从Swift将数据发布到PHP方法 [英] POST data to a PHP method from Swift

问题描述

我正在尝试从Swift向我的PHP文件中发布一些信息. 我的php文件已执行，但发布的变量只是不进入php文件.我在做什么错了?

I'm trying to post some info to my PHP file from Swift. My php file is executed, but the posted variables just don't get through to the php file. What am I doing wrong?

快捷代码:

@IBAction func buttonPress(sender: AnyObject) {

        let request = NSMutableURLRequest(URL: NSURL(string: "http://www.domain.com/php_swift_test/insert.php")!)
        request.HTTPMethod = "POST"

        let postString = "a=test&b=bla"
        request.HTTPBody = postString.dataUsingEncoding(NSUTF8StringEncoding)

        let task = NSURLSession.sharedSession().dataTaskWithRequest(request) {
            data, response, error in

            if error != nil {
                print("error=\(error)")
                return
            }

            print("response = \(response)")

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

PHP代码:

<?php
    @session_start();
    @ob_start();

    $host='localhost';
    $user='test';
    $password='Passw0rd99';
    $db_name="mysql_test"; 

    $connection = mysql_connect($host,$user,$password);

    $a = $_POST['a'];
    $b = $_POST['b'];

    if(!$connection){
        die('Connection Failed');
    }
    else{
        $dbconnect = @mysql_select_db($db_name, $connection);

        if(!$dbconnect){
            die('Could not connect to Database');
        }
        else{
            $query = "INSERT INTO res_club (FirstName, LastName) VALUES ('$a','$b')";
            mysql_query($query, $connection) or die(mysql_error());

            echo 'Successfully added.';
            echo $query;
            echo $a.$b;
        }
    }
?>

一个空行被添加到数据库中，没有名字和姓氏. PHP文件未获取$_Post['a']和b

An empty row is added to the database, with no first name and last name. The PHP file doesn't get the $_Post['a'] and b

echo语句echo $a.$b也保持空白.没有显示错误.

The echo statement, echo $a.$b stays blank too. No errors are shown.

推荐答案

这对我有用.

视频- https://youtu.be/wYkZ47Rz8iU

快速代码-示例

let request = NSMutableURLRequest(URL: NSURL(string: "http://www.kandidlabs.com/YouTube/SwiftToMySQL/insert.php")!)
        request.HTTPMethod = "POST"
        let postString = "a=\(usernametext.text!)&b=\(password.text!)&c=\(info.text!)&d=\(number.text!)"
        request.HTTPBody = postString.dataUsingEncoding(NSUTF8StringEncoding)

        let task = NSURLSession.sharedSession().dataTaskWithRequest(request) {
            data, response, error in

            if error != nil {
                print("error=\(error)")
                return
            }

            print("response = \(response)")

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

PHP代码-示例

 <?php
    $host='localhost';
    $user='root';
    $password='';

    $connection = mysql_connect($host,$user,$password);

    $usernmae = $_POST['a'];
    $pass = $_POST['b'];
    $info = $_POST['c'];
    $num = $_POST['d'];

    if(!$connection)
    {
        die('Connection Failed');
    }
    else
    {
        $dbconnect = @mysql_select_db('YoutubeTutorialDB', $connection);

        if(!$dbconnect)
        {
            die('Could not connect to Database');
        }
        else
        {
            $query = "INSERT INTO `YoutubeTutorialDB`.`Users` (`Username`, `Password`, `Info`, `FavoriteNumber`)
                VALUES ('$username','$pass','$info','$num');";
            mysql_query($query, $connection) or die(mysql_error());

            echo 'Successfully added.';
            echo $query;
        }
    }
?>

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