Cell D1 is "12 x 15KG" sometimes it might be "5 x 15KB"
this_packing = Val(Split(Worksheets("DA").Range("D1").Value, "X"))
The above return "Type Mismatch error 13"
How can I get the Integer value into 'this_packing' ?
Thanks!
Split returns an array (even if the delimiter isn't found, it'll return an array containing the original value as the first element) so you'll need to specify the index of the part to be converted:
' First element: 12 in "12 x 15KG"
first = Val(Split(Worksheets("DA").Range("D1").Value, "x")(0)) ' -> 12
' Second element: 15 in "12 x "15KG"
second = Val(Split(Worksheets("DA").Range("D1").Value, "x")(1)) ' -> 15
It can be kinda confusing to determine whether to use a 0-base or 1-base indexing, but Split uses 0-base by default.
1) Split returns an array, so Val(Split(..)) is invalid.
2) The Split is case sensitive, so you need "x" to match "X".
Option Explicit
Sub sub1()
Dim i1&, s0$, s1$, this_packing As Variant
this_packing = (Split("12 x 15KG", "x"))
s0 = this_packing(0)
s1 = this_packing(1)
i1 = Val(this_packing(0))
End Sub
Related
I have the following Text sample:
Ins-Si_079_GM_SOC_US_VI SI_SOC_FY1920_US_FY19/20_A2554_Si Resp_2_May
I want to get the number 079, So what I need is the first instance of digits of length 3. There are certain times the 3 digits are at the end, but they usually found with the first 2 underscores. I only want the digits with length three (079) and not 19, 1920, or 2554 which are different lengths.
Sometimes it can look like this with no underscore:
1920 O-B CLI 353 Tar Traf
Or like this with the 3 digit number at the end:
Ins-Si_GM_SOC_US_VI SI_SOC_FY1920_US_FY19/20_A2554_Si Resp_2_079
There are also times where what I need is 2 digits but when it's 2 digits its always at the end like this:
FY1920-Or-OLV-B-45
How would I get what I need in all cases?
You can split the listed items and check for 3 digits via Like:
Function Get3Digits(s As String) As String
Dim tmp, elem
tmp = Split(Replace(Replace(s, "-", " "), "_", " "), " ")
For Each elem In tmp
If elem Like "###" Then Get3Digits = elem: Exit Function
Next
If Get3Digits = vbNullString Then Get3Digits = IIf(Right(s, 2) Like "##", Right(s, 2), "")
End Function
Edited due to comment:
I would execute a 2 digit search when there are no 3 didget numbers before the end part and the last 2 digits are 2. if 3 digits are fount at end then get 3 but if not then get 2. there are times when last is a number but only one number. I would only want to get last if there are 2 or 3 numbers. The - would not be relevant to the 2 digets. if nothing is found that is desired then would return " ".
If VBA is not a must you could try:
=TEXT(INDEX(FILTERXML("<t><s>"&SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(A1,"_"," "),"-"," ")," ","</s><s>")&"</s></t>","//s[.*0=0][string-length()=3 or (position()=last() and string-length()=2)]"),1),"000")
It worked for your sample data.
Edit: Some explaination.
SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(A1,"_"," "),"-"," ")," ","</s><s>") - The key part to transform all three potential delimiters (hyphen, underscore and space) to valid XML node end- and startconstruct.
The above concatenated using ampersand into a valid XML construct (adding a parent node <t>).
FILTERXML can be used to now 'split' the string into an array.
//s[.*0=0][string-length()=3 or last() and string-length()=2] - The 2nd parameter of FILTERXML which should be valid XPATH syntax. It reads:
//s 'Select all <s> nodes with
following conditions:
[.*0=0] 'Check if an <s> node times zero
returns zero (to check if a node
is numeric. '
[string-length()=3 or (position()=last() and string-length()=2)] 'Check if a node is 3 characters
long OR if it's the last node and
only 2 characters long.
INDEX(.....,1) - I mentioned in the comments that usually this is not needed, but since ExcelO365 might spill the returned array, we may as well implemented to prevent spilling errors for those who use the newest Excel version. Now we just retrieving the very first element of whatever array FILTERXML returns.
