我的html-php网页包含表单,输入和sumbit按钮。使用html请求我试图填写一些字段并按下按钮,但我不能。 这是C#代码:
public static string PostData(string data)
{
HttpWebRequest request = (HttpWebRequest)WebRequest.Create("http://localhost/request.php"); //http://businesslist.com/search/clients/?m=userspace&d=addclassified
request.Method = "POST";
request.AllowAutoRedirect = true;
request.ContentType = "application/x-www-form-urlencoded";
byte[] EncodedPostParams = Encoding.UTF8.GetBytes(data);
request.ContentLength = EncodedPostParams.Length;
request.GetRequestStream().Write(EncodedPostParams, 0, EncodedPostParams.Length);
request.GetRequestStream().Close();
HttpWebResponse response = (HttpWebResponse)request.GetResponse();
string str = new StreamReader(response.GetResponseStream(), Encoding.UTF8).ReadToEnd();
return str;
}
static void Main(string[] args)
{
string data = PostData("name=" + HttpUtility.UrlEncode("lol") + "&btn=Clicked");
Console.WriteLine(data);
Console.ReadLine();
}
和2个php文件: request.php
<html>
<head>
<title>HTTP Request</title>
</head>
<body>
<form action ="http://localhost/response.php" method ="POST">
<input type="text" name="name">
<input type="password" name="pass">
<select name="country">
<option value="-1" selected="selected">Select State/Country</option>
<option value="82">Select 1</option>
<option value="83">Select 2</option>
</select>
<input type="submit" name="btn">
</form>
</body>
</html>
response.php
<?php
$data = $_POST["name"];
echo $data;
?>
Here是指向我网站的链接
那么,我怎么能按下这个按钮呢?
答案 0 :(得分:4)
我在LINQPad尝试了这个并且它有效:
void Main()
{
var request = (HttpWebRequest)WebRequest.Create("http://128.75.49.209/response.php");
request.Method = WebRequestMethods.Http.Post;
request.ContentType = "application/x-www-form-urlencoded";
using (var stream = request.GetRequestStream())
{
var buffer = Encoding.UTF8.GetBytes("name=asd&pass=asd&country=82&btn=Submit+Query");
stream.Write(buffer,0,buffer.Length);
}
var response = (HttpWebResponse)request.GetResponse();
string result = String.Empty;
using (var reader = new StreamReader( response.GetResponseStream()))
{
result = reader.ReadToEnd();
}
Console.WriteLine(result);
}
问题是您在代码中使用了request.php
而不是response.php
。