Php 输入记录时web服务出错

我已经创建了一个用于注册的Web服务，但当我运行Web服务时，它会给我错误信息：

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分析错误：语法错误，中出现意外的“INTO”（T_字符串） 第13行的C:\xampp\htdocs\create.php

我已经在这里添加了代码

<?php
/* include db.config.php */
include_once(‘db_config.php’);

if($_SERVER[‘REQUEST_METHOD’] == “POST”){
// Get post data`
$firstName = isset($_POST[‘firstname’]) ? mysql_real_escape_string($_POST[‘firstname’]) : “”;
$lastName = isset($_POST[‘lastname’]) ? mysql_real_escape_string($_POST[‘lastname’]) : “”;
$email = isset($_POST[’email’]) ? mysql_real_escape_string($_POST[’email’]) : “”;
$password = isset($_POST[‘password’]) ? mysql_real_escape_string($_POST[‘password’]) : “”;
 // Here we set by default status In-active.
// Save data into database
$query = “INSERT INTO users (firstname,lastname,email,password) VALUES (‘$firstName’,’$lastName’,’$email’,’$password’)”;
$insert = mysql_query($query);

if($insert){
$data = array(“result” => 1, “message” => “Successfully user added!”);
} else {
$data = array(“result” => 0, “message” => “Error!”);
}
} else {
$data = array(“result” => 0, “message” => “Request method is wrong!”);
}

mysql_close($conn);
/* JSON Response */
header(‘Content-type: application/json’);
echo json_encode($data);

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请将
'
替换为
'
，
“
替换为”

我不认为这些是真实的引用，请改为
“
和
”
！它们似乎是MsWord单引号和双引号。出于对佛陀的爱，请停止使用mysql。我在dbconfig.php中添加了codeFYI。它们不再被维护。请参阅？改为了解，并使用或-将帮助您决定哪一个最适合您。但它仍会给出错误它在db c中给出了错误onfig.phppasse错误：语法错误，在
db_config.php
第8行的C:\xampp\htdocs\db_config.php中出现意外的文件结尾。

<?php
$db_host = 'localhost'; //hostname
$db_user = 'root'; // username
$db_password = "; 
$db_name = 'test'; 
$conn = mysql_connect($db_host, $db_user, $db_password );
mysql_select_db($db_name, $conn);
?>

$query = "INSERT INTO users (firstname,lastname,email,password) VALUES ('$firstName','$lastName','$email','$password')";