Actual Record
"102","kal12 3# ","18009","10009","29","HR","del"
What I want (Expecting record)
"102","kal123","18009","10009","29","HR","del"
For above scenario,If you see the record "kal12 3#" which is actual record but I want transform the record as like "kal123" I were used "StringHandling.CHANGE("hello world!","world","guy")" function for remove the special characters and for space removing used StringHandling.TRIM(" hellow world! ")
But I want to remove the space and special characters at a time,so can anyone please help me to resolve this issue.
I am Using Talend open studio
Thanks& Regards,
Naresh
Supposing the field name is row1.myField, try this in a tMap:row1.myField.replaceAll("[# ]", "")
You just have to include each character you want to remove from row1.myField between the square brackets.
Hope this helps.
TRF
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.
Here is an example on my github profile - https://github.com/jack17529
I want to change this -
Silver Bullet in Issue KILLING.____
Master Mind to create Issues.______
My strongest language is Python not English.
I want to have newline instead of blanks.
like this -
Silver Bullet in Issue KILLING.
Master Mind to create Issues.
My strongest language is Python not English.
I have checked Bitbucket Bio is nowhere related to Github Bio.
Maybe they don't allow us to do it via the normal way, But It is possible to do of course. We can use the auto newline rule for the words which are too long for appending to the current line, for our need. All we need to do is putting other Unicode Spaces instead of normal space. And normal space between lines, for using newline rule against forbidden newline rule.
And if you want a free line, because of the character limitations, you can use the longer one;
" " instead of " " (Try selecting spaces between quotes with your mouse)
Also this trick allows me create unnecessary spaces in the Stack Overflow too, like above, in the quote box.
Here is the result: github.com/cosmicog:
I have tried other answers, html ways, but no, they handle html tricks of course.
Note: This causes a bad look in the list view and the profile overview tooltip:
Maybe that is why it is not allowed but I hope they will fix this in the future.
As told to me by github support there is no way !, see here -
According to Github Support
I just did it by simply copying and pasting the character corresponding to this codepoint | unicode-table.com | as many time as needed in order to align the text the way I wanted.
This is the procedure I followed: at the end of each line I pressed Enter, then I filled the new line with 7 instances of the character mentioned above; then I pressed Enter again and started the new line with its text.
This question is a little stale, but I found it before I solved this myself, so I thought I'd drop my solution.
The bio doesn't appear to honor markdown, but neither does it accept HTML entities or elements. I worked around this with non-breaking characters to create long "words" similar to how you've used "_".
You can see in my bio that I needed a " " and a "‑" to format mine. The long word will pop to the next line. If you have a real short line, you can extend it with a lot of non-breaking spaces, but this probably isn't necessary. Since you cannot enter " " you need to use copy/paste or ALT codes (not looked up, but someone might add these for you). Those are the real characters above, so you can take them from this answer.
Refer: How to create newline in Github Bio
Just use in HTML editor mode to new line is OK, This is my GitHub Bio
Do you know how to resolve the problem when one line is full, then the Chinese punctuations will be placed at the beginning of next line as shown in (1)? In fact we hope the punctuations to be placed at the end of each line as shown in (2).
（1）
你好你好
,你好你好
（2）
你好你好,
你好你好
Thank you very much for your help in advance!
You are placing a space between the last char and the punctuation and that is a split point. The simplest way is to remove the space before the puntuation and add it after. Other option is to replace the space with a non breaking space \u00a0 to avoid the split at that point.
I have an NSArray of lines (objective-c iphone), and I'm trying to find the line which starts with a number, followed by a dot and a space, but can have any number of spaces (including none) before it, and have any text following it eg:
1. random text
2. text random
3.
what regular expression would I use to get this? (I'm trying to learn it, and I needed the above expression anyway, so I thought I'd use it as an example)
With C#:
#"^ *[0-9]+\. "
It doesn't check for the presence of something after the ., so this is legal:
1.(space)
If you delete the # and escape the \ it should work with other languages (it is pretty "down-to-earth" as RegExpes go)
I may suggest (Perl-compatible regexp):
^\s*\d+\.\s
At the beginning of a line:
Any number (0-n) of spaces
One or more digits
A dot
A space
Something like
^\s*\d+\.
But it depends on the language.
/^\s*[0-9]+\.\s+/
would be my guess providing you don't have any space before the number
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.