php Guzzlehttp - 如何从 Guzzle 6 获取响应正文？
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Guzzlehttp - How get the body of a response from Guzzle 6?
提问by Greg
I'm trying to write a wrapper around an api my company is developing. It's restful, and using Postman I can send a post request to an endpoint like http://subdomain.dev.myapi.com/api/v1/auth/with a username and password as POST data and I am given back a token. All works as expected. Now, when I try and do the same from PHP I get back a GuzzleHttp\Psr7\Responseobject, but can't seem to find the token anywhere inside it as I did with the Postman request.
我正在尝试为我公司正在开发的 api 编写一个包装器。它很安静，使用 Postman 我可以向端点发送一个 post 请求，比如http://subdomain.dev.myapi.com/api/v1/auth/使用用户名和密码作为 POST 数据，然后我得到一个令牌。一切都按预期工作。现在，当我尝试从 PHP 执行相同操作时，我会返回一个GuzzleHttp\Psr7\Response对象，但似乎无法像处理 Postman 请求那样在其中的任何地方找到令牌。
The relevant code looks like:
相关代码如下所示：
$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...
The output of the code above looks something like (warning, incoming wall of text):
上面代码的输出看起来像（警告，传入的文本墙）：
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) {
}
}
}
The output from Postman was something like:
Postman 的输出类似于：
{
"data" : {
"token" "fasdfasf-asfasdfasdf-sfasfasf"
}
}
Clearly I'm missing something about working with the response objects in Guzzle. The Guzzle response indicates a 200 status code on the request, so I'm not sure exactly what I need to do to retrieve the returned data.
很明显，我缺少关于在 Guzzle 中使用响应对象的一些内容。Guzzle 响应在请求中指示 200 状态代码，所以我不确定我需要做什么来检索返回的数据。
回答by Federkun
Guzzle implements PSR-7. That means that it will by default store the body of a message in a Streamthat uses PHP temp streams. To retrieve all the data, you can use casting operator:
Guzzle 实现了PSR-7。这意味着默认情况下它会将消息正文存储在使用 PHP 临时流的Stream中。要检索所有数据，您可以使用强制转换运算符：
$contents = (string) $response->getBody();
You can also do it with
你也可以这样做
$contents = $response->getBody()->getContents();
The difference between the two approaches is that getContentsreturns the remaining contents, so that a second call returns nothing unless you seek the position of the stream with rewindor seek.
两种方法之间的区别在于getContents返回剩余的内容，因此除非您使用rewind或查找流的位置，否则第二次调用将不返回任何内容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
Instead, usings PHP's string casting operations, it will reads all the data from the stream from the beginning until the end is reached.
相反，使用 PHP 的字符串转换操作，它将从流中读取从头到尾的所有数据。
$contents = (string) $response->getBody(); // returns all the contents
$contents = (string) $response->getBody(); // returns all the contents
Documentation: http://docs.guzzlephp.org/en/latest/psr7.html#responses
回答by Maksim Ivanov
If expecting JSON back, the simplest way to get it:
如果期望返回 JSON，获取它的最简单方法是：
$data = json_decode($response->getBody()); // returns an object
// OR
$data = json_decode($response->getBody(), true); // returns an array
json_decode()will automatically cast the body to string, so there is no need to call getContents().
json_decode()会自动投身到string，所以没有必要调用getContents()。
