Question

我正在尝试在swift中发布请求。我在Objective-C中成功但是使用swift我无法发送请求。 PHP中的$emailOK无法获取信息。问题出在哪儿？谢谢你的帮助。这里是快速代码

func httpPost1(url:String, postData: String, completion: String -> Void) {
        let request = NSMutableURLRequest(URL: NSURL(string: url)!)
        request.HTTPMethod = "POST"
        let postString = postData
        request.HTTPBody = postString.dataUsingEncoding(NSUTF8StringEncoding)
        let task = NSURLSession.sharedSession().dataTaskWithRequest(request) { data, response, error in
            guard error == nil && data != nil else {                                                          // check for fundamental networking error
                print("error=\(error)")
                return
            }

            if let httpStatus = response as? NSHTTPURLResponse where httpStatus.statusCode != 200 {           // check for http errors
                print("statusCode should be 200, but is \(httpStatus.statusCode)")
                print("response = \(response)")
            }

            let responseString = NSString(data: data!, encoding: NSUTF8StringEncoding)
            print("responseString = \(responseString)")
        }
        task.resume()
        }

这里是执行代码

httpPost1("xxxxx", postData: "emailOK=Hello") { result in

                print(result)//result is your string-response from server

            }

这是我的PHP代码

<?php

$link = mysqli_connect($dbhost, $username, $dbpass, $database);

if (!$link) {
    echo "Error: Unable to connect to MySQL." . PHP_EOL;
    echo "Debugging errno: " . mysqli_connect_errno() . PHP_EOL;
    echo "Debugging error: " . mysqli_connect_error() . PHP_EOL;
    exit;
}

echo "Success: A proper connection to MySQL was made! The my_db database is great." . PHP_EOL;
echo "Host information: " . mysqli_get_host_info($link) . PHP_EOL;


$emailOK = isset($_GET["emailOK"]) ? $_GET["emailOK"] : '';

echo $emailOK;

$query = "INSERT INTO UserInfo VALUES ('', '$emailOK')";
mysqli_query($link, $query) or die (mysqli_error("error"));


mysqli_close($link);
?>

Answer 1

我找到了问题，感谢@OOPer评论。问题出在PHP $emailOK = isset($_GET["emailOK"]) ? $_GET["emailOK"] : '';中，应该是$emailOK = $_POST['emailOK'];

Answer 2

尝试使用此代码创建会话对象。我不确定这是否能完全解决你的问题，但我认为你应该试一试。

let config = NSURLSessionConfiguration.defaultSessionConfiguration()
let session = NSURLSession(configuration: config)

Answer 3

如果您需要将电子邮件作为网址的一部分发送，则应将其包含在网址中：

vm.type = getFromURL('what=') || 'Events';

function getFromURL(filterName){
            var url = $location.url();
            // if there is already a hash in the url
            if(url && url.indexOf('params?') >= 0){
                // If the search param already exists
                if(url.indexOf(filterName) >= 0){
                    // Split url at already existing search param
                    var searchParams = url.split(filterName)[1];
                    // if there is still an '&' in the url do something
                    if(searchParams.indexOf('&') >= 0){
                        // split at the '&' and get rid of anything after it(other search params)
                        var singledParam = searchParams.split('&')[0];
                        return singledParam;
                    }else{
                        return searchParams;
                    }
                // If the suggested search param doesn't exist
                }
            }
        }

但是由于您发送的是POST请求，因此通常会在HTTP正文中发送。相当常见的方式是以JSON格式发送它：

var requestURL = url + "?" + postData

let request = NSMutableURLRequest(URL: NSURL(string: url)!)

Answer 4

如果您不想花费大量时间无用地编写成功的获取/发布请求，那么您最好选择使用所需的所有方法来使用库。

以我的拙见，我建议你 Alamofire https://github.com/Alamofire/Alamofire

否则，要成功完成您的请求，您应指定 Content-Type ， Content-Type 标题信息

希望这会对你有所帮助。

Answer 5

    let dict = ["emailOK" : "Hello"]
    let jsonData = try! NSJSONSerialization.dataWithJSONObject(dict, options: NSJSONWritingOptions.PrettyPrinted)
    let jsonString = NSString(data: jsonData, encoding: NSUTF8StringEncoding)! as String
    let request = NSMutableURLRequest(URL: NSURL(string:"url")!)
    request.HTTPMethod = "POST"
    let postString = jsonString
    request.HTTPBody = postString.dataUsingEncoding(NSASCIIStringEncoding)
    let task = NSURLSession.sharedSession().dataTaskWithRequest(request) { data, response, error in
        guard error == nil && data != nil else {                                                          // check for fundamental networking error
            print("error=\(error)")
            return
        }

        if let httpStatus = response as? NSHTTPURLResponse where httpStatus.statusCode != 200 {           // check for http errors
            print("statusCode should be 200, but is \(httpStatus.statusCode)")
            print("response = \(response)")
        }

        let responseString = NSString(data: data!, encoding: NSUTF8StringEncoding)
        print("responseString = \(responseString)")
    }
    task.resume()
}

无法在swift中发布请求

5 个答案: