我正在尝试连接到 Yelp 的 API,目前使用的是 ZF2 和 ZendOAuth。我不知道为什么会收到 404。这是原始请求和响应标头。
POST /v2/search?term=tacos&location=sf HTTP/1.1
Host: api.yelp.com
Connection: close
Accept-Encoding: gzip, deflate
User-Agent: Zend\Http\Client
Content-Type: application/x-www-form-urlencoded
Authorization: OAuth realm="",oauth_consumer_key="<key>",oauth_nonce="<nonce>",oauth_signature_method="HMAC-SHA1",oauth_timestamp="1387401249",oauth_version="1.0",oauth_token="<token>",oauth_signature="<signature>"
HTTP/1.1 404 Not Found
Date: Wed, 18 Dec 2013 21:14:09 GMT
Server: Apache
X-Node: web41, api_com
Content-Length: 8308
Vary: User-Agent
Connection: close
Content-Type: text/html; charset=UTF-8
X-Mode: rw
X-Proxied: lb1
该请求是否看起来应该连接到某个地方?
这是一些代码。
$accessToken = new \ZendOAuth\Token\Access();
$accessToken->setToken('<token>');
$accessToken->setTokenSecret('<secret>');
$host = 'http://' . $_SERVER['HTTP_HOST'];
$config = array(
'consumerKey'=>'<key>',
'consumerSecret'=>'<secret>',
);
$client = $accessToken->getHttpClient($config);
$client->setUri('http://api.yelp.com/v2/search?term=tacos&location=sf');
$client->setMethod('POST');
$adapter = new \Zend\Http\Client\Adapter\Socket();
$client->setAdapter($adapter);
$response = $client->send();
$result = $response->getBody();
我见过的所有 OAuth 示例都使用请求令牌获取访问令牌,但 Yelp 已经给了我令牌和秘密,所以我正在尝试手动构建它。
更新: 改变
$client->setMethod('POST');
至
$client->setMethod('GET');
是第一步,但是参数不能手动添加到 URL 中,必须使用setParameterGet();
. 所以这是我更新的工作代码。
$accessToken = new \ZendOAuth\Token\Access();
$accessToken->setToken('<token>');
$accessToken->setTokenSecret('<secret>');
$host = 'http://' . $_SERVER['HTTP_HOST'];
$config = array(
'consumerKey'=>'<key>',
'consumerSecret'=>'<secret>',
);
$client = $accessToken->getHttpClient($config);
$client->setUri('http://api.yelp.com/v2/search');
$client->setMethod('GET');
$params = array('term'=>'tacos', 'location'=>'sf');
$client->setParameterGet($params);
$adapter = new \Zend\Http\Client\Adapter\Socket();
$client->setAdapter($adapter);
$response = $client->send();
$result = $response->getBody();