[英]convert curl to guzzle
i have this perfectly working curl command: 我有这个完美的curl命令:
curl -i --data "site=walletgroove.com&placement=above&device=desktop&source=*&campaign=*&url=*&country=*&active=1" http://10.0.0.38/adserver/src/public/api/rule
which i tried to execute with guzzle but for some reason i keep getting error from my code, an exception to be more accurate. curl -i --data "site=walletgroove.com&placement=above&device=desktop&source=*&campaign=*&url=*&country=*&active=1" http://10.0.0.38/adserver/src/public/api/rule
其中我试图用guzzle执行,但由于某种原因我不断从我的代码中得到错误,更准确的例外。 this exception is thrown when params are not passed properly. 当params未正确传递时抛出此异常。 this is one of few tries i had:
这是我的几次尝试之一:
public function testApi_postRule()
{
$client = new Client();
$client->post('http://10.0.0.38/adserver/src/public/api/rule',[ 'query' => [
'site' => 'walletgroove.com',
'placement' => 'guzzle_unique_placement',
'device' => 'desktop',
'source' => 'guzzource',
'campaign' => '*',
'country' => '*',
'url' => '*',
'active' => '1'
]]);
}
any idea what am i doing wrong here?? 知道我在这里做错了什么?
You are passing the query
parameter, which appends a query string instead of sending in the request body. 您正在传递
query
参数,该参数附加查询字符串而不是在请求正文中发送。 For a POST request, you probably want to use form_params
as shown in the documentation . 对于POST请求,你可能想使用
form_params
如图所示的文件 。
$client->post('http://10.0.0.38/adserver/src/public/api/rule', [
'form_params' => [
'site' => 'walletgroove.com',
'placement' => 'guzzle_unique_placement',
'device' => 'desktop',
'source' => 'guzzource',
'campaign' => '*',
'country' => '*',
'url' => '*',
'active' => '1'
]
]);
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