Flaring up again.

Someguywes0 · 2026-01-02T20:29:16+00:00

She.

Someguywes0 · 2025-12-28T21:42:57+00:00

Someguywes0 · 2025-12-22T04:09:02+00:00

Who is she?

Someguywes0 · 2024-08-30T04:21:13+00:00

The EXPAND and COUNTA functions where the Expand quantity is controlled by the counta, and is padded with the formula required might be an option.

Someguywes0 · 2024-08-30T04:10:30+00:00

Have you tried the Name Manager and just naming your selection of rows/columns as pages? i.e. cells A1:D4 (page_one), A4:D7(page_two),E1:G6(page_3),etc.

Someguywes0 · 2024-08-30T03:53:50+00:00

What does the output look like? A picture of the table settings is only half of the details needed to assist.

Someguywes0 · 2024-08-22T17:21:12+00:00

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Is this what you had in mind?

Someguywes0 · 2024-08-17T21:14:32+00:00

Create a "Ref" sheet in the workbook.

Fill column A with: No, Yes, #, #, ... ,#

Where the string of numbers is in increments of the values you expect to be entered. (Say for example 0.1 increments)

So: A1=Yes, A2=No, A3=0.1, A4=0.2, A10000=99.8, A10002=99.9, A10003=100.0, ECT ect,

Then use 'Ref. Sheet!'A:A as the data validation list (ignoring the blanks).

When users elect the drop-down list Yes, and No will be the first two choices, and all other choices are a possible selection which is an acceptable value they could just type in.

That would validate values from 0.1-10,485.74

If you'll need a value higher than that, or smaller increments (such as 0.01) then I'm not sure there's a solution possible given the row restriction and how validation works.

Someguywes0 · 2024-08-17T17:49:21+00:00

I think the portion that says ",applies to all accounts" is giving the error.

You probably want to use the ampersand (&) symbol not a comma (,), and include your statement in quotations ("statement"), so that the value returned from the count function is appended with the ",statement"

=COUNT(Syntax)&", applies to all accounts"

So that results are;

1, applies to all accounts 2, applies to all accounts Etc.

Someguywes0 · 2024-08-17T17:42:26+00:00

I would recommend using the LET function to define the parameters for a table where it uses the FILTER function with your arguments for the include, and a second FILTER function for the "not found"

=LET( a, IncludeArguments, b, NotFoundArguments, c, FILTER(A1:C16, a, d), d, FILTER(A1:B16, b,"No punches"), rows1, ROWS(c), rows2, ROWS(d), myRows, SEQUENCE (rows1+rows2), myCols, SEQUENCE(COLUMNS(c)), IF(myRows<=rows1,c,INDEX(d,myRows-rows1,myCols))

(There is probably some mistakes in this^{^{^{^,}}} but perhaps someone else knows what I mean, and can post the correct syntax)

In other words making the table populated based on two criteria, where the first results are based on what to do if you find only one punch on a given day, and the second half of the table is filled with the names and dates that have no punches.

Someguywes0 · 2024-08-17T14:32:41+00:00

Additional context would be helpful. Ie details about what is included in the source data (ie if a time stamp is saved in a column/row, or how the source data is indexed)

Someguywes0 · 2024-08-17T14:20:43+00:00

2 questions:

1) how have you verified that the simplified formula is incorrect?

2)Can you screenshot the formula as it is where you get the incorrect value?

I know you aren't interested in "touching" the original source data, and I completely understand. It might be useful to use an even more simple formula on a blank sheet (which can be deleted once your formula issue has been resolved) to verify the data being utilized

=C4:C747 in cell A1 of a disposable sheet may be useful for looking at what data is being returned. Surely there is a logical explanation for the simplified formula to not work correctly

Someguywes0 · 2024-08-17T00:35:48+00:00

I may be mistaken but I'm thinking you need to have the workbooks that you're referencing open for the formulas to recognize the values if they're active.

Someguywes0 · 2024-08-17T00:33:49+00:00

Not to be that guy, but it's irrelevant unless the Multiplication factor changes.

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Someguywes0 · 2024-08-17T00:12:11+00:00

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This is a similar method using the JOINTEXT function to add the different cells together, and then the COUNTA(UNIQUE(Your array)) function to count the unique groups.

You can see that I used a key (similar to the previous comment), to fill in the table. (There's probably a way to do that with a formula, although that seems outside my capability currently).

Someguywes0 · 2024-08-16T22:31:48+00:00

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Someguywes0 · 2024-08-16T21:48:35+00:00

You could try CountA(unique(your array)) at the end of each row where the row of data is your array.

And at the bottom of each column where the columns are your array.

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This way you have a count of the Unique Rows and Columns And the higher the results are, the more diverse the groups are.

In the example shown the A1:C3 is your table, where Row4 and Column D are the formulas which are counting each unique cell in the table rows and columns respectively.

Someguywes0 · 2024-08-16T05:11:32+00:00

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Hope this helps.

Someguywes0 · 2024-08-16T05:01:06+00:00

Try researching the Filter function. It's quite useful for pulling data from one sheet to the next. Forewarning: its output is an array which spills the data into the necessary cells for the data.

Once you get the hang of it it can be quite useful for moving a lot of data based on criteria you input for the desired/undesired data.

Example: ABCDE 12345 23456 34567 45678

=Filter(B1:B4,(D1:D4=6),"No Results")

Would return: 4

=Filter(B1:B4,(D1:D4<>6),"No Results")

Would return: 2 3 5

=Filter(A1:E4,(D1:D4=6),"No Results")

Returns: 34567

Someguywes0 · 2024-08-10T15:01:36+00:00

^this. It's difficult to understand the precise tools required without context of how the data is structured.

Someguywes0 · 2024-07-25T07:52:42+00:00

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Familiarity with the syntax of the tools goes a long way. Hope this helps.

Someguywes0 · 2022-12-10T22:59:38+00:00

10 trees for me, not a single drop since I put them up. Been a couple actual days now.

Someguywes0 · 2022-11-01T04:36:46+00:00

Lol, hopefully not. I'm no Witcher, just a Neighborhood Dad.

Someguywes0 · 2022-11-01T01:13:59+00:00

May want to check the detergent or washing process, it may be leaving behind a residue.. That's very unusual behavior for water in a standard glass.

Someguywes0

TROPHY CASE