【发布时间】:2020-10-07 23:20:35
【问题描述】:
我一直在尝试从 mysql 数据库表中选择值(学生数据)并循环访问数据库以使用 PHP CURL Post 请求发送到 API,但它不起作用。
这是 API 正文:
{
"students":[
{
"admissionNumber": "2010",
"class":"js one"
},
{
"admissionNumber": "2020",
"class":"ss one"
}
],
"appDomain":"www.schooldomain.com"
}
我要发送的参数是“admissionNumber”和“class”参数,而“appDomain”对所有参数都是一样的。这是我的代码:
if(isset($_POST['submit'])){
$body = "success";
$info = "yes";
class SendDATA
{
private $url = 'https://url-of-the-endpoint';
private $username = '';
private $appDomain = 'http://schooldomain.com/';
// public function to commit the send
public function send($admNo,$class)
{
$url_array= array('admissionNumber'=>$admNo,'class'=>$class,'appDomain'=>$this-> appDomain);
$url_string = $data = http_build_query($url_array);
// using the curl library to make the request
$curlHandle = curl_init();
curl_setopt($curlHandle, CURLOPT_URL, $this->url);
curl_setopt($curlHandle, CURLOPT_RETURNTRANSFER, true);
curl_setopt($curlHandle, CURLOPT_POSTFIELDS, $url_string);
curl_setopt($curlHandle, CURLOPT_POST, 1);
$responseBody = curl_exec($curlHandle);
$responseInfo = curl_getinfo($curlHandle);
curl_close($curlHandle);
return $this->handleResponse($responseBody,$responseInfo);
}
private function handleResponse($body,$info)
{
if ($info['http_code']==200){ // successful submission
$xml_obj = simplexml_load_string($body);
// extract
return true;
}
else{
// error handling
return false;
}
}
}
$sms = new SendDATA();
$result = mysqli_query( $mysqli, "SELECT * FROM school_kids");
while ($row = mysqli_fetch_array($result)) {
$admNo = $row['admNo'];
$class = $row['class'];
$sms->send($admNo,$class,"header");
echo $admNo. " ".$class;
}
}
【问题讨论】: