如何使用触发器发送firebase推送通知
如何使用触发器发送firebase推送通知
提问于 2018-04-11 00:23:00
我试图在数据库插入新值时发送通知，但在尝试此代码时不断收到“发送请求时出错”，无法找到问题所在，我已按照建议使用了服务器api密钥。我使用了一个带有两个文本框的表单，该文本框将被定向到上面的另一个文件。
<?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>
?>有人能在这方面帮我吗？
回答 1
Stack Overflow用户
发布于 2018-04-11 09:42:47
不确定是不是代码中的拼写错误，但您需要调整PHP代码-您已经将HTML包装在<?php ?>段中。
尝试在HTML启动之前退出PHP段，如下所示；
?>
<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>页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/49758563
复制相关文章
相似问题
腾讯云开发者