TEXT(....,"000") - Excel will try delete leading zeros of a numeric value so we use TEXT() to turn it into a string value of three digits.
Now, if no element can be found, this will return an error however a simple IFERROR could fix this.
Try this function, please:
Function ExtractThreeDigitsNumber(x As String) As String
Dim El As Variant, arr As Variant, strFound As String
If InStr(x, "_") > 0 Then
arr = Split(x, "_")
Elseif InStr(x, "-") > 0 Then
arr = Split(x, "-")
Else
arr = Split(x, " ")
End If
For Each El In arr
If IsNumeric(El) And Len(El) = 3 Then strFound = El: Exit For
Next
If strFound = "" Then
If IsNumeric(Right(x, 2)) Then ExtractThreeDigitsNumber = Right(x, 2)
Else
ExtractThreeDigitsNumber = strFound
End If
End Function
It can be called in this way:
Sub testExtractThreDig()
Dim x As String
x = "Ins-Si_079_GM_SOC_US_VI SI_SOC_FY1920_US_FY19/20_A2554_Si Resp_2_May"
Debug.Print ExtractThreeDigitsNumber(x)
End Sub
there. I made this code that replaces a character for two number (e.g. 0 = 10; 1 = 11; 2 = 12; ...) and everything works fine except for the first element (the zero element). So, if I put "010a4" string on cell A1 and use my formula "=GENERATECODE(A1)", my expected return value is "1011102014" but I've got a "110111102014" string. So, only zero value occur this error and I can't figured out why. Any thoughts?
My code:
Function GENERATECODE(Code As String)
Dim A As String
Dim B As String
Dim i As Integer
Const AccChars = "0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"
Const RegChars = "1011121314151617181920212223242526272829303132333435363738394041424344454647484950515253545556575859606162636465666768697071"
For i = 1 To Len(AccChars)
A = Mid(AccChars, i, 1)
B = Mid(RegChars, 2 * i - 1, 2)
Code = Replace(Code, A, B)
Next
GENERATECODE = Code
End Function
Your problem is that your code first change each 0 to 10 then each 1 to 11. So each 0 give you 10 then 110.
If you want to keep the same kind of algorithm (which might not be a good choice), you need to change AccChars and RegChars so that a character is never replaced by a string that can give a character found later on the AccChars String. In your case just replace Const AccChars = "012 ... per Const AccChars = "102 ... and Const RegChars = "101112 ... per Const RegChars = "111012 ...
But it might be better to change your algorithm altogether. I would first suggest to not use in place editing of the string, but rather to use 2 Strings.
In addition to being incorrect, your current code is inefficient since it involves scanning over the code string multiple times instead of just once. Simply scan over the string once, gathering the substitutions into an array which is joined at the end:
Function GENERATECODE(Code As String) As String
Dim codes As Variant
Dim i As Long, n As Long
Dim c As String
n = Len(Code)
ReDim codes(1 To n)
For i = 1 To n
c = Mid(Code, i, 1)
Select Case c
Case "0" To "9":
codes(i) = "1" & c
Case "a" To "z":
codes(i) = Asc(c) - 77
Case "A" To "Z":
codes(i) = Asc(c) - 19
Case Else:
codes(i) = "??"
End Select
Next i
GENERATECODE = Join(codes, "")
End Function
Example:
?generatecode("010a4")
1011102014
The point of the two offsets is that you want "a" to map to 20 and "A" to map to 46. Note Asc("a") - 77 = 97 - 77 and Asc("A") - 19 = 65-19 = 46.
I have the following input:
Dim str_format as string = "XXXXX00000"
Dim str as string = "INV"
Dim int as integer = "56"
How can I replace XXXXX with INV and replace 00000 with 56?
For the example above the result should be INVXX00056.
X can only replace with alphabet and 0 can only replace with integer, if str has more than five alphabet. The extra alphabets will be thrown away because str_format only has five X. The same algorithm is true for the integer.
Example 2
Dim str_format as string = "XXX00000"
Dim str as string = "ABCD"
Dim int as integer = 654321
Expected result: ABC54321
Process:
1. ABCD XXX00000 654321
2. ABC DXX000006 54321
3. AB CDX00065 4321
4. A BCD00654 321
5. ABC06543 21
6. ABC65432 1
7. ABC54321
As Spidey mentioned... show some code. That said the process you describe is a bit long-winded.
