So Im working on a Spreedsheet in Google Docs, my question is: I would like to combine say columns A,B,C,ect... in column I with a space between them. I'm currently using something like this =A1&" "&B1&" "&C1&" "&ect... This works fine and dandy but if the cell is blank I would like to ignore it. Should this be done via script or formula?
So in my head I'm thinking if cell A = value then grab it and combine it with B (if that contains a value if not leave blank or skip) But I'm not good at PHP So any help would be great!!! Happy NY to everyone ; )
Here's a custom function that will return a string with the given range values joined by the given separator. Any blanks are skipped.
function joinVals( rangeValues, separator ) {
function notBlank(element) {return element != '';}
return rangeValues[0].filter(notBlank).join(separator);
}
Examples:
A B C Formula Result
1 1 2 3 =joinVals(A1:C1," x ") 1 x 2 x 3
2 1 2 =joinVals(A1:C1," x ") 1 x 2
3 1 3 =joinVals(A1:C1," x ") 1 x 3
4 1 2 3 =joinVals(A1:C1) 1,2,3
By using IF and ISBLANK you can determine whether a cell should be included, or ignored. Like this:
=if(ISBLANK(A1),"",A1 & " ")
That reads "if the cell is blank, ignore it, otherwise echo it with a trailing space." You can daisy-chain a series of those together:
=if(ISBLANK(A1),"",A1 & " ")&if(ISBLANK(B1),"",B1 & " ")&if(ISBLANK(C1),"",C1 & " ")...
That gets pretty long and repetitive. Adding ARRAYFORMULA and JOIN, we can have that repetitive piece apply across a range of cells, A1:F1 in this case:
=JOIN("",ARRAYFORMULA(IF(ISBLANK(A1:F1),"",A1:F1&" ")))
Related
I would like to increment a cell in a column by 1 to 999. The thing is, it contains both a number and a date:
Item 1
001/2022
002/2022
003/2022
...
999/2022
Is there a way to do this? I assume it's possible to "hard code" the date as a string "2022" and increment the value beside it but, would it be possible to maintain the date format? If anyone can provide both answers and how to tackle them in each way, I'd be much obliged. Sorry if there is a reference to something like that, I just can't find it. I'm fairly new to this.
Place this in any cell with at least 998 open cells below it:
=ArrayFormula(TEXT(SEQUENCE(999),"000")&"/2022")
You could split the string by "/", parse & increment the left value and then rejoin the two parts like so:
function testIncrement() {
console.log(increment("005/2021"));
}
function increment(cellValue) {
const DIVIDER = "/";
let [a, b] = cellValue.split(DIVIDER)
return `${Utilities.formatString("%03d", parseInt(a) + 1)}${DIVIDER}${b}`;
}
I am from Bangladesh and in Bangladesh comma (,) is used as thousand separator. We use comma after 3 digit, 5 digit, 7 digit from right to left like 9,999, 99,999, 9,99,999,99,99,999,9,99,99,999 & 99,99,99,999. I was trying to accomplish this format by textbox format property. When I use #,##0 as format then it only format till 5 digit. When number is 6 digit or higher then it only shows one comma like 456,456 while expected is 4,56,456. I have tried to use #,##,##,##0 but it automatically goes to #,##0. So, how can I format the text box to get my desired result as below?
You can't. You'll have to run a custom format like this:
TextValue = Format(Fix(Value), Left("##\,##\,##\,##\,##\,##\,", -Int(-(Len(Abs(Fix(Value))) - 2) \ 2) * 4) & "##0")
Note, the negative values will be formatted correctly as well while decimals will be cut off.
If you have decimals, append these:
TextValue = Format(Fix(Value), Left("##\,##\,##\,##\,##\,##\,", -Int(-(Len(Abs(Fix(Value))) - 2) \ 2) * 4) & "##0") & LTrim(Str(Abs(CCur(Value)-Fix(Value))))
As used as ControlSource (read-only) in a form or a report:
=Format(Fix([Amount]),Left("##\,##\,##\,##\,##\,##\,",-Int(-(Len(Abs(Fix([Amount])))-2)\2)*4) & "##0")
Addendum:
To cover any situation with values:
Larger than 1, with or without decimals
Smaller than 1, a positive decimal value
Zero
Larger than -1, a negative decimal value
Smaller than -1, with or without decimals
an extended expression is needed:
TextValue = Format(Value, ";-") & _
Format(Abs(Fix(Value)), Left("##\,##\,##\,##\,##\,##\,", -Int(-(Len(CStr(Abs(Fix(Value)))) - 2) \ 2) * 4) & "##0") & _
IIf(Value - Fix(Value), LTrim(Str(Abs(Value - Fix(Value)))), "")
First part controls the sign
Second part controls the integer value
Third part controls decimals
This will output correctly for any value within the entire range of Currency.
Trying to attempt the following in MS Access.
Convert data in one field to an 18 digit number starting with 01 in another field.
There are also some conditions that have to be met:
the first dash should become double zeros
the second dash should be removed
the third and fourth dash should be a single zero
the decimal must also be replaced with a zero
My query works fine until the decimal is the 15th character in the data.
Here is the query:
SELECT MasterVacant.ParcelIdNumber,
"01" + Mid([ParcelIdNumber],1,2) + "00" + Mid([ParcelIdNumber],4,2) + Mid([ParcelIdNumber],7,1)
+ IIf(Mid([ParcelIDNumber],11,1) = "", "0"+Mid([ParcelIDNumber],9,2), Mid([ParcelIDNumber],9,3))
+ IIf(Mid([ParcelIDNumber],14,1) = ".", "0"+Mid([ParcelIDNumber],12,2), Mid([ParcelIDNumber],12,3))
+ Mid([ParcelIDNumber],15,3) AS ParcelNumber
FROM MasterVacant;
Here is a start and finish example...
