How to auto populate line number in vba editor? [closed]

问题:

Is there a macro that can auto populate the line number for a vba editor? I want to line number for debugging reasons as to know where the error occured.

回答1:

You don't want line numbers.

Not for "debugging reasons", not for anything. Line numbers are deprecated for a reason: they're a relic of an ancient time before procedures even existed, and GOTO the only way to get anywhere.

Erl only returns the last encountered line number before an error was raised. This can mean misleading error logs, if you're logging errors:

Sub DoSomething() 10 On Error GoTo 50    Debug.Print 42 / 0 30 Exit Sub  50 Debug.Print "Error raised on line " & Erl 'returns 10 End Sub 

Also, line numbers only have Integer resolution (a VBA module can have up to 65,535 lines, which is twice that resolution), and will silently fail and happily report wrong line numbers if you don't keep that in mind:

Sub DoSomething() 99997 On Error GoTo 99999 99998 Debug.Print 42 / 0 99999       Debug.Print Erl   'Prints 34462 - and which line is that? End Sub 

Any serious VBA application will use structured error handling instead.

Write small, specialized procedures (i.e. write code that follows modern-day best practices), and the line number becomes utterly meaningless.

Line numbers are a pain in the neck to maintain; they clutter up the code and make it overall harder to read (and therefore to debug).

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That said, IIRC MZ-Tools 3 had such a functionality.

Keep in mind, that BASIC looked like this when line numbers were a thing:

10 GOSUB 100 20 GOSUB 1000 99 END 100 REM CLEAR SCREEN 110 PRINT CHR$(147) 120 RETURN 200 REM MODULO 210 LET MOD% = V%-INT(V%/FB%)*FB% 220 RETURN 1000 REM INIT VARIABLES 1010 LET FIZZ$ = "FIZZ" 1011 LET BUZZ$ = "BUZZ" 1020 LET FIZZ% = 3 1021 LET BUZZ% = 5 1030 LET MIN% = 1 1031 LET MAX% = 15 1100 PRINT FIZZ$ + ":" + STR$(FIZZ%) 1101 PRINT BUZZ$ + ":" + STR(BUZZ%) 1102 PRINT FIZZ$ + BUZZ$ + ":" + STR$(FIZZ%*BUZZ%) 1105 PRINT 2000 REM ACTUAL FIZZBUZZ LOOP 2010 FOR X = MIN% TO MAX% 2015 LET RESULT$ = STR$(X) 2020 LET FB% = FIZZ%*BUZZ% 2021 LET V% = X 2024 GOSUB 200 2025 IF MOD%=0 THEN LET RESULT$=FIZZ$+BUZZ$ : GOTO 2050 2030 LET FB% = FIZZ% 2031 GOSUB 200 2035 IF MOD%=0 THEN LET RESULT$=FIZZ$ : GOTO 2050 2040 LET FB% = BUZZ% 2041 GOSUB 200 2045 IF MOD%=0 THEN LET RESULT$=BUZZ$ : GOTO 2050 2050 PRINT RESULT$ 2090 NEXT X 2099 RETURN 

The above is a working Commodore 64 BASIC 2.0 fizzbuzz program. VBA has fabulous backward-compatibility. With only slight modifications, it runs in VBA:

Sub Main() 10     GoSub 100 20     GoSub 1000 99     End 100     Rem CLEAR SCREEN 110     'Debug.Print Chr$(147) 'Chr$(147) was a special character on C64 120     Return 200     Rem MODULO 210     Let Modulo% = V% - Int(V% / FB%) * FB% 220     Return 1000     Rem INIT VARIABLES 1010     Let FIZZ$ = "FIZZ" 1011     Let BUZZ$ = "BUZZ" 1020     Let FZZ% = 3 1021     Let BZZ% = 5 1030     Let Min% = 1 1031     Let Max% = 15 1100     Debug.Print FIZZ$ + ":" + Str$(FZZ%) 1101     Debug.Print BUZZ$ + ":" + Str(BZZ%) 1102     Debug.Print FIZZ$ + BUZZ$ + ":" + Str$(FZZ% * BZZ%) 1105     Debug.Print 2000     Rem ACTUAL FIZZBUZZ LOOP 2010     For X = Min% To Max% 2015     Let RESULT$ = Str$(X) 2020     Let FB% = FZZ% * BZZ% 2021     Let V% = X 2024     GoSub 200 2025     If Modulo% = 0 Then Let RESULT$ = FIZZ$ + BUZZ$: GoTo 2050 2030     Let FB% = FZZ% 2031     GoSub 200 2035     If Modulo% = 0 Then Let RESULT$ = FIZZ$: GoTo 2050 2040     Let FB% = BZZ% 2041     GoSub 200 2045     If Modulo% = 0 Then Let RESULT$ = BUZZ$: GoTo 2050 2050     Debug.Print RESULT$ 2090     Next X 2099     Return End Sub 

Don't write 1980's code, we're 30 years later.

回答2:

The VBA editor has a built in way to see a line number, under the 'Standard' toolbar:

When you select a line of code, the line number will be represented here next to 'Ln'.

回答3:

This is not 100% tested, but using VBA extensibility you could do the following

Sub line_number(strModuleName As String)  Dim vbProj As VBProject Dim vbComp As VBComponent Dim cmCode As CodeModule Dim intLine As Integer  Set vbProj = Application.VBE.ActiveVBProject Set vbComp = vbProj.VBComponents(strModuleName) Set cmCode = vbComp.CodeModule  For intLine = 2 To cmCode.CountOfLines - 1    cmCode.InsertLines intLine, intLine - 1 &  cmCode.Lines(intLine, 1)    cmCode.DeleteLines intLine + 1, 1 Next intLine  End Sub 

This gives the results before and after as below, altering in this way is not recommended though.

文章来源: How to auto populate line number in vba editor? [closed]