(ASP) Duplicating a Simple HTML Form Submission (POST)

This example shows how to duplicate a simple HTML form submission. The HTML to be duplicated is shown below, and may be tested in a browser by going to this URL: http://www.chilkatsoft.com/postSample.html


<html>
<body>
<form action="echoPost.asp" method="post">
<p>First name: <input name="firstname"></p>
<p>Last name: <input name="lastname"></p>
<p>Your favorite color:
<select name="color">
<option>Blue</option>
<option>Green</option>
<option>Red</option>
<option>Yellow</option>
<option>Pink</option>
</select>
</p>
<input type="hidden" name="myHiddenField1", value="Hidden Value 1">
<input type="hidden" name="myHiddenField2", value="Hidden Value 2">
<p><input type="submit" value="Send POST"></p>
</form>

</body>
</html>

By default, a browser will send a form POST as an HTTP reqeuest using a Content-Type of "application/x-www-form-urlencoded". This is because the form parameters are sent in the body of the HTTP POST using the URL encoding. Therefore, one should use the PostUrlEncoded method to duplicate the form submission.

Each "input" or "select" tag found in the HTML is a form parameter having a name and value. These are duplicated in code by calling AddParam on the HTTP Request object for each parameter.

The target URL is indicated by the "action" attribute on the "form" tag. In the HTML form shown above, it is "echoPost.asp". The URL specified in the "action" attribute may be an absolute or relative URL. In this case, it is a relative URL, and since there is no directory path, the full URL of the POST target is "http://www.chilkatsoft.com/echoPost.asp". This will be the 1st argument passed to the PostUrlEncoded method.

The raw HTTP POST sent to the web server in this example is shown below. The form parameters are located in the body of the request (i.e. under the header fields), and they are URL encoded.

POST /echoPost.asp HTTP/1.1
Content-Type: application/x-www-form-urlencoded
Host: www.chilkatsoft.com
Content-Length: 87

firstname=John&lastname=Doe&myHiddenField1=Hidden+Value+1&myHiddenField2=Hidden+Value+2

Note: An HTTP file upload is a form submission having an "input" tag w/ "type=file". In this case, a browser will send the POST using a content-type of "multipart/form-data". The PostUrlEncoded method should not be called for HTTP file uploads. The SynchronousRequest would instead be called.

Download Chilkat HTTP ActiveX

<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
</head>
<body>
<%
set http = Server.CreateObject("Chilkat.Http")

'  Any string unlocks the component for the 1st 30-days.
success = http.UnlockComponent("Anything for 30-day trial")
If (success <> 1) Then
    Response.Write "<pre>" & Server.HTMLEncode( http.LastErrorText) & "</pre>"

End If

set request = Server.CreateObject("Chilkat.HttpRequest")

'  Add the form parameters to the HTTP request:
request.AddParam "firstname","John"
request.AddParam "lastname","Doe"
request.AddParam "myHiddenField1","Hidden Value 1"
request.AddParam "myHiddenField2","Hidden Value 2"

'  Send the HTTP POST by calling PostUrlEncoded.
'  The Content-Type used w/ the request will be application/x-www-form-urlencoded

' response is a Chilkat.HttpResponse
Set response = http.PostUrlEncoded("http://www.chilkatsoft.com/echoPost.asp",request)
If (response Is Nothing ) Then
    Response.Write "<pre>" & Server.HTMLEncode( http.LastErrorText) & "</pre>"

End If

'  Display the Body of the HTTP response
'  (This is the HTML that the browser would receive if it
'  had been an interactive form submission from the browser.)
Response.Write "<pre>" & Server.HTMLEncode( response.BodyStr) & "</pre>"


%>
</body>
</html>