The Letter part of the solution can be done by grabbing the first 3 characters of str using Left(str,3) this will bring in the leftmost 3 character (if there are less it will get what is there). Then check that you have 3 characters using str.Length(). If the length is less than 3 then append the appropriate number of 'X'.
The Numeric part can be done in a similar way. Your int is actually a string in your code above. If it was a real integer you can cast it to string. Use Right(int,5). Again check to see you have 5 digits and if not prepend with appropriate number of 0.
Have a go... if you run into problems post your code and someone is bound to help.
UPDATE
As there have been actual answers posted here is my solution
Function FormatMyString(str As String, num as String) As String
Dim result As String
result = Left(str,3).PadRight(3, "X"c).ToUpper() & Right(num,5).PadLeft(5, "0"c)
Return result
End Function
UPDATE 2
based on Wiktors answer... made an amendment to my solution to cope with different formats
Function FormatMyString(str As String, num as String, alpha as Integer, digits as Integer) As String
Dim result As String
result = Left(str, alpha).PadRight(alpha, "X"c).ToUpper() & Right(num, digits).PadLeft(digits, "0"c)
Return result
End Function
To use...
FormatMyString("ABCDE", "56",3 5) will return ABC00056
FormatMyString("ABCDE", "123456",4 3) will return ABCD456
FormatMyString("AB", "123456",4 3) will return ABXX456
Here is a possible solution that just uses basic string methods and PadLeft/PadRight and a specific method to count occurrences of specific chars in the string. It assumes the format string can only contain X and 0 in the known order.
Public Function CountCharacter(ByVal value As String, ByVal ch As Char) As Integer
Return value.Count(Function(c As Char) c = ch)
End Function
Public Sub run1()
Dim str_format As String = "XXXXX00000" '"XXX00000"
Dim str As String = "INV"
Dim int As Integer = 56 ' ABC54321
Dim xCnt As Integer = CountCharacter(str_format, "X")
Dim zCnt As Integer = CountCharacter(str_format, "0")
Dim result As String
If xCnt > str.Length Then
result = str.PadRight(xCnt, "X")
Else
result = str.Substring(0, xCnt)
End If
If zCnt > int.ToString().Length Then
result = result & int.ToString().PadLeft(zCnt, "0")
Else
result = result & int.ToString().Substring(int.ToString().Length-zCnt
End If
Console.WriteLine(result)
End Sub
Output for your both scenarios is as expected.
Take a look at this sample
Dim str_format As String = str_format.Replace("XXX", "ABC")
Msgbox(str_format )
As we assume that the X is 3 only. I dont want to give you more it is a start and everything will be easy.
If that kind of format is fix I mean the number of X will go or down then you can make a conditional statement based on the length of string
I'm trying to write a function in which a column contains one substring and does not contain another substring.
In the example bellow I would like my function to return 1 if my row contains "some project" AND DOES NOT CONTAIN "overhead".
row| example strings | desired return value
0 |some project,other project | 1
1 |some project | 1
2 |overhead | 0
3 |some project, overhead | 0
4 |some project, other, boo | 1
I was trying to formulate it first with exact strings such as:
=IF(AND((E3="some project"),NOT(E3="overhead")),1,0)
But this only gives correct results for row 1 and 2 because it only does exact mach for the string instead of matching on the substring.
What you need is some kind of Substring function. I think FIND might work. Check out this page: https://exceljet.net/excel-functions/excel-find-function
Your function would be like:
=IF(AND(ISERROR(FIND("some project", E3))=FALSE,ISERROR(FIND("overhead",E3))),1,0)
EDIT: Above function works after testing
Tricky part here is that FIND returns the starting position of the string, and if it fails it returns #VALUE, which I believe you can catch with the ISERROR() function. This is in no way a beautiful solution. I would try to utilize the code behind and write this is VBA, as I am certain there is a proper substring function in VBA.
