当我尝试向我的服务器上的 API 发送 POST 请求时遇到问题，我遵循了许多不同的教程，但它仍然不起作用。 我知道我的问题出在 POST 请求上，但我无法解决! 所以这是我在 Swift 中的代码和我在 php 中的 API:(是的，我已经用代码中的真实 ID 替换了 xxxx)

总而言之，服务器接收请求，例如，如果我手动输入一个伪指令，它就可以工作，实际上是 POST 方法不起作用。服务器没有接收到 POST 参数

快速代码:

var request = NSMutableURLRequest(URL: NSURL(string: "http://localhost:8888/academy/test.php")!)
    var session = NSURLSession.sharedSession()
    request.HTTPMethod = "POST"

    var params = ["pseudo":"test"] as Dictionary<String, String>

    var err: NSError?
    request.HTTPBody = NSJSONSerialization.dataWithJSONObject(params, options: nil, error: &err)
    request.addValue("application/json", forHTTPHeaderField: "Content-Type")
    request.addValue("application/json", forHTTPHeaderField: "Accept")

    var task = session.dataTaskWithRequest(request, completionHandler: {data, response, error -> Void in
        println("Response: \(response)")
        var strData = NSString(data: data, encoding: NSUTF8StringEncoding)
        println("Body: \(strData)")
        var err: NSError?
        var json = NSJSONSerialization.JSONObjectWithData(data, options: .MutableLeaves, error: &err) as? NSDictionary

        // Did the JSONObjectWithData constructor return an error? If so, log the error to the console
        if(err != nil) {
            println(err!.localizedDescription)
            let jsonStr = NSString(data: data, encoding: NSUTF8StringEncoding)
            println("Error could not parse JSON: '\(jsonStr)'")
        }
        else {
            // The JSONObjectWithData constructor didn't return an error. But, we should still
            // check and make sure that json has a value using optional binding.
            if let parseJSON = json {
                // Okay, the parsedJSON is here, let's get the value for 'success' out of it
                var success = parseJSON["success"] as? Int
                println("Succes: \(success)")
            }
            else {
                // Woa, okay the json object was nil, something went worng. Maybe the server isn't running?
                let jsonStr = NSString(data: data, encoding: NSUTF8StringEncoding)
                println("Error could not parse JSON: \(jsonStr)")
            }
        }
    })
    task.resume()*/

PHP 代码:

$BDD_hote = 'xxxxx';
$BDD_bd = 'xxxxx';
$BDD_utilisateur = 'xxxxx';
$BDD_mot_passe = 'xxxxx';

try{
$bdd = new PDO('mysql:host='.$BDD_hote.';dbname='.$BDD_bd, $BDD_utilisateur, $BDD_mot_passe);
$bdd->exec("SET CHARACTER SET utf8");
$bdd->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_WARNING);
}

catch(PDOException $e){
echo 'Erreur : '.$e->getMessage();
echo 'N° : '.$e->getCode();
}
$pseudo = addslashes($_POST["pseudo"]);
$req = $bdd->query("SELECT * from users WHERE pseudo='$pseudo'");
$resultArray = array();
$donnees = $req->fetch();
echo json_encode($donnees);

提前致谢:)

最佳答案

正如其他人所指出的，请求的编码不太正确。您的服务器代码不期望 JSON 请求，而是使用 $_POST 变量(这意味着该请求应该具有 application/x-www 的 Content-Type -form-urlencoded).所以这就是你应该创造的。例如在 Swift 3 及更高版本中:

var request = URLRequest(url: url)
request.httpMethod = "POST"

let parameters = ["somekey" : "valueforkey"]
request.setContent(with: parameters)

let task = session.dataTask(with: request) { data, response, error in
    guard let data = data, error == nil else {
        print(error ?? "Unknown error")
        return
    }
    // your response parsing code here
}
task.resume()

setContent 方法将采用 ["key1": "foo", "key2": "bar"] 形式的字典并填充 httpBody 带有类似 key1=foo&key2=bar 的内容。如果这样做，服务器将能够从请求中解析 $_POST。

extension URLRequest {
    /// Set body and header for x-www-form-urlencoded request
    ///
    /// - Parameter parameters: Simple string dictionary of parameters to be encoded in body of request

    mutating func setContent(with parameters: [String: String]) {
        setValue("application/x-www-form-urlencoded", forHTTPHeaderField: "Content-Type")

        let array = parameters.map { entry -> String in
            let key = entry.key.addingPercentEncoding(withAllowedCharacters: .urlQueryValueAllowed)!
            let value = entry.value.addingPercentEncoding(withAllowedCharacters: .urlQueryValueAllowed)!
            return key + "=" + value
        }
        httpBody = array.joined(separator: "&").data(using: .utf8)!
    }
}

请注意，这也对值进行百分比编码(以防万一，键也是)，这很关键。虽然其他人建议将 addingPercentEncoding 与 .urlQueryAllowed 一起使用，但遗憾的是这不会完成这项工作，因为它会让某些保留字符(特别是 +)通过未转义。大多数 Web 服务会将 + 解释为空格，符合 the x-www-form-urlencoded spec .因此，采用 Alamofire 采用的方法，我定义了一个字符集，它将使用各种分隔符对值进行编码:

extension CharacterSet {

    /// Returns the character set for characters allowed in the individual parameters within a query URL component.
    ///
    /// The query component of a URL is the component immediately following a question mark (?).
    /// For example, in the URL `http://www.example.com/index.php?key1=value1#jumpLink`, the query
    /// component is `key1=value1`. The individual parameters of that query would be the key `key1`
    /// and its associated value `value1`.
    ///
    /// According to RFC 3986, the set of unreserved characters includes
    ///
    /// `ALPHA / DIGIT / "-" / "." / "_" / "~"`
    ///
    /// In section 3.4 of the RFC, it further recommends adding `/` and `?` to the list of unescaped characters
    /// for the sake of compatibility with some erroneous implementations, so this routine also allows those
    /// to pass unescaped.

    static var urlQueryValueAllowed: CharacterSet = {
        let generalDelimitersToEncode = ":#[]@"    // does not include "?" or "/" due to RFC 3986 - Section 3.4
        let subDelimitersToEncode = "!$&'()*+,;="

        var allowed = CharacterSet.urlQueryAllowed
        allowed.remove(charactersIn: generalDelimitersToEncode + subDelimitersToEncode)
        return allowed
    }()
}

有关 Swift 2 版本，请参阅 previous revision of this answer .

关于php - 使用 swift 和 API 进行 POST，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/28008874/