Guzzlehttp-如何从Guzzle 6获得响应的主体?
我正在尝试为我公司正在开发的API编写包装器。这很安静,使用邮递员,我可以向http://subdomain.dev.myapi.com/api/v1/auth/
用户端发送一个邮寄请求,例如使用用户名和密码作为POST数据,并且还给我一个令牌。所有工作均按预期进行。现在,当我尝试从PHP中进行操作时,我得到了一个GuzzleHttp\Psr7\Response
对象,但似乎无法像在Postman请求中那样在其内部的任何地方找到令牌。
相关代码如下:
$client = new Client(['base_uri' => 'http://companysub.dev.myapi.com/']);$response = $client->post('api/v1/auth/', [
'form_params' => [
'username' => $user,
'password' => $password
]
]);
var_dump($response); //or $resonse->getBody(), etc...
上面代码的输出看起来像(警告,文本输入墙):
object(guzzlehttp\psr7\response)#36 (6) { ["reasonphrase":"guzzlehttp\psr7\response":private]=>
string(2) "ok"
["statuscode":"guzzlehttp\psr7\response":private]=>
int(200)
["headers":"guzzlehttp\psr7\response":private]=>
array(9) {
["connection"]=>
array(1) {
[0]=>
string(10) "keep-alive"
}
["server"]=>
array(1) {
[0]=>
string(15) "gunicorn/19.3.0"
}
["date"]=>
array(1) {
[0]=>
string(29) "sat, 30 may 2015 17:22:41 gmt"
}
["transfer-encoding"]=>
array(1) {
[0]=>
string(7) "chunked"
}
["content-type"]=>
array(1) {
[0]=>
string(16) "application/json"
}
["allow"]=>
array(1) {
[0]=>
string(13) "post, options"
}
["x-frame-options"]=>
array(1) {
[0]=>
string(10) "sameorigin"
}
["vary"]=>
array(1) {
[0]=>
string(12) "cookie, host"
}
["via"]=>
array(1) {
[0]=>
string(9) "1.1 vegur"
}
}
["headerlines":"guzzlehttp\psr7\response":private]=>
array(9) {
["connection"]=>
array(1) {
[0]=>
string(10) "keep-alive"
}
["server"]=>
array(1) {
[0]=>
string(15) "gunicorn/19.3.0"
}
["date"]=>
array(1) {
[0]=>
string(29) "sat, 30 may 2015 17:22:41 gmt"
}
["transfer-encoding"]=>
array(1) {
[0]=>
string(7) "chunked"
}
["content-type"]=>
array(1) {
[0]=>
string(16) "application/json"
}
["allow"]=>
array(1) {
[0]=>
string(13) "post, options"
}
["x-frame-options"]=>
array(1) {
[0]=>
string(10) "sameorigin"
}
["vary"]=>
array(1) {
[0]=>
string(12) "cookie, host"
}
["via"]=>
array(1) {
[0]=>
string(9) "1.1 vegur"
}
}
["protocol":"guzzlehttp\psr7\response":private]=>
string(3) "1.1"
["stream":"guzzlehttp\psr7\response":private]=>
object(guzzlehttp\psr7\stream)#27 (7) {
["stream":"guzzlehttp\psr7\stream":private]=>
resource(40) of type (stream)
["size":"guzzlehttp\psr7\stream":private]=>
null
["seekable":"guzzlehttp\psr7\stream":private]=>
bool(true)
["readable":"guzzlehttp\psr7\stream":private]=>
bool(true)
["writable":"guzzlehttp\psr7\stream":private]=>
bool(true)
["uri":"guzzlehttp\psr7\stream":private]=>
string(10) "php://temp"
["custommetadata":"guzzlehttp\psr7\stream":private]=>
array(0) {
}
}
}
Postman的输出如下:
{ "data" : {
"token" "fasdfasf-asfasdfasdf-sfasfasf"
}
}
显然,我缺少有关在Guzzle中使用响应对象的知识。Guzzle响应在请求中指示200状态代码,因此我不确定要检索返回的数据到底需要做什么。
回答:
食尸鬼执行PSR-7。这意味着默认情况下,它将在使用PHP临时流的Stream中存储消息的主体。要检索所有数据,可以使用强制转换运算符:
$contents = (string) $response->getBody();
您也可以使用
$contents = $response->getBody()->getContents();
两种方法之间的区别是getContents
返回剩余内容,因此第二次调用将不返回任何内容,除非您使用rewind
或查找流的位置seek
。
$stream = $response->getBody();$contents = $stream->getContents(); // returns all the contents
$contents = $stream->getContents(); // empty string
$stream->rewind(); // Seek to the beginning
$contents = $stream->getContents(); // returns all the contents
取而代之的是,使用PHP的字符串强制转换操作,它将从流中读取所有数据,从头到尾。
$contents = (string) $response->getBody(); // returns all the contents$contents = (string) $response->getBody(); // returns all the contents
文档:http :
//docs.guzzlephp.org/en/latest/psr7.html#responses
以上是 Guzzlehttp-如何从Guzzle 6获得响应的主体? 的全部内容, 来源链接: utcz.com/qa/409302.html