尝试从 php 脚本获取 Json 数据时出错

Question

我尝试获取数据时出现以下错误。在互联网上我读到这是因为 php 脚本无效并且没有 return json 数据。但是 php 脚本运行良好并输出正确的数据。

错误信息：

Error Domain=NSCocoaErrorDomain Code=3840 "JSON text did not start with array or object and option to allow fragments not set." UserInfo={NSDebugDescription=JSON text did not start with array or object and option to allow fragments not set.}

我试图允许片段，但后来我收到了另一条错误消息。

这是我尝试获取数据的 swift 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 return 当我打印它时：数组 ( [status] => 200 [userFirstName] => Paul [userLastName] => Heinemeyer [userEmail] => paul_heine [userId] => 63 )

Answer 1

当你正确格式化你的 PHP 代码时，你会看到，在其他部分你有

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

结果是

failed{...}

正如错误信息已经指出的，这个"JSON text did not start with array or object ..."

也许另一个 if 部分有类似的输出。

尝试从 php 脚本获取 Json 数据时出错

Error when trying to get Json data from php script

php

database

json

nserror

swift