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在laravel中使用GuzzleHttp发送原始的空json

[英]send raw empty json with GuzzleHttp in laravel

My solution my be simple but honestly spent almost 5 hour and still not found my answer. 我的解决方法很简单,但老实说花费了将近5个小时,但仍然找不到我的答案。 I need Send request with Post and retrieve my data. 我需要发送带有Post请求并检索我的数据。 With POSTMAN it's work just fine. 使用POSTMAN,一切正常。

在此处输入图片说明

and here my Code 这是我的代码

use GuzzleHttp\Client;

 $client =new Client();

     $client->post($url,[
            'form_params' => [
              '{}'
            ]
        ]);

with POSTMAN , I don't need to set Header and all i need to send {} . 使用POSTMAN ,我不需要设置Header ,也不需要发送{}

Get this Error : 得到这个错误:

Client error: `POST http://URL` resulted in a `400 
Bad Request` response: {"error":"Failed when parsing body as json"}

Well,How Can I do it now? 好吧,我现在该怎么办?

It looks like you want to send your POST data without application/x-www-form-urlencoded which is what automatically added by using 'form_params' 看起来您想发送不带application/x-www-form-urlencoded POST数据,这是通过使用“ form_params”自动添加的

If you wish to do this, you should do something more akin to the following: 如果您希望这样做,则应该执行以下操作:

use GuzzleHttp\Client;
$client = new Client();
$response = $client->post($url, [
    'body' => '{}'
]);

I think you can just do 我想你可以做

$client->post($url,[
    'parameter_one' => $value1,
    'param_two' => $value2,
]);

Not sure about how to pass object, i think you can pass PHP object with data as second parameter or array of objects, if you wish.. 不确定如何传递对象,我想您可以根据需要传递带有数据的PHP对象作为第二个参数或对象数组。

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