Question

我正在尝试将示例curl请求写入PHP

POST https://a.klaviyo.com/api/v1/list/{{ LIST_ID }}/members
curl https://a.klaviyo.com/api/v1/list/dqQnNW/members \
  -X POST \
  -d api_key=API_KEY \
  -d email=george.washington@example.com \
  -d properties='{ "$first_name" : "George", "Birthday" : "02/22/1732" }' \
  -d confirm_optin=true

这是我写的代码。我没有得到的部分是如何将电子邮件作为数据输入。

$data_string = 
            '{
            "email": "'.$email.'"
            }';



        $ch = curl_init("https://a.klaviyo.com/api/v1/list/$form_id/members?api_key=$mail_api_key");

        curl_setopt($ch, CURLOPT_CUSTOMREQUEST, "POST");                                                                     
        curl_setopt($ch, CURLOPT_POSTFIELDS, $data_string);                                                                  
        curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);                                                                      
        curl_setopt($ch, CURLOPT_HTTPHEADER, array(                                                                          
            'Content-Type: application/json'
            )                                                                       
        );

我得到的错误是

Array
(
    [status] => 400
    [message] => This request is missing the following required params: "email".
)

Answer 1

<?php
$email='test@gmmail.com';
$data_string =json_encode($email);


$form_id=5;
$mail_api_key=10;

$ch = curl_init("https://a.klaviyo.com/api/v1/list/$form_id/members?api_key=$mail_api_key");

curl_setopt($ch, CURLOPT_CUSTOMREQUEST, "POST");
curl_setopt($ch, CURLOPT_POSTFIELDS, $data_string);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
curl_setopt($ch, CURLOPT_HTTPHEADER, array(
        'Content-Type: application/json'
    )

);

form_id和api_key我将它们设置为硬编码，以便我可以看到响应。这段代码可以正常工作，只需删除$ form_id和mail_api_key并像设置它们一样设置它们。

Answer 2

来自php doc：＆＃34;此参数可以作为urlencoded字符串传递，例如＆＃39; para1 = val1＆amp; para2 = val2＆amp; ...＆＃39;或者作为一个数组，字段名称为键，字段数据为值。＆＃34; 因此，您应该将json字符串更改为数组或urlencode数据。

我在PHP中使用curl请求做错了什么？

2 个答案: