I have an input box that I use to enter a alphanumeric account numbers in a database. The box accepts up to 25 characters. However, for data entry, each account number may not be as long as 25 characters. In such a case, the account numbers are saved with blank spaces before it instead of being saved to the left of the column. How can I solve this?
I would like each number to be saved like the two hyphenated numbers and not with a space like the first record.
Code summary:
Set objDB = New db.Detail_Data
objDB.ConnectionString = CONNECTSTRING
With objDB
.summary_code = CDbl(mvarSumcode)
.charge_code = UCase$(Me.txtChargeCode)
.clientID = UCase$(Me.txtClientID)
.JobID = UCase$(Me.txtJobID)
.Invno = UCase$(Me.txtInvno.Text)
.TransAmt = CCur(Me.txtTransAmt)
.Gl_accno = Format(Me.txtGL, "#########################")
.Description = Me.txtDescription
blnStatus = .AddDetail
End With
Looks like it works as coded. Your line:
.Gl_accno = Format(Me.txtGL, "#########################")
Format with the # symbol right justifies the string, filling in spaces on the left. Unless you add a ! like so (source).
.Gl_accno = Format(Me.txtGL, "!#########################")
Related
I have a simple table that has a column with a date in this format:
MM/DD/YYYY.
Unfortunately, there are some folks who are working without leading zeros.
Therefore I would like to add a leading zero into the Month and Day element using Power Query to have a common format.
But how? Does someone have any function to share?
Again, not sure why you want to do this, but
Assuming all of the entries are text that looks like dates, you can use the following M-Code:
Split the string on the delimiter
Change each entry in the list to a number
Add 2000 to the last number
Change the numbers back to text with a "00" format
Recombine with the delimiter
let
Source = Excel.CurrentWorkbook(){[Name="Table29"]}[Content],
//set type = Text
#"Changed Type" = Table.TransformColumnTypes(Source,{{"TextDate", type text}}),
xform = Table.TransformColumns(#"Changed Type",
{"TextDate", each
let
x = Text.Split(_,"/"),
y = List.Transform(x,each Number.From(_)),
z = List.ReplaceRange(y,2,1, {2000+y{2}}),
a= List.Transform(z,each Number.ToText(_,"00")),
b = Text.Combine(a,"/")
in b})
in
xform
I am thinking a better solution might be to set up your data entry method so that all dates are entered as dates rather than text
My code currently looks like this:
FormatNumber((CDbl(0.05935)),4)
The returned value is 0.0594 rather than 0.0593 which is what I need.
You can try parsing this number to string then trimming it and again parsing back to float.
example:
v = 100.0097
x = Str$(v) ' Gives " 100.0097"
//This adds a leading space for positive numbers
or
x = CStr(v) ' Gives "100.0097"
and then trim it as your need
finalstr = LEFT(variable, (LEN(variable)-4))
then parse it to float
finaltrimed = CDbl(finalstr)
I have a string called indicators, that the original developer of this application used to store single characters to indicate certain components of a model. I need to change the 7th character in the string, which I tried to do with the following code:
indicators[6] = "R"
The problem, I discovered quickly, was that the string is not always 7 characters long. For example, I have one set of values with U 2, that I need to convert to U 2 R (adding an additional space after the 2). Is there an easy way to force character count with Ruby?
use String.ljust(integer, padstr=' ')
If integer is greater than the length of [the receiver], returns a new String of
length integer with [the return value] left justified and padded with padstr;
otherwise, returns [an unmodified version of the receiver].
indicators = indicators.ljust(7)
indicators[6] = "R"
string = 'xabcdexfghijk'
In the example above, 'x' appears twice. I want to capture everything between the first 'x' and the next 'x'. Thus, the desired result is a new string that equals 'xabcdex'. Any ideas?
You could use a simple regular expression: /x.*?x/. This basically means "match any characters in between two x characters, as few times as possible (non-greedy)".
The matched text can be extracted with String#[regexp]
string = 'xabcdexfghijk'
string[/x.*?x/] # => "xabcdex"
I am using the Spreadsheet gem.
My code is:
book = Spreadsheet.open 'excel-file.xls'
sheet = book.worksheet 0
book.write 'output-file.xls'
I want to remove data that comes after ";" in a column:
FULTON BANK NA;FULTON BANK
I just want it to be FULTON BANK NA for example.
Also, I want to leave price data like this: $78,000.00 and want to strip
all other data from a specific column:
MORTGAGE - CORPORATE;($78,000.00)
I just want it to be $78,000.00 for example.
You could do it this way:
s = 'FULTON BANK NA;FULTON BANK'
s = s[/[^;]+/]
that will leave every before the first semicolon in s. Or you could it like this:
s = s.split(';')[0]
Or
s.gsub!(/;.*/, '') # This modifies s in place
For the second one, it depends on the format of your data but you could start with this:
s = 'MORTGAGE - CORPORATE;($78,000.00)'
s = s[/\(([^)]+)\)/, 1]
Or, if the last component may or may not have parentheses, you could do something like this:
s = s.split(';')[-1].tr('()', '')
That will split s into pieces at the semicolons (split(';')), take the last component ([-1]), and then remove any parentheses that there (.tr('()', '')).