Question

当我尝试获取数据时出现以下错误。在互联网上我读它，因为PHP脚本无效，不返回json数据。但php脚本运行正常并输出正确的数据。

错误讯息：

Error Domain = NSCocoaErrorDomain Code = 3840“JSON文本没有以数组或对象开头，并且选项允许未设置片段。” UserInfo = {NSDebugDescription = JSON文本不以数组或对象开头，并且选项允许未设置片段。}

我试图允许片段，但后来我只得到另一条错误信息。

以下是我尝试获取数据的快速代码：

let myUrl = NSURL(string: "http://xxxxxxxxxxx.xxx/xxxxxxxx.php")

let request = NSMutableURLRequest(URL: myUrl!)
request.HTTPMethod = "POST"

let postString = "userEmail=\(userEmail!)&userPassword=\(userPassword!)"

request.HTTPBody = postString.dataUsingEncoding(NSUTF8StringEncoding)

NSURLSession.sharedSession().dataTaskWithRequest(request, completionHandler: { (data:NSData?, response:NSURLResponse?, error:NSError?) -> Void in

    dispatch_async(dispatch_get_main_queue())
    {
        if(error != nil)
        {
            var alert = UIAlertController(title: "Achtung", message: error?.localizedDescription, preferredStyle: UIAlertControllerStyle.Alert)

            let action = UIAlertAction(title: "Ok", style: UIAlertActionStyle.Default, handler: nil)

            alert.addAction(action)

            self.presentViewController(alert, animated: true, completion: nil)
        }
        print("1")
        do {
            let json = try NSJSONSerialization.JSONObjectWithData(data!, options: .MutableContainers) as? NSDictionary

            if let parseJSON = json {

                let userId = parseJSON["userId"] as? String
                if( userId != nil)
                {
                    print("SUCESS FUCKER")
                    let mainView = self.storyboard?.instantiateViewControllerWithIdentifier("main") as! FlickrPhotosViewController

                    let mainPageNavi = UINavigationController(rootViewController: mainView)
                    //open mainView
                    let appdele = UIApplication.sharedApplication().delegate
                    appdele?.window??.rootViewController = mainPageNavi


                } else {
                    let userMassage = parseJSON["message"] as? String
                    let myAlert = UIAlertController(title: "Alert", message: userMassage, preferredStyle: UIAlertControllerStyle.Alert);

                    let okAction = UIAlertAction(title: "OK", style: UIAlertActionStyle.Default, handler: nil)
                    myAlert.addAction(okAction);
                    self.presentViewController(myAlert, animated: true, completion: nil)

                }

            }
        } catch{
            print(error)
            print("FAILED CATCHED")
        }

    }
}).resume()

这是php文件的重要部分：

$userSecuredPassword = $userDetails["user_password"];

$userSalt = $userDetails["salt"];

if($userSecuredPassword === sha1($userPassword . $userSalt))
{
    $returnValue["status"]="200";

    $returnValue["userFirstName"] = $userDetails["first_name"];

    $returnValue["userLastName"] = $userDetails["last_name"];

    $returnValue["userEmail"] = $userDetails["email"];

    $returnValue["userId"] = $userDetails["user_id"];
} else {
    $returnValue["status"]="403";

    $returnValue["message"]="User not found";

     echo "failed";

    echo json_encode($returnValue);

    return;
}



echo json_encode($returnValue);

$ returnValue在我打印时返回：数组（[status] =＆gt; 200 [userFirstName] =＆gt;保罗[userLastName] =＆gt; Heinemeyer [userEmail] =＆gt; paul_heine [userId] =＆gt; 63）

Answer 1

正确格式化PHP代码后，您会看到，在其他部分中有

echo "failed";
echo json_encode($returnValue);

导致

failed{...}

正如错误消息所说，这个“JSON文本没有以数组或对象开头......”

如果是其他部分，也许有相似的输出。

尝试从php脚本获取Json数据时出错

1 个答案: