I've developed a workaround since crystal reports doesn't seem to have a substring function with the following formula:
right({_v_hardware.groupname},
truncate(instr(replace({_v_hardware.groupname},".",
","), ","))
What I'm trying to do is search for the period (".") in a string and replace it with a comma. Then find the comma position in the string and print all characters following after the comma. This is assuming the string will only have 1 period in the entire string.
Now when I attempt to do this, I get some weird characters which look like wingdings. Any ideas?
thanks in advance.
I don't know the entire issue that you are attempting to accomplish, but for this question alone, the step of replacing the period with a comma seems to be unnecessary. If you know that there is only one period in the string and you only want the characters right of the period then you should be able to do something like the following (this is #first_formula):
right({_v_hardware.groupname}, len({_v_hardware.groupname}) - instr({_v_hardware.groupname},"."))
If for some reason you want to show the comma then I'd do that in a separate formula. If you need the entire screen with the comma replaced then just do:
replace({_v_hardware.groupname},".",",")
And if you need the comma plus included in the string then it might just be easier to do something like:
"," + {#first_formula}
Hope this helps.
Related
I've run into an interesting problem I'm hoping someone can shed some light on.
I'm trying to pull a unique list of names from an MS SQL Database - but the company has been sloppy with their names. They were tacking on a code to the end of last name for some users. I need to remove that code.
Example:
firstname lastname
John Doe
Mary Smith AST
Mike Jackson AST
Brian Astor
Jackie Masterson
In the example, "AST" is the code they tack on. It's not tacked on to all last names either. I need to get an output of just the last names without the code.
I would have expected this is a simple use of REPLACE. I tried:
select REPLACE(lastname, ' AST', '') from table
Note the leading space in the quotes for the search phrase... this does work to remove the "AST" appended to the last names.
However - my problem is that it will also remove anywhere AST appears at the BEGINNING of the field. So Brian Astor comes out as "Brian or" since the field started with AST. However... it correctly does not remove ast from the middle, so Jackie Masterson is fine.
Any ideas why it is ignoring the leading space in my search phrase for the beginning of the field? I've tried ltrim to eliminate the possibility the field has leading spaces.
Thanks!
Replace with an empty string will eliminate the searched string anywhere in your source string. So the behaviour is as expected.
If you only need to replace ' ast' at the end of your searched string, try something like this:
select replace(lastname + '$$$', ' AST$$$', '') from table
Of course you need to be sure that the $$$ appended don't appear by chance in your source string (lastname). Which I guess is not that likely.
I'm working to build an import tool that utilizes a quoted CSV file. However, several of the fields in the CSV file are reported as such:
"=""38000"""
Where 38000 is the data I need. The data integration software I use (Talend 6.11) already strips the leading and trailing double quotes for me (so, "38000" becomes 38000), but I can't find a way to get rid of those others.
So, essentially, I need "=""38000""" to become "38000" where the leading "=" is removed and the trailing "" is removed.
Is there a TRIM function that can accomplish this for me? Perhaps there is a method in Talend that can do this?
As the other answer stated, you could do that operation in SQL. Or, you could do it in Java, Groovy, etc, within Talend. However, if there is an existing Talend component which does the job, my preference is to use it. That leads to faster development, potentially less testing, and easier maintenance. Having said that, it is important to review all the components which are available, so you know what's available to you.
You can use the Talend component tReplace, to inspect each of the input columns you want to trim of quotes and equal signs. A single tReplace component can do search and replace operations on multiple input columns. If all the of the replaces are related to each other, I would keep them within a single tReplace. When it gets to the point of doing unrelated replacements, I might place those within a new tReplace so that logical operations are organized and grouped together.
tReplace
For a given Input Column
search for "=", replace with ""
search for "\"", replace with ""
Something like that:
SELECT format( '"%s"', trim( both '"=' from '"=""38000"""' ) );
-[ RECORD 1 ]---
format | "38000"
1st: trim() function removes all " and = chars. Result is simply 38000
2nd: with format can add double quote back to get wishful end result
Alternatively, can use regexp and other Postgres string functions.
