Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
587 views
in Technique[技术] by (71.8m points)

post - How to check form submission ASP classic

I'm setting up a form in ASP classic and it will reload after submission (action self)

But this time it shows results of previous submissions, so how can I check that a POST submission has been made?

Like in PHP:

if($_POST['submit']) {
  show results...
}
See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

You have several options:

Method 1 - Check the request method:

If Request.ServerVariables("REQUEST_METHOD") = "POST" Then
    'Show Results...
End If

Method 2 - add a hidden field to your form with a value then check if that value has been posted:

If Request.form("HiddenValue") = "1" Then
    'Show Results...
End If

Method 3 - Check if the request.form collection contains items:

If Request.Form.Count > 0 Then
    'Show Results...
End If

Method 4 - Post to a querystring (i.e. set action of <form> to ?post=yes)

If Request.QueryString("post") = "yes" Then
    'Show Results...
End If

Which one to pick?

My preferred option is method 4 – as it’s easily visible in the address bar as to what’s going on – if for some reason I want to avoid presenting this level of detail in the url, I tend to use option 3 as it’s easy to implement, requires no changes on the source forms & is reliable. As for the other two methods:

  • Method 1 – I tend to avoid relying on server variables if I don’t have 100% control over the server – no real justification for that, just a general habit I tend to work with.
  • Method 2 – You could substitute a hidden field for another field that will always contain a value.

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...