【问题标题】:POST data to a PHP page from C# WinForm从 C# WinForm POST 数据到 PHP 页面
【发布时间】:2008-11-20 04:30:52
【问题描述】:

我有一个 winForms NET3.5SP1 应用程序,想将数据发布到 PHP 页面。

我也打算将它作为 JSON 传递,但想先让 POST 直接工作。

代码如下:

    Person p = new Person();
    p.firstName = "Bill";
    p.lastName = "Gates";
    p.email = "asdf@hotmail.com";
    p.deviceUUID = "abcdefghijklmnopqrstuvwxyz";

    JavaScriptSerializer serializer = new JavaScriptSerializer();
    string s;
    s = serializer.Serialize(p);
    textBox3.Text = s;
    // s = "{\"firstName\":\"Bill\",\"lastName\":\"Gates\",\"email\":\"asdf@hotmail.com\",\"deviceUUID\":\"abcdefghijklmnopqrstuvwxyz\"}"
    HttpWebRequest request = (HttpWebRequest)WebRequest.Create("http://www.davemateer.com/ig/genius/newuser.php");
    //WebRequest request = WebRequest.Create("http://www.davemateer.com/ig/genius/newuser.php");
    request.Method = "POST";
    request.ContentType = "application/x-www-form-urlencoded";
    //byte[] byteArray = Encoding.UTF8.GetBytes(s);
    byte[] byteArray = Encoding.ASCII.GetBytes(s);
    request.ContentLength = byteArray.Length;
    Stream dataStream = request.GetRequestStream();
    dataStream.Write(byteArray, 0, byteArray.Length);
    dataStream.Close ();

    WebResponse response = request.GetResponse();
    textBox4.Text = (((HttpWebResponse)response).StatusDescription);
    dataStream = response.GetResponseStream ();

    StreamReader reader = new StreamReader(dataStream);
    string responseFromServer = reader.ReadToEnd ();
    textBox4.Text += responseFromServer;

    reader.Close ();
    dataStream.Close ();
    response.Close ();

而PHP5.2的代码是:

<?php
echo "hello world";
var_dump($_POST);
?>

这会返回:

array(0) {}

有什么想法吗?我希望它返回我刚刚传递给它的值,以证明我可以从服务器端访问数据。

【问题讨论】:

    标签: c# php winforms


    【解决方案1】:

    我认为您需要正确编码和发送实际的帖子内容。看起来您只是在序列化为 JSON,PHP 不知道如何处理(即,它不会将其设置为 $_POST 值)

    string postData = "firstName=" + HttpUtility.UrlEncode(p.firstName) +
                      "&lastName=" + HttpUtility.UrlEncode(p.lastName) +                    
                      "&email=" + HttpUtility.UrlEncode(p.email) +
                      "&deviceUUID=" + HttpUtility.UrlEncode(p.deviceUUID);
    byte[] byteArray = Encoding.ASCII.GetBytes(postData);
    // etc...
    

    这应该会在 PHP 集中获得您的 $_POST 变量。稍后当您切换到 JSON 时,您可以执行以下操作:

    string postData = "json=" + HttpUtility.UrlEncode(serializer.Serialize(p) );
    

    并从 PHP 中获取:

    $json_array = json_decode($_POST['json']);
    

    【讨论】:

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