I know this is how it's supposed to be done however, in VB.NET it gives me an overflow exception crashing my program. If there is any "VB.NET equivalent" of doing this, help is much appreciated.
Dim hActiveWeapon = _mem.rdInt(GetLocalBase() + &H2EE8)
Dim wepEntity = _mem.rdInt(_client + Offsets.EntityList + ((hActiveWeapon & &HFFF) - 1) * &H10) ' Crashes here # hActiveWeapon & &HFFF
Dim wepIndex = _mem.rdInt(wepEntity + Offsets.iItemDefinitionIndex)
I am doing this to get the lower 12 bytes of m_hActiveWeapon (0xFFF = 0b111111111111 = 4095), so I can get the index of the active weapon.
Meh the answer was simple, changing "&" to "And" fixed it.
Related
I just installed FileHelper version 3.15 and I'm having issues. I used some code from their examples.
Dim lDetectedFormat As FileHelpers.Detection.RecordFormatInfo = formats(0)
Dim lDetectedClass As Type = lDetectedFormat.ClassBuilderAsDelimited.CreateRecordClass()
Dim lFieldInfoList As New List(Of FieldInfo)(lDetectedFormat.ClassBuilderAsDelimited.FieldCount)
For Each lField As FileHelpers.Dynamic.DelimitedFieldBuilder In lDetectedFormat.ClassBuilderAsDelimited.Fields
lFieldInfoList.Add(lDetectedClass.GetField(lField.FieldName))
Next
Dim lFileEngine As New FileHelperAsyncEngine(lDetectedClass)
Dim lRecNo As Integer = 0
lFileEngine.BeginReadFile(cReadingsFile)
Try
While True
Dim lRec As Object = lFileEngine.ReadNext()
If lRec Is Nothing Then
Exit While
End If
Trace.WriteLine("Record " + lRecNo)
lFieldInfoList.ForEach(Function(f) Trace.WriteLine(" " + f.Name + " = " + f.GetValue(lRec)))
lRecNo += 1
End While
Finally
lFileEngine.Close()
End Try
I'm getting errors on the following: formats(0), FieldInfo and FileHelperAsyncEngine.
I'm pretty dumb about the FileHelper engine and I need a little help to get started. What I'm trying to do is read a CVS file that has multiple delimiters, quotes and comma.
One more thing: I used Nuget package manager to get FileHelper. Also, if there's a better way to do it in regular code, I'm open to that. Thanks for looking.
Okay, so I'm working in VB.NET, manually writing error logs to log files (yes, I know, I didn't make the call). Now, if the files are over an arbitrary size, when the function goes to write out the new error data, it should start a new file with a new file name.
Here's the function:
Dim listener As New Logging.FileLogTraceListener
listener.CustomLocation = System.Configuration.ConfigurationManager.AppSettings("LogDir")
Dim loc As String = DateTime.UtcNow.Year.ToString + DateTime.UtcNow.Month.ToString + DateTime.UtcNow.Day.ToString + DateTime.UtcNow.Hour.ToString + DateTime.UtcNow.Minute.ToString
listener.BaseFileName = loc
Dim logFolder As String
Dim source As String
logFolder = ConfigurationManager.AppSettings("LogDir")
If ex.Data.Item("Source") Is Nothing Then
source = ex.Source
Else
source = ex.Data.Item("Source").ToString
End If
Dim errorFileInfo As New FileInfo(listener.FullLogFileName)
Dim errorLengthInBytes As Long = errorFileInfo.Length
If (errorLengthInBytes > CType(System.Configuration.ConfigurationManager.AppSettings("maxFileSizeInBytes"), Long)) Then
listener.BaseFileName = listener.BaseFileName + "1"
End If
Dim msg As New System.Text.StringBuilder
If String.IsNullOrEmpty(logFolder) Then logFolder = ConfigurationManager.AppSettings("LogDir")
msg.Append(vbCrLf & "Exception" & vbCrLf)
msg.Append(vbTab & String.Concat("App: AppMonitor | Time: ", Date.Now.ToString) & vbCrLf)
msg.Append(vbTab & String.Concat("Source: ", source, " | Message: ", ex.Message) & vbCrLf)
msg.Append(vbTab & "Stack: " & ex.StackTrace & vbCrLf)
listener.Write(msg.ToString())
listener.Flush()
listener.Close()
I have this executing in a loop for testing purposes, so I can see what happens when it gets (say) 10000 errors in all at once. Again, I know there are better ways to handle this systemically, but this was the code I was told to implement.