See more:
https://www.postgresql.org/docs/current/static/functions-string.html
I'm typing up a table with org mode, where the equal sign(=) if the first character in the cell and it want to start a formula. how do I get it to display the symbol without it being a formula, of a way to use formulas to display it. I get errors when I use single quotes, and I see the Unicode decimal value when using double quotes.
I have tried the following
='=+'
="=+"
they give
#ERROR
[61, 43]
Use an escaped entity, \equal{} and it should display as you wish. See the variable org-entities for others you can use.
I'm a bit late :D
There may be a better way, but you can try with :='(format "=+")
Source: https://emacs.stackexchange.com/questions/19183/can-i-use-formula-to-manipulate-text-in-org-mode-table
When I ran into this problem just now, I found that I was able to get around it by replacing the equals sign with some other similar-looking character. Two which come to mind are ꞊ ‘U+A78A MODIFIER LETTER SHORT EQUALS SIGN’ and ⹀ ‘U+2E40 DOUBLE HYPHEN’.
I'm trying to do a simple mail merge in Word 2010 but when I insert an excel field that's supposed to represent a zip code from Connecticut (ie. 06880) I am having 2 problems:
the leading zero gets suppressed such as 06880 becoming 6880 instead. I know that I can at least toggle field code to make it so it works as {MERGEFIELD ZipCode # 00000} and that at least works.
but here's the real problem I can't seem to figure out:
A zip+4 field such as 06470-5530 gets treated like an arithmetic expression. 6470 - 5530 = 940 so by using above formula instead it becomes 00940 which is wrong.
Perhaps is there something in my excel spreadsheet or an option in Word that I need to set to make this properly work? Please advise, thanks.
See macropod's post in this conversation
As long as the ZIP codes are reaching Word (with or without "-" signs in the 5+4 format ZIPs, his field code should sort things out. However, if you are mixing text and numeric formats in your Excel column, there is a danger that the OLE DB provider or ODBC driver - if that is what you are using to get the data - will treat the column as numeric and return all the text values as 0.
Yes, Word sometimes treats text strings as numeric expressions as you have noticed. It will do that when you try to apply a numeric format, or when you try to do a calculation in an { = } field, when you sum table cell contents in an { = } field, or when Word decides to do a numeric comparison in (say) an { IF } field - in the latter case you can get Word to treat the expression as a string by surrounding the comparands by double-quotes.
in Excel, to force the string data type when entering data that looks like a number, a date, a fraction etc. but is not numeric (zip, phone number, etc.) simply type an apostrophe before the data.
=06470 will be interpreted as a the number 6470 but ='06470 will be the string "06470"
The simplest fix I've found is to save the Excel file as CSV. Word takes it all at face value then.
I asked a question earlier today and got a really quick answer from llbrink. I really should have asked that question before I spent several hours trying to find an answer.
So - here's another question that I have never found an answer for (although I have created a work-around which seems very cludgy).
My AHK program asks the user for a login name. The program then compares the login name with an existing list of names in a file.
The login name in the file may contain spaces, but there are never spaces at the beginning of the name. When the user enters the name, he may include spaces at the beginning. This means that when my program compares the name with those in the file, it can not find a match (because of the extra spaces).
I want to find a way of stripping the spaces from the beginning of the input.
My work-round has been to split the input string into an array (which does ignore leading spaces) and then use the first element of the array. This is my code :
name := DoStrip(name)
DoStrip(xyz) ; strip leading and trailing spaces from string
{
StringSplit, out, xyz, `,, %A_Space%
Return out1
}
This seems to be a very laboured way to do it - is there a better way ?
I don't see a problem with your example if it works on all cases.
There is a much simpler way; just use Autotrim which works like this.
AutoTrim, On ; not required it is on by default
my_variable = %my_variable%
There are also many other different ways to trim string in autohotkey,
which you can combine into something useful.
You can also use #LTrim and #RTrim to remove white spaces at the beginning and at the end of the string.