我有一个c#desktop application尝试以编程方式将行插入到基于Web的MySQL数据库中。我需要帮助来解决这个问题。
在网络端,我使用的网页包含要插入的列的表单。表单的提交转到PHP文件,该文件从发布的数据中插入行。页面/表单如下所示,当我从网站手动提交时,插入工作正常。
<html>
<head>
<meta content="text/html; charset=utf-8" http-equiv="Content-Type" />
<title>test</title>
</head>
<body>
<form action="insertrow.php" method="POST">
<input type="text" name="Id"><br>
<input type="text" name="Name"><br>
<input type="submit" name="submit">
</form>
</body>
</html>
在桌面上,我使用WebRequest将数据发布到表单。这是我从MSDN复制的c#代码:
private static void UploadRow(int myId, string myName)
{
// Create a request using a URL that can receive a post.
WebRequest request = WebRequest.Create("http://www.example.com/myform.html");
// Set the Method property of the request to POST.
request.Method = "POST";
// Create POST data and convert it to a byte array.
string postData = "Id=" + myId
+ "&Name=" + myName;
byte[] byteArray = Encoding.UTF8.GetBytes(postData);
// Set the ContentType property of the WebRequest.
request.ContentType = "application/x-www-form-urlencoded";
// Set the ContentLength property of the WebRequest.
request.ContentLength = byteArray.Length;
// Get the request stream.
Stream dataStream = request.GetRequestStream();
// Write the data to the request stream.
dataStream.Write(byteArray, 0, byteArray.Length);
// Close the Stream object.
dataStream.Close();
// Get the response.
WebResponse response = request.GetResponse();
// Display the status.
Console.WriteLine("Post Status=" + ((HttpWebResponse)response).StatusDescription);
// Get the stream containing content returned by the server.
dataStream = response.GetResponseStream();
// Open the stream using a StreamReader for easy access.
StreamReader reader = new StreamReader(dataStream);
// Read the content.
string responseFromServer = reader.ReadToEnd();
// Display the content.
Console.WriteLine(responseFromServer);
// Clean up the streams.
reader.Close();
dataStream.Close();
response.Close();
}
当我运行时,帖子状态是&#34; OK&#34;并且responseFromServer似乎是页面的HTML源。但是,没有任何内容插入数据库。如果我错过了如何使用WebRequest的问题,那我正在思考。我无法在线找到这个问题的任何答案。我很感激任何帮助。感谢。