How can I reliably get the size of the log file before writing to it, as I try to do above?
Well, as with many things, the answer to this turned out to be "did you read your own code closely" with a side order of "eat something, you need to fix your blood sugar."
On review, I saw that I was always checking BaseFileName and, if it was over the arbitrary limit, appending a character and writing to that file. What I didn't do was check to see if that file or, indeed, other more recent files existed. I've solved the issue be grabbing a directory list of all the files matching the "BaseFileName*" argument in Directory.GetFiles and selecting the most recently accessed one. That ensures that the logger will always select the more current file to write to or -if necessary- use as the base-name for another appended character.
Here's that code:
Dim directoryFiles() As String = Directory.GetFiles(listener.Location.ToString(), listener.BaseFileName + "*")
Dim targetFile As String = directoryFiles(0)
For j As Integer = 1 To directoryFiles.Count - 1 Step 1
Dim targetFileInfo As New FileInfo(targetFile)
Dim compareInfo As New FileInfo(directoryFiles(j))
If (targetFileInfo.LastAccessTimeUtc < compareInfo.LastAccessTimeUtc) Then
targetFile = directoryFiles(j)
End If
Next
Dim errorFileInfo As New FileInfo(listener.Location.ToString() + targetFile)
Dim errorLengthInBytes As Long = errorFileInfo.Length
I have a macro that changes single quotes in front of a number to an apostrophe (or close single curly quote). Typically when you type something like "the '80s" in word, the apostrophe in front of the "8" faces the wrong way. The macro below works, but it is incredibly slow (like 10 seconds per page). In a regular language (even an interpreted one), this would be a fast procedure. Any insights why it takes so long in VBA on Word 2007? Or if someone has some find+replace skills that can do this without iterating, please let me know.
Sub FixNumericalReverseQuotes()
Dim char As Range
Debug.Print "starting " + CStr(Now)
With Selection
total = .Characters.Count
' Will be looking ahead one character, so we need at least 2 in the selection
If total < 2 Then
Return
End If
For x = 1 To total - 1
a_code = Asc(.Characters(x))
b_code = Asc(.Characters(x + 1))
' We want to convert a single quote in front of a number to an apostrophe
' Trying to use all numerical comparisons to speed this up
If (a_code = 145 Or a_code = 39) And b_code >= 48 And b_code <= 57 Then
.Characters(x) = Chr(146)
End If
Next x
End With
Debug.Print "ending " + CStr(Now)
End Sub
Beside two specified (Why...? and How to do without...?) there is an implied question – how to do proper iteration through Word object collection.
Answer is – to use obj.Next property rather than access by index.
That is, instead of:
For i = 1 to ActiveDocument.Characters.Count
'Do something with ActiveDocument.Characters(i), e.g.:
Debug.Pring ActiveDocument.Characters(i).Text
Next
one should use:
Dim ch as Range: Set ch = ActiveDocument.Characters(1)
Do
'Do something with ch, e.g.:
Debug.Print ch.Text
Set ch = ch.Next 'Note iterating
Loop Until ch is Nothing
Timing: 00:03:30 vs. 00:00:06, more than 3 minutes vs. 6 seconds.
Found on Google, link lost, sorry. Confirmed by personal exploration.
Modified version of #Comintern's "Array method":
Sub FixNumericalReverseQuotes()
Dim chars() As Byte
chars = StrConv(Selection.Text, vbFromUnicode)
Dim pos As Long
For pos = 0 To UBound(chars) - 1
If (chars(pos) = 145 Or chars(pos) = 39) _
And (chars(pos + 1) >= 48 And chars(pos + 1) <= 57) Then
' Make the change directly in the selection so track changes is sensible.
' I have to use 213 instead of 146 for reasons I don't understand--
' probably has to do with encoding on Mac, but anyway, this shows the change.
