Apparently the report seems to explicitly trimming any trailing spaces on report fields. But I need those spaces because it's a bank feed. I got an expression something like :
=PadRight( [BAccount.AcctName] , 50, ' ' )
or
=' BLAH '
Still no spaces
Does anyone has any tricks to achieve it ?
TIA
I believe the report engine will trim all whitespace and non-printable characters.
Does anyone has any tricks to achieve it ?
Yes, based on assumption above; one possible trick is to append a printable Unicode character at the end of the string that renders as whitespace but isn't technically considered white space. This type of trick is very common in phishing attacks.
I choose the 'Braille Pattern Blank' character which renders like a space character but isn't considered whitespace.
Reference:
https://www.compart.com/en/unicode/U+2800
You can copy it directly from character box in link above (highlighted blue):
And paste it in Acumatica report designer to append it after the blank space:
I tested this by selecting the field in report, it appended the desired spaces:
Related
I have a current saved search formula that groups values from two fields in lowercase, and It also replaces any spaces with hyphen. Currently it also removes two special characters including period and forward slash. I am now trying to modify this formula to also remove 2 additional characters including Apostrophe (‘) and quotation mark (“)
Following is the current formula: TRANSLATE(LOWER(CONCAT({custitem38}||'-', {custitem16})), ' /.', '--')
This worked for me: TRANSLATE(LOWER(CONCAT({custitem38}||'-', {custitem16})), ' "/.', '--')
enter code hereTRANSLATE(LOWER(CONCAT({custitem38}||'-', {custitem16})), ' "/.', '--')
This might work
TRANSLATE(LOWER(CONCAT({custitem38}||'-', {custitem16})),'‘"/"/.', '--')
I have a string:
result = " is programming that manages the execution of programs written in any of several supported languages."
I want to remove white space at the beginning of a sentence.
How I can do it?
You can do this by using Trim(String) in VBA (for trimming left and right spaces) or use LTrim(String) for only trimming leading spaces.
You can use Trim(result) it will remove all white spaces at the beginning and at the end of a string
I am trying to parse a CSV file which has single quote as text qualifier. The problem here is that some values with single quote text qualifier itself contains single quote
e-g:
'Fri, 24 Feb 2017 17:44:57 +0700','th01ham000tthxs','/','','Writer's Tools Data','7.1.0.0',
I am struggling to parse the file as after this row, all of the remaining rows get displaced.
I tried working with OpenCSV, UnivocityParsers but didn't get any luck.
If I place the above row in excel (Excel Image) and provide text qualifier as single quote, it give correct result without any displacement of rows.
If using java, the JRecord library should handle the File.
How it works: if a field starts with a quote (e.g. ,') specifically look for ', or ''', or ''''', or ' etc (an odd number of quotes followed by either a comma or end-of-line marker). This approach breaks down if:
The embedded quote is the last character in a field i.e. 'Field with quote '',
White space between the quote and comma i.e. 'Field' , or , '
Here is the line in ReCsvEditor
Also in the ReCsvEditor when editing the file, if you select Generate >>> Java Code >>> ... it will generate Java/JRecord Code to read the file.
Disclaimer: I am the author of JRecord / ReCvEditor. Also the ReCsvEditor Generate function is new and needs more work
Try configuring univocity-parsers to handle the unescaped quote according to your scenario. 'Writer's Tools Data' has an unescaped quote. From your input, I can see you want to use STOP_AT_CLOSING_QUOTE as the strategy to work around these values.
Add this line to your code and it should work fine:
parserSettings.setUnescapedQuoteHandling(UnescapedQuoteHandling.STOP_AT_CLOSING_QUOTE);
Hope this helps.
I am translating a game, and the game's text box only supports 50 characters max per line. Is there a way to use a formula to split the entire sentence every 50 characters or whole word (49, 48, 47, etc)?
I am currently working with this formula.
=JOIN(CHAR(10),SPLIT(REGEXREPLACE(A1, "(.{50})", "/$1"),"/"))
The problem with this code, is that it splits at exactly 50 characters (one time), and will split in the middle of the word.
So again, my goal is to have it not split on the 50th character IF the 50th character is in the middle of the word, and for the rule to apply for the rest of the lines too because it only applies on the first line.
Please take a look at this test google sheet to get an example of what I am talking about.
If it's impossible to do it on Google Sheets, I don't mind moving to Excel provided I get a functioning code.
For the record, I did ask in Google's product forums 2 days ago, and still haven't received an answer.
=REGEXREPLACE(A1, "(.{1,50})\b", "$1" & CHAR(10))
{50} matches exactly 50 times, but what you need is 50 or less.
\b is word boundary that matches between alphanumeric and non-alphanumeric character.
= REGEXEXTRACT(A1,"(?ism)^"&REPT("([\w\d'\(\),. ]{0,49}\s)", ROUNDUP(LEN(A1)/50,0))&"([\w\d'\(\),. ]{0,49})$")
Tested with various expressions and works as intended. Note that only these characters [a-zA-Z0-9_'(),.] are allowed, Which means - and other characters not mentioned will not work. If you need them, add them inside the REPT expression and finishing regexp formula. Otherwise, This will work perfectly.
You are pretty close. I'm not an expert in Sheets, so not sure if this is the best way, but your Regex is wrong for what you want.
Also, you need to be certain that you don't use a split character that might appear in the phrase itself. However, using CHAR(10) for the replace character allows you to insert LF without going through the JOIN SPLIT sequence.
replace any line feeds, carriage returns and spaces with a single space
Match strings that start with a non-Space character followed by up to 49 more characters which are followed by a space or the end of the string.
replace the capture group with the capturing group followed by the CHAR(10) (and delete the space following).
There will be extra CHAR(10) at the end which you can strip off.
EDIT Regex changed slightly due to a difference in behavior between Google's RE and what I am used to (probably has to do with how a non-backtracking regex works). The problem showed up on your example:
=regexreplace(REGEXREPLACE(REGEXREPLACE(A1 & " ","[\r\n\s]+"," "),"(\S.{0,49})\s","$1" & char(10)),"\n+\z","")
I would like to trim off the text which is after the bracket in the cell Value
The current formula I'm using keeps giving me the error not being able to extract the targetted string.
=LEFT([#[Name ]],FIND("(",[#[Name ]]))
I want to go shopping (Today)
Goal: Is to remove
(Today)
Expected Result:
I want to go shopping
One of these should do.
=TRIM(LEFT([name], FIND("(", [name]&"(")-1))
=TRIM(REPLACE([name], FIND("(", [name]&"("), LEN([name]), TEXT(,)))
Note that I suffixed the original text with the character that the FIND is looking for. In this manner, it will always be found even if it is not in the original text.
You may find that you have a rogue trailing space in the Name header label.