Is there any function to replace the special characters by null in informatica
if we used replacestr function, i think we should specify all special characters
as follows replacestr(input,'!','~','#','#','$','%','^','&','*',null)
But we dont know what are teh special characters will be coming as input.
can u please let me know that which function will be suitable.
Did you try, REG_MATCH(input,'[^a-zA-Z0-9]') ?
Sorry for the late answer, I know you might not need it anymore .. I just seen this question. I suggest that you use a regular expression that looks for anything but chars/numbers/spaces/newlines and replace them with null. It will be a Replace transformation with the patternSearch "your regular Expression" and null for the field ReplaceWith .
I am assuming by all special characters are all characters besides what I mentioned in the Regular expression.
Try this:
REG_REPLACE('[^[:print:]]',NULL)
This will find all non-print characters and replace them with null (I think it'd be better to replace with '', but thats just my opinion. Informatica uses POSIX-standard, so you can use others like alnum, punct, whatever you need.
http://en.wikipedia.org/wiki/Regular_expression#POSIX_Extended_Regular_Expressions
Related
Does someone know about some way how to extract allowed characters from regular expression and construct user friendly message?
For example, by providing regular expression
^[a-zA-Z0-9&\-\+_\.\s]{1,10}$
to get something like
a-z A-Z 0-9 & - + _ . with spaces
I am using java. I can imagine that it could be too complicated or even impossible to cover all types of regular expressions, but maybe you know about some library, tool or algorithm that could help.
Thanks
Yes. It can be done.
What you need is:
Turn your regexp body into a string.
Parse that string (with a regex for instance) that will output the desired list.
Apply possible regexp options (such as ignore case to the result).
This is tedious work if you're not VERY familiar with Regexp. I actually have code in production doing just that, but it's proprietary so I can't post it here and it's not in Java.
I guess you should first ask yourself whether there is no simpler solution for your problem. If for instance your regexp is a constant, you could associate it with a by-hand list of accepted characters.
If your input is a character-class like the one you provided, you could match it with the expression
([^\\]-[^\\]|\\.|[^^$[\]])
that will give you a list of elements like "a-z", "\+", "_" that you could then tidy up a little further, e.g., removing the "\", and then print it nicely formatted.
And you could extract the length information using
{([0-9]+)(,([0-9]+))?}
that accepts {1,10} as well as {10} with the "from" and "to" values being captured each in their own group.
That should get you started.
I want to create a simple regex check to simply check the length of the text, nothing else. Closest I got was [^.]{1,6} however this will not allow . characters.
How can I make sure that only thing that regex expression does is checks length?
Example: http://regex101.com/r/eJ6pK5#pcre
Use this regex:
[\s\S]{1,6}
The above will work with multiline.
Otherwise . represents any character other than new line.
So, .{1,6} will also do the trick.
I'm sure this is a simple question for someone at ease with regular expressions:
I need to match everything up until the character #
I don't want the string following the # character, just the stuff before it, and the character itself should not be matched. This is the most important part, and what I'm mainly asking. As a second question, I would also like to know how to match the rest, after the # character. But not in the same expression, because I will need that in another context.
Here's an example string:
topics/install.xml#id_install
I want only topics/install.xml. And for the second question (separate expression) I want id_install
First expression:
^([^#]*)
Second expression:
#(.*)$
[a-zA-Z0-9]*[\#]
If your string contains any other special characters you need to add them into the first square bracket escaped.
I don't use C#, but i will assume that it uses pcre... if so,
"([^#]*)#.*"
with a call to 'match'. A call to 'search' does not need the trailing ".*"
The parens define the 'keep group'; the [^#] means any character that is not a '#'
You probably tried something like
"(.*)#.*"
and found that it fails when multiple '#' signs are present (keeping the leading '#'s)?
That is because ".*" is greedy, and will match as much as it can.
Your matcher should have a method that looks something like 'group(...)'. Most matchers
return the entire matched sequence as group(0), the first paren-matched group as group(1),
and so forth.
PCRE is so important i strongly encourage you to search for it on google, learn it, and always have it in your programming toolkit.
Use look ahead and look behind:
To get all characters up to, but not including the pound (#): .*?(?=\#)
To get all characters following, but not including the pound (#): (?<=\#).*
If you don't mind using groups, you can do it all in one shot:
(.*?)\#(.*) Your answers will be in group(1) and group(2). Notice the non-greedy construct, *?, which will attempt to match as little as possible instead of as much as possible.
If you want to allow for missing # section, use ([^\#]*)(?:\#(.*))?. It uses a non-collecting group to test the second half, and if it finds it, returns everything after the pound.
Honestly though, for you situation, it is probably easier to use the Split method provided in String.
More on lookahead and lookbehind
first:
/[^\#]*(?=\#)/ edit: is faster than /.*?(?=\#)/
second:
/(?<=\#).*/
For something like this in C# I would usually skip the regular expressions stuff altogether and do something like:
string[] split = exampleString.Split('#');
string firstString = split[0];
string secondString = split[1];
I need a regular expression to list accepted Version Numbers. ie. Say I wanted to accept "V1.00" and "V1.02". I've tried this "(V1.00)|(V1.01)" which almost works but then if I input "V1.002" (Which is likely due to the weird version numbers I am working with) I still get a match. I need to match the exact strings.
Can anyone help?
The reason you're getting a match on "V1.002" is because it is seeing the substring "V1.00", which is part of your regex. You need to specify that there is nothing more to match. So, you could do this:
^(V1\.00|V1\.01)$
A more compact way of getting the same result would be:
^(V1\.0[01])$
Do this:
^(V1\.00|V1\.01)$
(. needs to be escaped, ^ means must be on the beginning of the text and $ must be on the end of the text)
I would use the '^' and '$' to mark the beginning and end of the string, like this:
^(V1\.00|V1\.01)$
That way the entire string must match the regex.
I have to two strings that I want to match everything that doesn't equal them, the first string can be followed by a number of characters. I tried something like this, negating two ors and negating that result.
?!(?!^.*[^Factory]$|?![^AppName])
Any ideas?
Try this regular expression:
(?!.*Factory$|.*AppName)^.*
This matches every string that does not end with Factory and does not contain AppName.
dfa's answer is by far the best option. But if you can't use it for some reason, try:
^(?!.*Factory|AppName)
It's very difficult to determine from your question and your regex what you're trying to do; they seem to imply opposite behaviors. The regex I wrote will not match if Factory appears anywhere in the string, or AppName appears at the beginning of it.
what about
if (!match("(Factory|AppName)")) {
// your code
}
Would it work if you looked for the existence of those two strings and then negated the regex?