Selection.Characters(pos + 1) = Chr(213)
End If
Next pos
End Sub
Maybe this?
Sub FixNumQuotes()
Dim MyArr As Variant, MyString As String, X As Long, Z As Long
Debug.Print "starting " + CStr(Now)
For Z = 145 To 146
MyArr = Split(Selection.Text, Chr(Z))
For X = LBound(MyArr) To UBound(MyArr)
If IsNumeric(Left(MyArr(X), 1)) Then MyArr(X) = "'" & MyArr(X)
Next
MyString = Join(MyArr, Chr(Z))
Selection.Text = MyString
Next
Selection.Text = Replace(Replace(Selection.Text, Chr(146) & "'", "'"), Chr(145) & "'", "'")
Debug.Print "ending " + CStr(Now)
End Sub
I am not 100% sure on your criteria, I have made both an open and close single quote a ' but you can change that quite easily if you want.
It splits the string to an array on chr(145), checks the first char of each element for a numeric and prefixes it with a single quote if found.
Then it joins the array back to a string on chr(145) then repeats the whole things for chr(146). Finally it looks through the string for an occurence of a single quote AND either of those curled quotes next to each other (because that has to be something we just created) and replaces them with just the single quote we want. This leaves any occurence not next to a number intact.
This final replacement part is the bit you would change if you want something other than ' as the character.
I have been struggling with this for days now. My attempted solution was to use a regular expression on document.text. Then, using the matches in a document.range(start,end), replace the text. This preserves formatting.
The problem is that the start and end in the range do not match the index into text. I think I have found the discrepancy - hidden in the range are field codes (in my case they were hyperlinks). In addition, document.text has a bunch of BEL codes that are easy to strip out. If you loop through a range using the character method, append the characters to a string and print it you will see the field codes that don't show up if you use the .text method.
Amazingly you can get the field codes in document.text if you turn on "show field codes" in one of a number of ways. Unfortunately, that version is not exactly the same as what the range/characters shows - the document.text has just the field code, the range/characters has the field code and the field value. Therefore you can never get the character indices to match.
I have a working version where instead of using range(start,end), I do something like:
Set matchRange = doc.Range.Characters(myMatches(j).FirstIndex + 1)
matchRange.Collapse (wdCollapseStart)
Call matchRange.MoveEnd(WdUnits.wdCharacter, myMatches(j).Length)
matchRange.text = Replacement
As I say, this works but the first statement is dreadfully slow - it appears that Word is iterating through all of the characters to get to the correct point. In doing so, it doesn't seem to count the field codes, so we get to the correct point.
Bottom line, I have not been able to come up with a good way to match the indexing of the document.text string to an equivalent range(start,end) that is not a performance disaster.
Ideas welcome, and thanks.
This is a problem begging for regular expressions. Resolving the .Characters calls that many times is probably what is killing you in performance.
I'd do something like this:
Public Sub FixNumericalReverseQuotesFast()
Dim expression As RegExp
Set expression = New RegExp
Dim buffer As String
buffer = Selection.Range.Text
expression.Global = True
expression.MultiLine = True
expression.Pattern = "[" & Chr$(145) & Chr$(39) & "]\d"
Dim matches As MatchCollection
Set matches = expression.Execute(buffer)
Dim found As Match
For Each found In matches
buffer = Replace(buffer, found, Chr$(146) & Right$(found, 1))
Next
Selection.Range.Text = buffer
End Sub
NOTE: Requires a reference to Microsoft VBScript Regular Expressions 5.5 (or late binding).
EDIT:
The solution without using the Regular Expressions library is still avoiding working with Ranges. This can easily be converted to working with a byte array instead:
Sub FixNumericalReverseQuotes()
Dim chars() As Byte
chars = StrConv(Selection.Text, vbFromUnicode)
Dim pos As Long
For pos = 0 To UBound(chars) - 1
If (chars(pos) = 145 Or chars(pos) = 39) _
And (chars(pos + 1) >= 48 And chars(pos + 1) <= 57) Then
chars(pos) = 146
End If
Next pos
Selection.Text = StrConv(chars, vbUnicode)
End Sub
Benchmarks (100 iterations, 3 pages of text with 100 "hits" per page):
Regex method: 1.4375 seconds
Array method: 2.765625 seconds
OP method: (Ended task after 23 minutes)
About half as fast as the Regex, but still roughly 10ms per page.
