php将数据存储在mysql中作为可选(“a”)而不是普通字符串
我的swift应用程序有一个问题,每当我注册一个新用户登录到我的系统时,字符串就会被解析成不同的内容,并存储用户名,例如 我不知道它为什么这样做。以下是我在xcode swift中的代码:php将数据存储在mysql中作为可选(“a”)而不是普通字符串,php,mysql,json,swift,Php,Mysql,Json,Swift,我的swift应用程序有一个问题,每当我注册一个新用户登录到我的系统时,字符串就会被解析成不同的内容,并存储用户名,例如 我不知道它为什么这样做。以下是我在xcode swift中的代码: let myURL = NSURL(string: "http://localhost:8888/userRegister.php"); let request = NSMutableURLRequest(URL: myURL!); request.HTTPMethod = "PO
let myURL = NSURL(string: "http://localhost:8888/userRegister.php");
let request = NSMutableURLRequest(URL: myURL!);
request.HTTPMethod = "POST";
let postString = "email=\(userEmail)&password=\(userPassword)";
request.HTTPBody = postString.dataUsingEncoding(NSUTF8StringEncoding);
let task = NSURLSession.sharedSession().dataTaskWithRequest(request){
data, response, error in
if error != nil {
print("error=\(error)")
return
}
do{
let json = try NSJSONSerialization.JSONObjectWithData(data!, options: .MutableContainers) as? NSDictionary
if let parseJSON = json{
let resultValue = parseJSON["status"] as! String
print("result:\(resultValue)")
var isUserRegistered: Bool = false;
if(resultValue == "success"){
isUserRegistered = true;
}
var messageToDisplay:String = parseJSON["message"] as! String;
if (!isUserRegistered)
{
messageToDisplay = parseJSON["message"] as! String;
}
dispatch_async(dispatch_get_main_queue(), {
let myAlert = UIAlertController(title: "Alert", message: messageToDisplay, preferredStyle: UIAlertControllerStyle.Alert);
let okAction = UIAlertAction(title: "Ok", style: UIAlertActionStyle.Default){ action in
self.dismissViewControllerAnimated(true, completion: nil);
};
myAlert.addAction(okAction);
self.presentViewController(myAlert, animated: true, completion: nil);
}
)};
} catch { print(error)}
}
task.resume();
我的php文件(userRegister.php)
最后是存储查询的mysqlDao.php文件
<?php
class MySQLDao {
var $dbhost = null;
var $dbuser = null;
var $dbpass = null;
var $conn = null;
var $dbname = null;
var $result = null;
function __construct() {
$this->dbhost = Conn::$dbhost;
$this->dbuser = Conn::$dbuser;
$this->dbpass = Conn::$dbpass;
$this->dbname = Conn::$dbname;
}
public function openConnection() {
$this->conn = new mysqli($this->dbhost, $this->dbuser, $this->dbpass, $this- >dbname);
if (mysqli_connect_errno())
echo new Exception("Could not establish connection with database");
}
public function getConnection() {
return $this->conn;
}
public function closeConnection() {
if ($this->conn != null)
$this->conn->close();
}
public function getUserDetails($email)
{
$returnValue = array();
$sql = "select * from users where username='" . $email . "'";
$result = $this->conn->query($sql);
if ($result != null && (mysqli_num_rows($result) >= 1)) {
$row = $result->fetch_array(MYSQLI_ASSOC);
if (!empty($row)) {
$returnValue = $row;
}
}
return $returnValue;
}
public function getUserDetailsWithPassword($email, $userPassword)
{
$returnValue = array();
$sql = "select id,username from users where username='" . $email . "' and password='" .$userPassword . "'";
$result = $this->conn->query($sql);
if ($result != null && (mysqli_num_rows($result) >= 1)) {
$row = $result->fetch_array(MYSQLI_ASSOC);
if (!empty($row)) {
$returnValue = $row;
}
}
return $returnValue;
}
public function registerUser($email, $password)
{
$sql = "insert into users set username=?, password=?";
$statement = $this->conn->prepare($sql);
if (!$statement)
throw new Exception($statement->error);
$statement->bind_param(ss, $email, $password);
$returnValue = $statement->execute();
return $returnValue;
}
}
?>
这是我的前后工作代码。只是打开了可选变量的包装
之前:
let postString = "email=\(email)&password=\(password)&phone=\ (phone)&name=\(name)"
例如,名称存储如下:可选(“Wissa”)
之后:
let postString = "email=\(email!)&password=\(password!)&phone=\ (phone!)&name=\(name!)"
这种方式对我很有效。看起来您的电子邮件和密码字段是可选的,您应该在发布前将它们插入字符串时打开它们。$\u POST[“email”]
输出为什么?为什么要将电子邮件和密码实体化(密码应该散列)?我不认为我的字段是可选的。您能否指出,PBUSH25看起来像是您将可选(“a”)
作为一个值发布到PHP中,因此该错误最有可能出现在您的swift应用程序中htmlentities
将“
替换为”
,这就是为什么您会得到可选(“a”)
。您可能还想看看?@user3423164我不知道您的电子邮件和密码字段声明在哪里,但因为您将它们放入字符串“\(emailField)和一些比\(passwordField)“
更多的文本中,而这些变量被错误地放入了您的数据库中,我的第一个想法是它们可能是可选的。这里没有足够的代码让我可以确定,这就是为什么我建议使用它。
let postString = "email=\(email!)&password=\(password!)&phone=\ (phone!)&name=\(name!)"