我已经在vba中编写了一些代码,以在某些网站中针对某些名称查找某些身份。如果一切都按正确的顺序执行,则代码运行良好,这意味着如果链接有效,名称与a
标记匹配,最后正则表达式可以找到身份。如果这三个中的任何一个或所有三个都是错误的搜索,则脚本将引发错误。我已经在下面的脚本中指定了发生错误的位置。
我希望您的专家为我提供有关如何处理错误的任何解决方案,并让我的脚本继续运行,直到所有链接都用尽。
由于我对VBA不太了解,因此我尝试使用On error resume next
跳过错误。但是,当我查看结果时,结果显然是一团糟。我正在粘贴一个粗略的示例,当我使用On error resume next
时会得到什么。
Sub Identity_Finder()
Dim http As New XMLHTTP60, html As New HTMLDocument
Dim post As Object, link As Variant, refined_links As String
Dim rxp As New RegExp, identity As Object
For Each link In [{"http://spltech.in/","http://www.unifrostindia.com/","http://www.unitfrostindia.com/","http://www.greenplanet.in/"}]
With http
.Open "GET", link, False
.send '''throws here the first error if the link is invalid
html.body.innerHTML = .responseText
End With
For Each post In html.getElementsByTagName("a")
If InStr(post.innerText, "certain_name") > 0 Then refined_links = post.href: Exit For
Next post
With http
.Open "GET", refined_links, False
.send ''throws another error here if no such link is found
End With
With rxp
.Pattern = "some_regex"
.Global = True
Set identity = .Execute(http.responseText)
End With
r = r + 1: Cells(r, 1) = link
Cells(r, 2) = identity(0) ''''throws another error here if no such identity is noticed
Next link
End Sub
使用
On error resume next
后我得到了什么:John executive
Mac lawyer
lulu lawyer
Robin lawyer
Cathy student
预期产量:
John executive
Mac lawyer
lulu
Robin
Cathy student
当我使用
On error resume next
时,空字段(当未找到它们时)将被以前的值填充。我如何解决这个误导性的结果?提前致谢。
最佳答案
在VBA中最有效的error trap
方法是
1)实际上是在通过定制功能或内置编码概念或二者结合来运行输入/结果之前测试输入/结果。
2)如果绝对需要,请使用VBA内置的错误处理
示例1
例如。您可以使用自定义函数包装此语句,以测试URL是否有效。
With http
.Open "GET", link, False
.send '''throws here the first error if the link is invalid
html.body.innerHTML = .responseText
End With
If ValidURL Then
With http
.Open "GET", link, False
.send
html.body.innerHTML = .responseText
End With
End If
其中ValidURL是定义为:
Function ValidURL(URL as String) as Boolean
Dim result as Boolean
'I don't know how you would specify a valid link in your specific case
'but that code goes here
'a dummy example follows
result = Left(URL,7) = "http://"
ValidURL = result 'True or False
End Function
示例2
我在此声明中假设:
With http
.Open "GET", refined_links, False
.send ''throws another error here if no such link is found
End With
当找不到这样的链接时,会生成一个特定的错误号(代码)。发现该号码并使用此代码绕过。
With http
.Open "GET", refined_links, False
On Error Resume Next
.Send
On Error GoTo 0
End With
If err.Number <> 9999 'replace with correct number
'continue with regex test
End If
将所有内容放到一起
最后,将所有内容放在一起,您可以像这样来构建,只需最少使用
On Error Resume Next
且不使用GoTo
语句。For Each link In [{"http://spltech.in/","http://www.unifrostindia.com/","http://www.unitfrostindia.com/","http://www.greenplanet.in/"}]
If ValidURL(link) Then
With http
.Open "GET", link, False
.send
html.body.innerHTML = .responseText
End With
For Each post In html.getElementsByTagName("a")
If InStr(post.innerText, "certain_name") > 0 Then refined_links = post.href: Exit For
Next post
With http
.Open "GET", refined_links, False
On Error Resume Next
.Send
On Error GoTo 0
End With
If err.Number <> 9999 'replace with correct number
With rxp
.Pattern = "some_regex"
.Global = True
Set identity = .Execute(http.responseText)
End With
'i will leave it to you on how to account for no pattern match
r = r + 1: Cells(r, 1) = link
Cells(r, 2) = identity(0) ''''throws another error here if no such identity is noticed
End If
End If
Next link
关于vba - 如何有效地处理错误以防止产生误导性的结果?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/47601429/