我只想要在 php 文件中通过 AngularJS 调用 Ajax 后的成功/失败消息.但是我也得到了带有标签的完整HT

Question

>我只想要在php文件中通过AngularJS调用Ajax后的成功/失败消息。 但是我也得到了带有标签的完整HTML页面代码

我的 AngularJS 代码：

$http.post('ajax_Location.php',{
'uname': $scope.user.name, 
'uid': $scope.user.id,
'uemail':$scope.user.email}).
success(function(data, status, headers, config) {$scope.myData.fromServer = data;});

我的 php 代码：

<?php
$data = json_decode(file_get_contents("php://input"));
$uname = mysql_real_escape_string($data->uname);
$uid = mysql_real_escape_string($data->uid);
$uemail = mysql_real_escape_string($data->uemail);
$servername = "localhost";
$username = "root"; $password = "";
$conn = mysqli_connect($servername, $username, $password); 
$db=mysqli_select_db($conn,'mydbtest');
$sql='INSERT INTO formdetails (firstName,lastName,enrollNumber) VALUES ("'.$uname.'","'.$uemail.'",'.$uid.')';
$inserts=mysqli_query($conn,$sql);
echo "Record inserted successfully ";
?>

Answer 1

在你的php文件中试试这个 -

<?php 
$data = json_decode(file_get_contents("php://input")); 
$uname = mysql_real_escape_string($data->uname); 
$uid = mysql_real_escape_string($data->uid); 
$uemail = mysql_real_escape_string($data->uemail); 
$servername = "localhost"; 
$username = "root"; 
$password = ""; 
$conn = mysqli_connect($servername, $username, $password); 
$db=mysqli_select_db($conn,'mydbtest'); 
$sql='INSERT INTO formdetails (firstName,lastName,enrollNumber) VALUES ("'.$uname.'","'.$uemail.'",'.$uid.')'; 
$inserts=mysqli_query($conn,$sql);
echo json_encode(array("successfully")); 
exit;
?>