If you can insert a little VBA code, then you can use a custom function like so:
=StrContains(E3, "some project", "overhead")
And this will return True if the value in E3 contains both of those substrings. This function relies mainly on VBA's Instr function, which
Function code:
Public Function StrContains(ByRef cl As Excel.Range, ParamArray strings() As Variant)
'Function returns TRUE if the range contains ALL of the passed substring items
' uses ParamArray to allow varying number of substring items
' Ex:
' =StrContains(A1, "something", "else", "foo")
'
Dim val$
Dim s
Dim i As Integer
Dim ret As Boolean
Dim length As Integer
length = UBound(strings) + 1
val = cl.Value2
For Each s In strings
If InStr(1,val, s) <> 0 Then
i = i + 1
End If
Next
ret = (i = length)
StrContains = ret
End Function
You could modify this relatively easily to be case-insensitive, or to accept partial matches optionally, etc. Here is what it looks like extended for both of those concepts:
=StrContains(E3, False, True, "some project", "overhead")
Function Code:
Public Function StrContains(ByRef cl As Excel.Range, MatchCase As Boolean, MatchAll as Boolean, ParamArray strings() As Variant)
'MatchAll is matching switch, use True to require ALL matching items, or False to allow for fewer.
'MatchCase is the Case-sensitive switch, use False to ignore case.
' uses ParamArray to allow varying number of substring items
' Ex:
' =StrContains(A1, "something", "else", "foo")
'
Dim val$
Dim s
Dim i As Integer
Dim ret As Boolean
Dim length As Integer
length = UBound(strings) + 1
val = cl.Value2
If Not MatchCase Then
val = LCase(val)
For i = lBound(strings) to UBound(strings)
strings(i) = lcase(strings(i))
Next
Next
For Each s In strings
If InStr(val, s) <> 0 Then
i = i + 1
End If
Next
ret = (i = IIF(MatchAll, length, 1))
StrContains = ret
End Function
You can actually use the SEARCH function, like this:
=IF(ISNUMBER(SEARCH("some project",B2)),NOT(ISNUMBER(SEARCH("overhead",B2)))*1,0)
Considerations:
SEARCH function: returns either a number or error (that's why I used ISNUMBER, you could actually have used ISERROR too). This function is case insensitive. For case sensitiveness you might just use FIND, in the previous formula woud be =IF(ISNUMBER(FIND("some project",B2)),NOT(ISNUMBER(FIND("overhead",B2)))*1,0)
ISNUMBER function: returns either FALSE or TRUE. I converted the result to 1 or 0 simply by multiplying TRUE or FALSE by 1.
Hope helps
This is my Excel table with 3 input columns. The last column is the output column with the result I need.
Excel Input and Output data
Here is the sample macro I have:
Function month(x As string, y As string, z As String) As String
If (((x = "January.Winter") Or (y = "January.Winter")) And (z = "jan")) Then
month= "true"
Else If (((x = "January.2016") Or (y = "January.2016")) And (z = "jan")) Then
month= "true"
Else If (((x = "January.today") Or (y = "January.today")) And (z = "jan")) Then
month= "true"
Else
month= False
End If
End Function
My worksheet contains thousands of rows which includes "january" as a substring as a text in the cells. Instead of writing multiple checks like "if "x=January.winter"", I would like to use simplify the macro by checking if the string "x" or string "y" contains the string "January". Is there a way I could change the macro to do that?
Three ways that spring to mind:
If LCase$(x) Like "january*" Or LCase$(y) Like "january*" Then
...
If InStr(LCase$(x), "january") Or InStr(LCase$(x), "january") Then
...
If LCase$(Left$(x, 7)) = "january" Or LCase$(Left$(y, 7)) = "january" Then
...
It really just depends how inventive you want to get.
Note that I've used LCase$() to force text to lower case and prevent any issues with capitilisation - always something worth thinking about when comparing strings.
You can use the InStr function. Example use:
If (InStr(x,"January") > 0 Or InStr(y,"January")>0) And (z = "Jan") Then
...
It returns 0 if your substring isn't found, otherwise it returns the position of your substring.
More info here
Also careful with upper and lower casing. "January" will not match "january". Use LCase and UCase functions to force upper or lower casing, as Macro Man did in his answer.