PHP：如何从 JSON 中获取变量（从 Swift iOS 应用程序发送）并以 JSON 响应

Posted 2023-03-08

【中文标题】PHP：如何从 JSON 中获取变量（从 Swift iOS 应用程序发送）并以 JSON 响应

【英文标题】：PHP: how to get variables from a JSON(sent from a Swift iOS app) and respond with a JSON

【发布时间】：2015-05-14 08:24:12

【问题描述】：

我正在使用 Swift 开发一个 ios 应用程序，它应该根据用户的位置从 mysql 数据库中获取一些数据。我不懂 php，也找不到解释如何从应用程序接收数据的资源。

我有这个 PHP 代码：

<?php

// Create connection
$con=mysqli_connect("localhost","*******","*******","*******");

// Check connection
if (mysqli_connect_errno())

  echo "Failed to connect to MySQL: " . mysqli_connect_error();


// This SQL statement selects ALL from the table 'Locations'
$sql = "SELECT * FROM *******";

// Check if there are results
if ($result = mysqli_query($con, $sql))


    // If so, then create a results array and a temporary one
    // to hold the data
    $resultArray = array();
    $tempArray = array();

    // Loop through each row in the result set
    while($row = $result->fetch_object())
    
        // Add each row into our results array
        $tempArray = $row;
        array_push($resultArray, $tempArray);
    

    // Finally, encode the array to JSON and output the results
    echo " \"posts\": ";
    echo json_encode($resultArray);
    echo "";


// Close connections
mysqli_close($con);
?>

正如你所看到的，当它被调用时，它会从一个表中获取所有数据并将其作为 JSON 返回。我要做的下一步是使用以下代码从 Swift 应用程序发送我的位置：

@IBAction func submitAction(sender: AnyObject) 

            //declare parameter as a dictionary which contains string as key and value combination.
            var parameters = ["name": nametextField.text, "password": passwordTextField.text] as Dictionary<String, String>

            //create the url with NSURL 
            let url = NSURL(string: "http://myServerName.com/api") //change the url

            //create the session object 
            var session = NSURLSession.sharedSession()

            //now create the NSMutableRequest object using the url object
            let request = NSMutableURLRequest(URL: url!)
             request.HTTPMethod = "POST" //set http method as POST

            var err: NSError?
            request.HTTPBody = NSJSONSerialization.dataWithJSONObject(parameters, options: nil, error: &err) // pass dictionary to nsdata object and set it as request body

            request.addValue("application/json", forHTTPHeaderField: "Content-Type")
            request.addValue("application/json", forHTTPHeaderField: "Accept")

            //create dataTask using the session object to send data to the server
            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()
        

感谢http://jamesonquave.com/blog/making-a-post-request-in-swift/

我不知道如何“获取”（接受，使用什么函数）这个 JSON：

"items": [
            
                "minLat": "43.000000",
                "maxLat": "44.000000",
                "minLon": "-79.000000",
                "maxLon": "-78.000000",
            
          ]
    

为了在 PHP 中有这样的东西：

$minLat = $json['minLat'];
$maxLat = $json['maxLat'];
$minLon = $json['minLon'];
$maxLon = $json['maxLon'];

$sql = "SELECT * FROM ******* WHERE latitude BETWEEN".$minLat." AND ".$maxLat." AND longitude BETWEEN ".$minLon." AND ".$maxLon;

谢谢

【问题讨论】：

嗨 Andrei，最简单的方法是您必须在服务器中创建 index.php 文件，并在您的 swift 应用程序中指定 myServerName.com/index.php url，在 php 文件中获取请求，如 if (isset($_POST["your_key_here"])) $json = json_decode($_POST["your_key_here"]);

你需要使用 Alamofire 和 SwiftyJSON。

我的 Swift 代码没有问题！！我有它的 PHP

@styopdev 感谢您的回复，但是如果 (isset($_POST["your_key_here"])) 中的“your_key_here”应该是什么？我不使用表格。

【参考方案1】：

答案实际上非常简单：

在我评论这两行之前，首先没有任何效果：

request.addValue("application/json", forHTTPHeaderField: "Content--Type")
request.addValue("application/json", forHTTPHeaderField: "Accept")

然后我使用字符串而不是 JSON 来发送 POST 数据（它当然也可以与 JSON 一起使用，但这就是目前的工作方式）：

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

let postString = "minLat=43.0&maxLat=44.0&minLon=26.0&maxLon=27.0";

request.HTTPBody = postString.dataUsingEncoding(NSUTF8StringEncoding);

在服务器端：

$minLat = $_REQUEST["minLat"];
$maxLat = $_REQUEST["maxLat"];
$minLon = $_REQUEST["minLat"];
$maxLon = $_REQUEST["maxLat"];

:|