Re: help extract this string
This is probably the easiest way of doing it:
Code:
List1.AddItem "<a href=""http://www.xxx.org"">;xx.Org Human Edited Web Directory</a>"
If that does not work for you, then you need to be more specific about your requirements.
BTW, are you sure your code isn't doing anything illegal? :rolleyes:
Re: help extract this string
bonnie i need to extract the string highlighted in blue , as u se all that data is in text1.text and by clicking command1 i want that string extracted from text1 to list1
yes this is the data i need
List1.AddItem "<a href=""http://www.xxx.org"">;xx.Org Human Edited Web Directory</a>"
but this will only just add it not extract it from text1 , i wish to extract this data dear "<a href=""http://www.xxx.org"">;xx.Org Human Edited Web Directory</a>" from the string above to list1
Re: help extract this string
ok here it is working , reference for future use someone may need this piece of function is awsome
Code:
Public Function InStrBetween(ByRef Search As String, ByRef StringBegin As String, ByRef StringEnd As String, Optional ByVal Start As Long = 1, Optional ByVal Compare As VbCompareMethod = vbBinaryCompare) As String
Dim lngLenSea As Long, lngLenBeg As Long, lngLenEnd As Long
Dim lngBegin As Long, lngEnd As Long
Dim strSearch As String, strBegin As String, strEnd As String
' get string lengths
lngLenSea = LenB(Search)
lngLenBeg = LenB(StringBegin)
lngLenEnd = LenB(StringEnd)
' make sure we have lengths and a valid starting position
If (lngLenSea <> 0) And (lngLenBeg <> 0) And (lngLenEnd <> 0) And (Start > 0) Then
' make start a byte position
Start = ((Start - 1) * 2) + 1
' case sensitive?
If Compare = vbBinaryCompare Then
' find the starting position
lngBegin = InStrB(Start, Search, StringBegin, vbBinaryCompare)
' because InStrB finds byte positions, we have to ensure we do not get "mid character" positions
Do While ((lngBegin And 1) = 0) And (lngBegin > 0)
lngBegin = InStrB(lngBegin + lngLenBeg, Search, StringBegin, vbBinaryCompare)
Loop
If lngBegin Then
lngBegin = lngBegin + lngLenBeg
' find the ending position
lngEnd = InStrB(lngBegin, Search, StringEnd, vbBinaryCompare)
' because InStrB finds byte positions, we have to ensure we do not get "mid character" positions
Do While ((lngEnd And 1) = 0) And (lngEnd >= lngBegin)
lngEnd = InStrB(lngEnd + lngLenEnd, Search, StringEnd, vbBinaryCompare)
Loop
' make sure we have something
If lngEnd > lngBegin Then
' return the result
InStrBetween = MidB$(Search, lngBegin, lngEnd - lngBegin)
End If
End If
Else
' make upper case copies
strSearch = UCase$(Search)
strBegin = UCase$(StringBegin)
' find the starting position
lngBegin = InStrB(Start, strSearch, strBegin, vbBinaryCompare)
' because InStrB finds byte positions, we have to ensure we do not get "mid character" positions
Do While ((lngBegin And 1) = 0) And (lngBegin > 0)
lngBegin = InStrB(lngBegin + lngLenBeg, strSearch, strBegin, vbBinaryCompare)
Loop
If lngBegin Then
lngBegin = lngBegin + lngLenBeg
' make upper case copy
strEnd = UCase$(StringEnd)
' find the ending position
lngEnd = InStrB(lngBegin, strSearch, strEnd, vbBinaryCompare)
' because InStrB finds byte positions, we have to ensure we do not get "mid character" positions
Do While ((lngEnd And 1) = 0) And (lngEnd >= lngBegin)
lngEnd = InStrB(lngEnd + lngLenEnd, strSearch, strEnd, vbBinaryCompare)
Loop
' make sure we have something
If lngEnd > lngBegin Then
' return the result
InStrBetween = MidB$(Search, lngBegin, lngEnd - lngBegin)
End If
End If
End If
End If
End Function
Private Sub Command4_Click()
Text2.Text = InStrBetween(Text1.Text, Text3.Text, Text4.Text, , False)
End Sub
text2.text ' results
text1.text = </td></tr><tr><td class="label">Reciprocal Link URL:</td><td class="field"><input type="text" name="RECPR_URL" value="" size="40" maxlength="255" class="text" /><br /><p class="small">To validate the reciprocal link please include the<br />following HTML code in the page at the URL<br />specified above, before submiting this form:</p><textarea name="RECPR_TEXT" rows="2" readonly="readonly" cols="37" class="text"><a href="http://www.xxx.org">;xx.Org Human Edited Web Directory</a>;</textarea></td></tr><tr><td class="label"><span class='req'>*</
text3.text has readonly="readonly" cols="37" class="text">
text4.text has </textarea>
no dear not illigal i am extracting data out of my site making a aio tool for seo :p:) thanks tho bonnie:p
bonus , i like the top function betta
Code:
Private Function GetBetween(ByRef sSearch As String, _
ByRef sStart As String, _
ByRef sStop As String, _
Optional ByVal lSearch As Long = 1, _
Optional ByVal bCaseSensitive As Boolean = True) As String
Dim lonS As Long, lonE As Long
Dim strLCSearch As String
Dim strLCStart As String, strLCStop As String
'Case-sensitive...
