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dgameman1 *BANNED* · #1 03-26-2011, 07:54 PM

I am trying to learn how to parse information.
For my first time, I want a messagebox to be able to say
"Example of a simple HTML page"
(http://help.nucleus.com/example_of_a..._html_page.htm)

So this is my code

Code:

Option Explicit

Private Function GetBetween(ByVal Start As Long, Data As String, _
    StartString As String, EndString As String, _
    Optional ByVal CompareMethod As VbCompareMethod = vbBinaryCompare) As String
    
    Dim lonStart As Long, lonEnd As Long
    
    '1. Find start string.
    lonStart = InStr(1, HTML, "<h1>", vbTextCompare)
    
    If lonStart > 0 Then
        '2. Move to end of start string.
        lonStart = lonStart + Len("<h1>")

        '3. Find end string.
        lonEnd = InStr(lonStart, HTML, "</h1>", vbTextCompare)
        
        If lonEnd > 0 Then
            '4. Extract data between start and end strings.
            GetBetween = Mid$(HTML, lonStart, lonEnd - lonStart)
            
            'Done!
        MsgBox GetBetween

        End If
    End If
    
End Function

Now how do I make it so that it know what site to go to and how do I also make it so that when I press "Command1" it will parse that information?

Covey · #2 03-26-2011, 11:30 PM

You will need to add an Inet Control to your form and make sure it's called 'Inet1'.

Code:

Option Explicit

private sub command1_click()

    dim strhtml as string
    strhtml = inet1.openurl("http://help.nucleus.com/example_of_a_simple_html_page.htm")
    msgbox getbetween(1,strhtml,<h1>,</h1>)

end sub

Private Function GetBetween(ByVal Start As Long, Data As String, _
    StartString As String, EndString As String, _
    Optional ByVal CompareMethod As VbCompareMethod = vbBinaryCompare) As String
    
    Dim lonStart As Long, lonEnd As Long
    
    '1. Find start string.
    lonStart = InStr(1, data, startstring, comparemethod)
    
    If lonStart > 0 Then
        '2. Move to end of start string.
        lonStart = lonStart + Len(startstring)

        '3. Find end string.
        lonEnd = InStr(lonStart, data, endstring, comparemethod)
        
        If lonEnd > 0 Then
            '4. Extract data between start and end strings.
            GetBetween = Mid$(data, lonStart, lonEnd - lonStart)
        End If
    End If
    
End Function

dgameman1 *BANNED* · #3 03-27-2011, 06:44 AM

Ok so now this is my whole code

Code:

Option Explicit

Private Sub command1_click()

    Dim strhtml As String
    strhtml = Inet1.OpenURL("http://help.nucleus.com/example_of_a_simple_html_page.htm")
    msgbox getbetween(1,strhtml,<h1>,</h1>)

End Sub
Private Function GetBetween(ByVal Start As Long, Data As String, _
    StartString As String, EndString As String, _
    Optional ByVal CompareMethod As VbCompareMethod = vbBinaryCompare) As String
    
    Dim lonStart As Long, lonEnd As Long
    
    '1. Find start string.
    lonStart = InStr(1, HTML, "<h1>", vbTextCompare)
    
    If lonStart > 0 Then
        '2. Move to end of start string.
        lonStart = lonStart + Len("<h1>")

        '3. Find end string.
        lonEnd = InStr(lonStart, HTML, "</h1>", vbTextCompare)
        
        If lonEnd > 0 Then
            '4. Extract data between start and end strings.
            GetBetween = Mid$(HTML, lonStart, lonEnd - lonStart)
            

        End If
    End If
    
End Function

Syntax error on

Code:

Private Sub command1_click()

    Dim strhtml As String
    strhtml = Inet1.OpenURL("http://help.nucleus.com/example_of_a_simple_html_page.htm")
    msgbox getbetween(1,strhtml,<h1>,</h1>)

End Sub

msgbox getbetween(1,strhtml,<h1>,</h1>) is red

Covey · #4 03-27-2011, 10:58 AM

umm, VB6 might need an equal sign (=) after msgbox? i forget....

Code:

msgbox = getbetween(1,strhtml,<h1>,</h1>)

try that.

dgameman1 *BANNED* · #5 03-27-2011, 08:10 PM

Rawrrr
This is my whole code

Code:

Option Explicit

Private Sub command1_click()

    Dim strhtml As String
    strhtml = Inet1.OpenURL("http://help.nucleus.com/example_of_a_simple_html_page.htm")
    msgbox = getbetween(1,strhtml,<h1>,</h1>)

End Sub
Private Function GetBetween(ByVal Start As Long, Data As String, _
    StartString As String, EndString As String, _
    Optional ByVal CompareMethod As VbCompareMethod = vbBinaryCompare) As String
    
    Dim lonStart As Long, lonEnd As Long
    
    '1. Find start string.
    lonStart = InStr(1, HTML, "<h1>", vbTextCompare)
    
    If lonStart > 0 Then
        '2. Move to end of start string.
        lonStart = lonStart + Len("<h1>")

        '3. Find end string.
        lonEnd = InStr(lonStart, HTML, "</h1>", vbTextCompare)
        
