laravel GuzzleHttp\Psr7\Request 类的对象无法转换为字符串
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Object of class GuzzleHttp\Psr7\Request could not be converted to string
提问by php dev
I've got an issue for laravel 5.4 when I trying to using guzzleHttp. here is my code.
我在尝试使用 guzzleHttp 时遇到了 laravel 5.4 的问题。这是我的代码。
use GuzzleHttp\Client;
$url = 'http://example.com';
$client = new Client();
$parameter = ['query' => ['name' => 'xxx', 'address' => 'yyy'], 'headers' => [ 'User-Agent' => 'xxxx', 'exceptions' => false, 'timeout' => 10 ]];
$res = $client->request('GET', $url, $parameter);
if ($res->getStatusCode() == 200)
{
$json = (string)$res->getBody();
return $json;
}
and I've got this error on log: Error Exception: Object of class GuzzleHttp\Psr7\Request could not be converted to string
我在日志中遇到了这个错误：错误异常：GuzzleHttp\Psr7\Request 类的对象无法转换为字符串
what is wrong with my code? please kindly help me. fyi, this error not always happen. sometimes it show this error, sometimes success.
我的代码有什么问题？请帮助我。仅供参考，此错误并不总是发生。有时会显示此错误，有时会显示成功。
thank you
谢谢你
回答by Exprator
$json = $res->getBody()->getContents();
try this
尝试这个
回答by Farhad Hossen
Try this.....
尝试这个.....
try {
$parameter = ['query' => ['name' => 'xxx', 'address' => 'yyy'], 'headers' => [ 'User-Agent' => 'xxxx', 'exceptions' => false, 'timeout' => 10 ]];
$res = $client->request('GET', $url, $parameter);
if ($res->getStatusCode() == 200)
{
return $res->getBody()->getContents();
}
}catch(Exception $e){
echo 'Caught exception: ', $e->getMessage();
}
回答by James Mark Saphani
$response = $client->post('http:yanjye.com3', ['phone' => '00','password' => '5555',]);
if ($response->getStatusCode() == 200){
$json = (string)$response->getBody();
return $json;
}
var_dump( $response);
die();
Hello, brother, I think this is the Good way to which is Working On both laravel 5.2[larave 5.2]
i have removed Httm
[laraver 5.2][1]
and use this code :
[1]: https://laravel.com/docs/7.x/http-client`