If bCaseSensitive Then
lonS = InStr(lSearch, sSearch, sStart)
If lonS Then
lonS = lonS + Len(sStart)
lonE = InStr(lonS, sSearch, sStop)
If lonE Then
GetBetween = Mid$(sSearch, lonS, lonE - lonS)
End If
End If
'Not case-sensitive
'Faster to convert string to lowercase rather than vbTextCompare
Else
strLCSearch = LCase$(sSearch)
strLCStart = LCase$(sStart)
strLCStop = LCase$(sStop)
lonS = InStr(lSearch, strLCSearch, strLCStart)
If lonS Then
lonS = lonS + Len(strLCStart)
lonE = InStr(lonS, strLCSearch, strLCStop)
If lonE > 0 Then
GetBetween = Mid$(sSearch, lonS, lonE - lonS)
End If
End If
End If
End Function
:p:p
Re: help extract this string
Code:
Private Sub Command1_Click()
Dim StartPos As Long, EndPos As Long, sText As String
sText = Text1
StartPos = InStr(1&, sText, "<a href", vbTextCompare)
If StartPos Then
EndPos = InStr(StartPos + 7&, sText, "</a>", vbTextCompare)
If EndPos Then
List1.AddItem Mid$(sText, StartPos, (EndPos + 4&) - StartPos)
End If
End If
End Sub
Re: help extract this string
@ladoo: Weren't you shown how to do this sort of thing here: http://www.vbforums.com/showthread.p...t=#post4431681 ?
Re: help extract this string
simple clean code , very affective very good thanks allot!!!!!!!!!!!!!! bonnie
i was searching for it @doogle didnt find it this is why i posted
Re: help extract this string
Very good, Bonnie. Just curious about one thing. I never added the ampersand in my code after the 1, 7, and 4 that you show in your code example. Why is it there in yours?
Re: help extract this string
@Doc: By default literal numbers in VB are of type Integer, suffixing a literal with a '&' forces it to be of type Long. Bonnie's code is avoiding an implied CLng when the code is executed.
Re: help extract this string
Quote:
Originally Posted by
Doogle
@Doc: By default literal numbers in VB are of type Integer, suffixing a literal with a '&' forces it to be of type Long. Bonnie's code is avoiding an implied CLng when the code is executed.
That's what I figured. In this thread question, I doubt that any number larger than integer would ever be encountered. Years ago we always used integer for speed whenever possible. Forcing it to long seems unnecessary here. Just MHO.
Re: help extract this string
try this :
the text is in a textbox control : Text
The search text gets sent to the listbox : List
The function is executed by pressing a button control : Command
Code:
Private Sub Command_Click()
Dim ThisText As String
Dim SearchText As String
SearchText = Text.Text
ThisText = SearchID(SearchText, Chr$(34) & "text" & Chr$(34) & ">", ";</text", True)
If ThisText <> "" Then Form1.List.AddItem ThisText
End Sub
Public Function SearchID(ScanString As String, StartMarker As String, FinishMarker As String, Optional CaseRequired As Boolean = False) As String
Dim a As Integer
Dim B As Integer
Dim OriginalString As String
SearchID = ""
OriginalString = ScanString
If CaseRequired = False Then
ScanString = LCase$(ScanString)
StartMarker = LCase$(StartMarker)
FinishMarker = LCase$(FinishMarker)
End If
a = InStr(1, ScanString, StartMarker)
If a <> 0 Then
B = InStr(a + 1, ScanString, FinishMarker)
If B <> 0 Then
SearchID = Mid$(OriginalString, a + Len(StartMarker), B - a - Len(StartMarker))
End If
End If
End Function
Re: help extract this string
Quote:
Originally Posted by
Code Doc
I doubt that any number larger than integer would ever be encountered.
I don't know, some Web pages get very large in terms of number of characters, especially those with embedded Java and the like. The home page of these forums is about 8,500 words and at an average of 4 characters (including spaces) per word gives 34,000 characters, enough to explode an Integer.
(Murphy's Law states that the items required will be within the last 50 characters or so on the page if Integers have been used and within the first 50 if Longs have been used)
It's my understanding that using Longs rather than Integers is more efficient.
Re: help extract this string
However, if StartPos and EndPos are both defined as Long as she did in her code, then using an integer in between the two has got to be safe in my book. In this problem, you would never have a string length of over 32K characters between the two.
I do not see how using longs rather than integers could ever result in more efficient code unless the upper bound is exceeded and stops the program dead in its tracks.
Re: help extract this string
Quote:
Originally Posted by
Code Doc
I do not see how using longs rather than integers could ever result in more efficient code unless the upper bound is exceeded and stops the program dead in its tracks.
Hi Doc! :)
I was just following this tip from the manual:
Quote:
Originally Posted by MSDN
Use Long Integer Variables and Integer Math
For arithmetic operations avoid Currency, Single, and Double variables. Use Long integer variables whenever you can, particularly in loops. The Long integer is the 32-bit CPU's native data type, so operations on them are very fast; if you can’t use the Long variable, Integer or Byte data types are the next best choice.
I usually tend to favor speed over memory. That's why I typically use the appropriate type for my variables, constants and numeric literals even if they consume more memory. As mentioned already by Doogle, I prefer to avoid run-time data type conversions whenever possible.
