我想向Guzzle Http请求添加一些数据。有文件名、文件内容和带有授权密钥的头。
$this->request = $this->client->request('POST', 'url', [
'multipart' => [
'name' => 'image_file',
'contents' => fopen('http://localhost:8000/vendor/l5-swagger/images/logo_small.png', 'r'),
'headers' =>
['Authorization' => 'Bearer uCMvsgyuYm0idmedWFVUx8DXsN8QzYQj82XDkUTw']
]]);
但是我得到错误
可捕获的致命错误:传递给GuzzleHttp''Psr7''MultipartStream::addElement()的参数2的类型必须为数组、字符串给定,在vendor ''guzzlehttp''psr7''src''MultipartStream.php第70行调用,并在vendor '' guzzlehttp''psr7''''src''MultipartStream.php第79行中定义
在Guzzle 6中,文档是这样的:http://docs.guzzlephp.org/en/latest/request-options.html#multipart
谁知道我哪里搞错了?
这是解决方案。具有访问令牌的标头应位于多部分节之外。
$this->request = $this->client->request('POST', 'request_url', [
'headers' => [
'Authorization' => 'Bearer access_token'
],
'multipart' => [
[
'Content-type' => 'multipart/form-data',
'name' => 'image_file',
'contents' => fopen('image_file_url', 'r')
]
]
]);
根据文档,"multipart的值是关联数组的数组",因此您需要嵌套更深一层:
$this->request = $this->client->request('POST', 'url', [
'multipart' => [
[
'name' => 'image_file',
'contents' => fopen('http://localhost:8000/vendor/l5-swagger/images/logo_small.png', 'r'),
'headers' => ['Authorization' => 'Bearer uCMvsgyuYm0idmedWFVUx8DXsN8QzYQj82XDkUTw']
]
]
]);
试试这个适用于我
use GuzzleHttp'Client;
use GuzzleHttp'Psr7'Utils;
$this->client = new Client([
'base_uri' => 'https://baseurl'
]);
$body = Utils::tryFopen($tempPath . $fileName, 'r');
$res = $this->client->request(
'POST',
'url',
[
'headers' => [
...
],
'body' => $body
]
);