我试图发送通知，每当数据库插入新的值，但我不断得到“错误，在发送请求”反复而尝试这段代码，不能找到什么问题，我已经使用服务器api密钥的建议。我使用了一个表单，它包含两个文本框，这两个文本框将指向上面的另一个文件。
<?php
$host='localhost';
$username='';
$pwd="";
$db="";
$con=mysqli_connect($host,$username,$pwd,$db);
if($_SERVER['REQUEST_METHOD']=='POST'){
$full_name = $_POST['full_name'];
$contact_number = $_POST['contact_number'];
//require_once('dbConnect.php');
$sql = "INSERT INTO notification (full_name,contact_number) VALUES (
'$full_name'
'$contact_number')";
$check = "SELECT * from notification where full_name='$full_name' AND contact_number='$contact_number'";
$checkData = mysqli_query($con,$check);
if (mysqli_num_rows($checkData) > 0) {
echo "Request already posted";
}else{
if(mysqli_query($con,$sql)){
$notiTitle = "notification request";
$notiMessage ="by".$full_name;
sendNotification($notiTitle, $notiMessage);
echo "sucessfully added";
}else{
echo "error in sending request";
}
}
}else{
echo 'error';
}
function sendNotification($title, $msg) {
$titlee = $title;
$message = $msg;
$path_to_fcm = 'https://fcm.googleapis.com/fcm/send';
$server_key = "AIz#################################";
$sql = "SELECT app_id FROM user_app_id";
$result = mysqli_query($con,$sql);
// fetch all key of devices
$finalKey=array();
while($row= mysqli_fetch_array($result)){
$finalKey[]=$row['app_id'];
}
$headers = array(
'Authorization:key=' .$server_key,
'Content-Type : application/json');
$fields = array('registration_ids'=>$finalKey, 'notification'=>array('title'=>$title, 'body'=>$message));
$payload = json_encode($fields);
$curl_session = curl_init();
curl_setopt($curl_session, CURLOPT_URL, $path_to_fcm);
curl_setopt($curl_session, CURLOPT_POST, true);
curl_setopt($curl_session, CURLOPT_HTTPHEADER, $headers);
curl_setopt($curl_session, CURLOPT_RETURNTRANSFER, true);
curl_setopt($curl_session, CURLOPT_SSL_VERIFYPEER, false);
curl_setopt($curl_session, CURLOPT_IPRESOLVE, CURL_IPRESOLVE_V4);
curl_setopt($curl_session, CURLOPT_POSTFIELDS, $payload);
$result = curl_exec($curl_session);
curl_close($curl_session);
echo $result;
mysqli_close($con);
}
<html>
<body>
<form action="init.php" method="post">
Full Name: <input type="text" name="full_name"><br>
Contact No: <input type="text" name="contact_number"><br>
<input type="submit" value="submit">
</form>
</body>
</html>
?>有人能帮我吗？
1条答案
按热度按时间qybjjes11#
不确定这是否是您的代码中的一个输入错误，但是您需要调整php代码—您已经 Package 了
HTML在在html开始之前尝试退出php段，如下所示；