'12-06-1-00-50.000-RR' should become '011200061000050000'
'12-06-3-07-09.000-RR' should become '011200063007009000'
'13-35-1-01-129.000-RR' should become '011300035100112900'
However, instead of getting `0113000351001129000' I get '013000351001129.00'.
The issue is how do I remove the decimal when the decimal is the 15th character like in the third set of example?
I receive the data as a single column. Some of it is below....
1. 13-35-1-07-29.000-RR
2. 13-35-1-01-112.000-RR (Removing the decimal when the data is like this is the issue)
3. 13-35-4-01-01.000-RR
4. 13-35-4-02-04.000-RR
5. 13-35-1-13-17.000-RR
The output for the above data should be
1. 011300351007029000
2. 011300351001112000
3. 011300354001001000
4. 011300354002004000
5. 011300351013017000
Use a custom function:
Public Function Make18(ByVal Value As String) As String
Const Head As String = "01"
Const Tail As String = "000"
Const Lead As String = "00"
Dim Parts As Variant
Dim Part As Integer
Dim Result As String
Parts = Split(Split(Value, ".")(0), "-")
For Part = LBound(Parts) To UBound(Parts)
Select Case Part
Case 0
Parts(Part) = Head & Parts(Part)
Case 1
Parts(Part) = Lead & Parts(Part)
Case 3, 4
Parts(Part) = Right(Lead & Parts(Part), 3)
End Select
Next
Result = Join(Parts, "") & Tail
Make18 = Result
End Function
and your query becomes:
SELECT
MasterVacant.ParcelIdNumber,
Make18([ParcelIdNumber]) AS ParcelNumber
FROM
MasterVacant;
I am assuming you meant the opposite where:
12-06-1-00-50.000-RR should become 011200061000050000
12-06-3-07-09.000-RR should become 011200063007009000
13-35-1-01-129.000-RR should become 0113000351001129.00
I would recommend the REPLACE() in MSACCESS to strip the dashes out. Once you have the dashes out you can MID()
Unfortunately your attempted code does something different with the 3rd row because 3 zeros are being put in when there should be only two in my opinion.
Try in a text box:
=Replace("13-35-1-01-129.000-RR","-","")
will return 1335101129.000RR
and see if that assists you in making your code.
Maybe go one step further and put it in a function.
I'm struggeling with a part of code that I want to implement in Microsoft Access.
The required code is used for project asignments.
The code format contains the last 2 numbers of the year + 4 digits which add up until a new year, then the last 2 numbers of the year add up with 1 and the 4 digits start at 1 again.
For example:
2019:
190001 = first task;
190002 = second task;
etc...
2020:
200001 = first task;
200002 = second task;
etc...
Could anybody help me out how to code this in Microsoft Access, preferably by VBA?
This way I can asign the code to a "submit" button to avoid similar numbers.
Thanks!
Formatting your code given an integer could be achieved in several ways, here is one possible method:
Function ProjectCode(ByVal n As Long) As Long
ProjectCode = CLng(Format(Date, "yy") & Format(n, "0000"))
End Function
?ProjectCode(1)
200001
?ProjectCode(2)
200002
?ProjectCode(100)
200100
You probably need to assign the next task id to a project.
So, look up the latest id in use and add 1 to obtain the next task id:
NextTaskId = (Year(Date()) \ 100) * 10000 + Nz(DMax("TaskId", "ProjectTable", "TaskId \ 10000 = Year(Date()) \ 100"), 0) Mod 10000 + 1
Nz ensures that a task id also can be assigned the very first task of a year.
I am having an issue where a field is stored in our database as '##ABC' with no space between the number and letters. The number can be anything from 1-100 and the letters can be any combination, so no consistency of beginning letter or numeric length.
I am trying to find a way to insert a space between the number and letters.
For example, '1DRM' would transform to '1 DRM'
'35PLT' would transform to '35 PLT'
Does anyone know of a way to accomplish this?
You can use regular expressions like the one below (assuming your pattern is digits-characters)
= System.Text.RegularExpressions.Regex.Replace( Fields!txt.Value, "(\d)(\D)", "$1 $2")
Unfortunately, there's no built in function to do this.
Fortunately, Visual Studio lets you create functions to help with things like this.
You can add Visual BASIC custom code by going to the Report Properties and going to the Custom Code tab.
You would just need to write some code to go through some text input character by character. If it finds a number and a letter in the next character, add a space.
Here's what I wrote in a few minutes that seems to work:
Function SpaceNumberLetter(ByVal Text1 AS String) AS String
DIM F AS INTEGER
IF LEN(Text1) < 2 THEN GOTO EndFunction
F = 1
CheckCharacter:
IF ASC(MID(Text1, F, 1)) >= 48 AND ASC(MID(Text1, F, 1)) <=57 AND ASC(MID(Text1, F + 1, 1)) >= 65 AND ASC(MID(Text1, F + 1, 1)) <=90 THEN Text1 = LEFT(Text1, F) + " " + MID(Text1, F+1, LEN(Text1))
F = F + 1
IF F < LEN(Text1) THEN GOTO CheckCharacter
EndFunction:
SpaceNumberLetter = Text1
End Function
Then you call the function from your text box expression:
=CODE.SpaceNumberLetter("56EF78GH12AB34CD")
Result:
I used text to test but you'd use your field.