I haven't so much tried to answer your question directly, instead I
attempted to solve the same problem you did. For me, this gives a
simpler answer. Given:
Col A: The initial bank value as text
Col B: =IF(RIGHT(A1,1)="-","Debit","Credit")
Col C: =LEN(A1)
Col D: =LEFT(A1,C1-1)
Col E: =FIND(",",D1)
Col F: =SUBSTITUTE(D1,".",",")
Col G: =REPLACE(F1,E1,1,".")
Col H: =VALUE(G1)
Col I: =IF(B1="Debit", 0-H1, H1)
Col A holds the initial value as given by your bank, as text.
Col B then looks at the last character of that value, and displays
either "Debit", if the character was a minus sign, or "Credit" for
Col C gives the total length of the initial text value
Col D gives the initial value stripped of the last character, the plus
or minus sign.
Col E gives us the position of the comma, so that we can turn it into a
Col F gives us the stripped value (from col D) with the periods
replaced by commas, for correct thousand separators.
Col G gives us the value from col F with the final comma changed to a
period, for correct decimal separator. This text string should now be
correctly formatted for interpreting as a number.
Col H gives us the value from col G interpreted as a number.
Col I gives us the final value we want. This is derived from col
H, which has the initial value as a number, without the correct plus or
minus sign, and we use the value in col B to determine if we should use
the number as is, or subtract it from zero to get a negative number. I
formatted this column as currency.
I tested it with the following:
"11.200,33+" ==> "R 11,200.3"
"2.500.236,65-" ==> "R-2,500,236.65"
"11,25+" ==> "R 11.25"
"0,96-" ==> "R-0.96"
If I understand you correctly, this is exactly what you want. No need
to mess around with numbers under or over "9.999,99".
Of course, you can combine some of those formulas to use less extra
cells, but the more you combine stuff, the more complex it becomes, and
I don't know where you'd like to draw the line for complexity of
formulas vs complexity of page layout, so I leave that part up to you.
Hope this helps you out.
On Sun, 12 Jan 2014 17:55:17 +0200
Pertti Rönnberg <firstname.lastname@example.org> wrote:
Happy New Year Dear LibO experts,
Since years back I have copied my bank's digital listing of my bank
account(s); first using MSOWord and later on MSOExcel.
My intension is to transform these listings so I can calculate with
the currency values in LibO-Calc.
The bank's table has four cols: colA=date, colB&colC= text and colD
is the currency as text.
Each listing consists of several hundreds of events (rows).
The problem is that the damn bank -- against all standards -- gives
the currency values with a dot (".") as thousand separator and "+" or
"-" chars (plus or minus) in the right end of each number (e.g.
"987,65+", "1.234,56-", "23.456,78+") -- which is against Calc's will.
By now I have managed to get LibO/Calc to accept all values less than
> first dragged (copied) the table from MSWord => MSExcel;
saved in LibO/Calc as "ods"
> in Calc => function TRIM(D5) to get rid of non-printable chars
> formula (in E5) :
removes the separator/dot and the "+" sign and changes the value from
text to number
> formula (in F5):
removes the separator/dot and the "-" sign and changes the value from
text to number
> copy(dragged) down the 100-300 rows
> Copy(cols E&F) => PasteSpecial to colG:H gives the desired list of
incomes and expences separated in their own cols as acceptable numbers
What formula/function gives the same result for the bank's currency
values bigger than 9.999,99?
e.g. "11.222,33+", "11.222.333,44-"
I have tried to define the "." using the Find() - but have had
trouble with it when nested in another function/formula (e.g. an IF()
function); the definition of it in Calc/Help is obviously at least
unclear if not wrong.
Cannot get the Fixed() -- when trying to eliminate the separator/dot
that way -- working in this 'project' either.
Any advice and help is greatly appreciated.
computer: PC, win7prof/64bit; LO188.8.131.52
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