如何检查表单提交的ASP经典 [英] How to check form submission ASP classic

问题描述

我正在用ASP Classic建立一个表单，提交后会重新加载(动作自我)

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

但是这次显示的是以前提交的结果，那么如何检查已提交的POST?

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

类似于PHP:

if($_POST['submit']) {
  show results...
}

推荐答案

您有几种选择:

方法1 -检查请求方法:

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

方法2 -向表单中添加一个带有值的隐藏字段，然后检查该值是否已发布:

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

方法3 -检查request.form集合中是否包含项目:

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

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

方法4 -发布到查询字符串(即，将<form>的action设置为?post=yes)

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

选择哪个?

我的首选选项是方法4 –因为它很容易在地址栏中显示发生的情况–如果出于某种原因我想避免在url中显示此详细级别，我倾向于使用选项3，因为它很容易实施，无需更改源表单&是可靠的. 至于其他两种方法:

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:

• 方法1 –如果不这样做，我倾向于避免依赖服务器变量 对服务器有100％的控制权–没有任何正当理由， 只是我倾向于与之共处的一个习惯.
• 方法2 –您可以将隐藏字段替换为始终包含以下内容的其他字段 一个值.

• 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.

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