basics set up, but according to the documentation of the website, there is an action that's required for the username and password. How can I pass the 'action' along with the credentials needed to log in and get access?
The website states:
curl [-k] –dump-header <header_file> -F “action=login” -F “username=<username>” -F “password=<password>” https://<website_URL>
My controller:
$client = new \GuzzleHttp\Client();
$response = $client->get('http://website.com/page/login/', array(
'auth' => array('username', 'password')
));
$xml = $response;
echo $xml;
The website will load on the echo, but it will only pull up the login screen. I need those credentials to bypass the login screen (with a successful login) to get to the portion of information I need for cURL.
curl -F submits a POST request instead of a GET request. So you'll need to modify your code accordingly, something like
$client = new \GuzzleHttp\Client();
$response = $client->post('http://website.com/page/login/', [
'body' => [
'username' => $username,
'password' => $password,
'action' => 'login'
],
'cookies' => true
]
);
$xml = $response;
echo $xml;
See http://guzzle.readthedocs.org/en/latest/quickstart.html#post-requests, http://curl.haxx.se/docs/manpage.html#-F
Edit:
Just add ['cookies' => true] to requests in order to use the auth cookie associated with this GuzzleHttp\Client(). http://guzzle.readthedocs.org/en/latest/clients.html#cookies
$response2 = $client->get('http://website.com/otherpage/', ['cookies' => true]);
I was having trouble getting #JeremiahWinsley's answer to work on newer version of Guzzle so I've updated their code to work as of Guzzle 5.x.
Three major changes are required
Using form_params instead of body to prevent the error "Passing in the "body" request option as an array to send a POST request has been deprecated."
Changing the cookies to use the CookieJar object
Use ->getBody()->getContents() to get the body of the request
Here is the updated code:
$client = new \GuzzleHttp\Client();
$cookieJar = new \GuzzleHttp\Cookie\CookieJar();
$response = $client->post('http://website.com/page/login/', [
'form_params' => [
'username' => $username,
'password' => $password,
'action' => 'login'
],
'cookies' => $cookieJar
]
);
$xml = $response->getBody()->getContents();
echo $xml;
And to continue using cookies in future requests, pass in the cookieJar to the request:
$response2 = $client->get('http://website.com/otherpage/', ['cookies' => $cookieJar]);
I was having trouble getting #JeremiahWinsley's and #Samsquanch's answer to work on newer version of Guzzle. So I've updated the code to work as of Guzzle 6.x.
Guzzle 6.x. documents: http://docs.guzzlephp.org/en/stable/index.html
Here is the updated code:
use GuzzleHttp\Client;
use GuzzleHttp\Cookie\CookieJar;
try {
$client = new Client();
$cookieJar = new CookieJar();
$response = $client->request('POST', 'http://website.com/page/login/', [
'form_params' => [
'username' => 'test#example.com',
'password' => '123456'
],
'cookies' => $cookieJar
]);
$response2 = $client->request('GET', 'http://website.com/otherpage/', [
'cookies' => $cookieJar
]);
if ($response2->getStatusCode() == 200) {
return $response2->getBody()->getContents();
} else {
return "Oops!";
}
} catch (\Exception $exception) {
return 'Caught exception: ', $exception->getMessage();
}

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