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Php Guzzle中异步请求的主体_Php_Http_Guzzle - Fatal编程技术网

Php Guzzle中异步请求的主体

Php Guzzle中异步请求的主体,php,http,guzzle,Php,Http,Guzzle,我想使用Guzzle PHP HTTP客户端发送一个异步请求,但它似乎只允许body是字符串 我的头变量是 $headers = [ "Authorization" : $token ]; 同样地,我想让body也作为数组 $body = [ "x"=>$y, "y"=>$z, ] 我将请求变量设置为 $request = new \GuzzleHttp\Psr7\Request( 'POST', 'API_URL', $head

我想使用Guzzle PHP HTTP客户端发送一个异步请求,但它似乎只允许body是字符串

我的头变量是

$headers = [
    "Authorization" : $token
];
同样地,我想让body也作为数组

$body = [
    "x"=>$y,
    "y"=>$z,
]
我将请求变量设置为

$request = new  \GuzzleHttp\Psr7\Request(
    'POST',
    'API_URL',
    $headers,
    $body
 );
然而,我得到了InvalidArgumentException无效的资源类型:数组错误,但在尝试时,
$body=“一些无用的字符串”
,请求被发送到服务器,但由于body没有适当的参数,所以得到了错误


我如何在这里使用所需的键将Body设置为数组/(嵌套数组,如果需要)。

使用
json\u encode
函数,将Body数组传递给

$request = new \GuzzleHttp\Psr7\Request(
    'POST',
    'API_URL',
    $headers,
    json_encode($body)
);