        If lonEnd > 0 Then
            '4. Extract data between start and end strings.
            GetBetween = Mid$(HTML, lonStart, lonEnd - lonStart)
            

        End If
    End If
    
End Function

Syntax error on

Code:

Private Sub command1_click()

    Dim strhtml As String
    strhtml = Inet1.OpenURL("http://help.nucleus.com/example_of_a_simple_html_page.htm")
    msgbox = getbetween(1,strhtml,<h1>,</h1>)

End Sub

msgbox = getbetween(1,strhtml,<h1>,</h1>) is red

Covey · #6 03-27-2011, 09:04 PM

uhh whoops do this:

Code:

msgbox = getbetween(1,strhtml,"<h1>","</h1>")

dgameman1 *BANNED* · #7 03-27-2011, 09:22 PM

lolwtf

"Compile Error:
Function call on left-hand side of assignment must return Variant or Object"

My whole code

Code:

Option Explicit

Private Sub command1_click()

    Dim strhtml As String
    strhtml = Inet1.OpenURL("http://help.nucleus.com/example_of_a_simple_html_page.htm")
    MsgBox = GetBetween(1, strhtml, "<h1>", "</h1>")

End Sub
Private Function GetBetween(ByVal Start As Long, Data As String, _
    StartString As String, EndString As String, _
    Optional ByVal CompareMethod As VbCompareMethod = vbBinaryCompare) As String
    
    Dim lonStart As Long, lonEnd As Long
    
    '1. Find start string.
    lonStart = InStr(1, HTML, "<h1>", vbTextCompare)
    
    If lonStart > 0 Then
        '2. Move to end of start string.
        lonStart = lonStart + Len("<h1>")

        '3. Find end string.
        lonEnd = InStr(lonStart, HTML, "</h1>", vbTextCompare)
        
        If lonEnd > 0 Then
            '4. Extract data between start and end strings.
            GetBetween = Mid$(HTML, lonStart, lonEnd - lonStart)
            

        End If
    End If
    
End Function

Code:

Private Sub command1_click()

    Dim strhtml As String
    strhtml = Inet1.OpenURL("http://help.nucleus.com/example_of_a_simple_html_page.htm")
    MsgBox = GetBetween(1, strhtml, "<h1>", "</h1>")

End Sub

Private Sub command1_click() is yellow

Covey · #8 03-28-2011, 02:23 AM

Remove the equal sign after the msgbox.... Dude you need to at least try instead of posting your errors and expecting others to fix them

dgameman1 *BANNED* · #9 03-28-2011, 02:28 AM

Quote:

Originally Posted by Covey

Remove the equal sign after the msgbox.... Dude you need to at least try instead of posting your errors and expecting others to fix them

But I don't exactly know what the error means.. so I don't know what to fix =O

This is variable unidentified

Code:

Private Function GetBetween(ByVal Start As Long, Data As String, _
    StartString As String, EndString As String, _
    Optional ByVal CompareMethod As VbCompareMethod = vbBinaryCompare) As String

is highlighted =O

Covey · #10 03-28-2011, 12:02 PM

Quote:

Originally Posted by dgameman1

But I don't exactly know what the error means.. so I don't know what to fix =O

This is variable unidentified

Code:

Private Function GetBetween(ByVal Start As Long, Data As String, _
    StartString As String, EndString As String, _
    Optional ByVal CompareMethod As VbCompareMethod = vbBinaryCompare) As String

is highlighted =O

I can see the problem, i suggest you try looking for it.
I'll give you a hit, change one of your variable names to something the system doesn't already use.

dgameman1 *BANNED* · #11 03-29-2011, 12:06 AM

So you're telling me that somewhere in

Code:

Private Function GetBetween(ByVal Start As Long, Data As String, _
    StartString As String, EndString As String, _
    Optional ByVal CompareMethod As VbCompareMethod = vbBinaryCompare) As String

is a variable that is not defined?

Covey · #12 03-29-2011, 02:13 AM

Incorrect

dgameman1 *BANNED* · #13 03-29-2011, 02:45 AM

Then why is that code highlighted?

Covey · #14 03-29-2011, 10:31 AM

Quote:

Originally Posted by dgameman1

Then why is that code highlighted?

Because you are using a variable in that "GetBetween" function that isn't defined.....

You need to change one of the following variables:
Start
Data
StartString
EndString

to a variable you are using in your code, that you HAVEN'T declared....
If you can't work that out for yourself, i'm sorry but i'm not going to bother helping you...As i've already done 100% of what you were asking,..;

Hiatus · #15 03-30-2011, 12:23 AM

He pretty much just fixed the problem for you. Another thing, why are you using vb6? Its so old.

/facepalm

_trevaSaurus · #16 06-14-2011, 06:54 AM

This is such A simple task.

Parsing one item from a site:

Code:

dim spSplit as string
spSplit = split(split(returnedData,"Tags Before")(Your delimeter example, 1"),"Tags After")(0)

Parsing more then one item:

Code:

dim sp() as string, i as long
sp() = split(returnedData,"Before Tags")
for i = 0 to ubound(sp())
sp(i) = split(sp(i),"After Tags")(0)

msgbox sp(i)

next i

Use:
Put it in your command button, make sure you name your objects.

Should work, I didn't open vb6 to test so...