当我尝试使用guzzleHttp时,我遇到了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;
}
我在日志中遇到此错误: 错误异常:类GuzzleHttp \ Psr7 \ Request的对象无法转换为字符串
我的代码出了什么问题?请帮助我。 fyi,这个错误并不总是发生。有时它会显示此错误,有时会成功。
谢谢
答案 0 :(得分:8)
$json = $res->getBody()->getContents();
试试这个
答案 1 :(得分:0)
试试这个......
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();
}
答案 2 :(得分:0)
$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`