EDIT 2: Apparently the methods above are not format safe, so method 3:
Sub FixNumericalReverseQuotesVThree()
Dim full_text As Range
Dim cached As Long
Set full_text = ActiveDocument.Range
full_text.Find.ClearFormatting
full_text.Find.MatchWildcards = True
cached = full_text.End
Do While full_text.Find.Execute("[" & Chr$(145) & Chr$(39) & "][0-9]")
full_text.End = full_text.Start + 2
full_text.Characters(1) = Chr$(96)
full_text.Start = full_text.Start + 1
full_text.End = cached
Loop
End Sub
Again, slower than both the above methods, but still runs reasonably fast (on the order of ms).
I'm learning for loop and I cannot get this problem fixed.
The problems are in the following codes.
dim rt as integer = 2
dim i As Integer = 0
dim currentpg as string = "http://homepg.com/"
For i = 0 To rt
currentpg = currentpg & "?pg=" & i
messagebox.show(currentpg)
next
'I hoped to get the following results
http://homepg.com/?pg=0
http://homepg.com/?pg=1
http://homepg.com/?pg=2
'but instead I'm getting this
http://homepg.com/?pg=0
http://homepg.com/?pg=0?pg=0
http://homepg.com/?pg=0?pg=0?pg=0
Please help me
Thank you.
You probably need something like this:
Dim basepg as string = "http://homepg.com/"
For i = 0 To rt
Dim currentpg As String = basepg & "?pg=" & i
messagebox.show(currentpg)
Next
Although a proper approach would be to accumulate results into a List(Of String), and then display in a messagebox once (or a textbox/file, if too many results). You don't want to bug user for every URL (what if there are 100 of them?). They would get tired of clicking OK.
First of all, you went wrong while copying the output of the buggy code. Here is the real one.
http://homepg.com/?pg=0
http://homepg.com/?pg=0?pg=1
http://homepg.com/?pg=0?pg=1?pg=2
It does not work because currentpg should be a constant but it is changed on each iteration.
Do not set, just get.
MessageBox.Show(currentpg & "?pg=" & i)
Or you can use another variable to make it more readable.
Dim newpg As String = currentpg & "?pg=" & i
MessageBox.Show(newpg)
Also, your code is inefficient. I suggest you to change it like this.
Dim iterations As Integer = 2
Dim prefix As String = "http://homepg.com/?pg="
For index As Integer = 0 To iterations
MessageBox.Show(prefix & index)
Next
At the moment, I have three different times that the following code is ran, back-to-back just with the variables changed:
txtCourseName.LoadFile(strRootLocation + "\subject\" + strSubject + "\" + "\class\" + cmbCourses.SelectedItem, RichTextBoxStreamType.PlainText)
aData = txtCourseName.Text
i = aData.IndexOf("<h3 class=""panel-title"">") + "<h3 class=""panel-title"">".Length
j = aData.IndexOf("</h3>") - i
txtCourseName.Text = aData.Substring(i, j)
For every time it is ran, the rich-text box that is being used is changed, aData is changed to bData, cData, etc., and the data that i and j are indexing is changed. It will run properly for the first two iterations, returning what it is supposed to into the text box, however on the third one, it gives me a System.ArgumentOutOfRangeException with the additional information of Length cannot be less than zero.
My only assumption for what could be causing this is that the third iteration, which I included below, is only supposed to find a 7-letter long string of characters and this is causing some math issues.
I have no idea how to fix this.
txtCourseNumber.LoadFile(strRootLocation + "\subject\" + strSubject + "\" + "\class\" + cmbCourses.SelectedItem, RichTextBoxStreamType.PlainText)
cData = txtCourseNumber.Text
i = cData.IndexOf("Course Number: </b>") + "Course Number: </b>".Length
j = cData.IndexOf("</li>") - i
txtCourseNumber.Text = cData.Substring(i, j)
Example Data That Is Returned By Each Iteration
aData - "English 4"
bData - "insert some really long course description here"
cData - "10045C"