I tried searching for an answer to this question and also reading the Regex Wiki but I couldn't find what I'm looking for exactly.
I have a program that validates a document. (It was written by someone else).
If certain lines or characters don't match the regex then an error is generated. I've noted that a few false errors are always generated and I want to correct this. I believe I have narrowed down the problem to this:
Here is an example:
This error is flagged by the program logic:
ERROR: File header immediate origin name is invalid: CITIBANK, N.A.
Here is the code that causes that error:
if(strLine.substring(63,86).matches("[A-Z,a-z,0-9, ]+")){
}else{
JOptionPane.showMessageDialog(null, "ERROR: File header immediate origin name is invalid: "+strLine.substring(63,86));
errorFound=true;
fileHeaderErrorFound=true;
bw.write("ERROR: File header immediate origin name is invalid: "+strLine.substring(63,86));
bw.newLine();
I believe the reason that the error is called at runtime is because the text contains a period and comma.. I am unsure how to allow these in the regex.
I have tried using this
if(strLine.substring(63,86).matches("[A-Z,a-z,0-9,,,. ]+")){
and it seemed to work I just wanted to make sure that is the correct way because it doesn't look right.
You're right in your analysis, the match failed because there was a dot in the text that isn't contained in the character class.
However, you can simplify the regex - no need to repeat the commas, they don't have any special meaning inside a class:
if(strLine.substring(63,86).matches("[A-Za-z0-9,. ]+"))
Are you sure that you'll never have to match non-ASCII letters or any other kind of punctuation, though?
Alphabets and digits : a-zA-Z0-9 can effectively be replaced by \w denoting 'words'.
The period and comma don't need escaping and can be used as is. Hence this regex might come in handy:
"[\w,.]"
Hope this helps. :)
Related
I do have a Java Web Application, where I get some inputs from the user. Once I got this input I have to parse it and the parsing part depends on what kind of input I'll get. I decided to use the Pattern class of java for some of predefined user inputs.
So I need the last 2 regex patterns:
a)Enumaration:
input can be - A03,B24.1,A25.7
The simple way would be to check if there are a comma in there ([^,]+) but it will end up with a lot of updates in to parsing function, which I would like to avoid. So, in addition to comma it should check if it starts with
letter
minimum 3 letters (combined with numbers)
can have one dot in the word
minimum 1 comma (updated it)
b) Mixed
input can be A03,B24.1-B35.5,A25.7
So all of what Enumuration part got, but with addition that it can have a dash minimum one.
I've tried to use multiple online regex generators but didnt get it correct. Would be much appreciated if you can help.
Here is what I got if its B24.1-B35.5 if its just a simple range.
"='.{1}\\d{0,2}-.{1}\\d{0,2}'|='.{1}\\d{1,2}.\\d{1,2}-.{1}\\d{1,2}.\\d{1,2}'";
Edit1: Valid and Invalid inputs
for a)Enumaration
A03,B24.1,A25.7 Valid
A03,B24.1 Valid
A03,B24.1-B25.1 -Invalid because in this case (enumaration) it should not contain dash
A03 invalid because no comma
A03,B24.1 - Valid
A03 Invalid
for b)Mixed
everything that a enumeration has with addition that it can have dash too.
You can use this regex for (a) Enumeration part as per your rules:
[A-Za-z][A-Za-z0-9]{2,}(?:\.[A-Za-z0-9]{1,})?(?:,[A-Za-z][A-Za-z0-9]{2,}(?:\.[A-Za-z0-9]{1,})?)+
Rules:
Verifies that each segment starts with a letter
Minimum of three letters or numbers [A-Za-z][A-Za-z0-9]{2,}
Optionally followed by decimal . and one or more alphabets and numbers i.e (?:\.[A-Za-z0-9]{1,})?
Same thing repeated, and seperated by a comma ,. Also must have atleast one comma so using + i.e (?:,[A-Za-z][A-Za-z0-9]{2,}(?:\.[A-Za-z0-9]{1,})?)+
?: to indicate non-capturing group
Using [A-Za-z0-9] instead of \w to avoid underscores
Regex101 Demo
For (b) Mixed, you haven't shared too many valid and invalid cases, but based on my current understanding here's what I have:
[A-Za-z][A-Za-z0-9]{2,}(?:\.[A-Za-z0-9]{1,})?(?:[,-][A-Za-z][A-Za-z0-9]{2,}(?:\.[A-Za-z0-9]{1,})?)+
Note that , from previous regex has been replaced with [,-] to allow - as well!
Regex101 Demo
// Will match
A03,B24.1-B35.5,A25.7
A03,B24.1,A25.7
A03,B24.1-B25.1
Hope this helps!
EDIT: Making sure each group starts with a letter (and not a number)
Thanks to #diginoise and #anubhava for pointing out! Changed [A-Za-z0-9]{3,} to [A-Za-z][A-Za-z0-9]{2,}
As I said in the comments, I would chop the input by commas and verify each segment separately. Your domain ICD 10 CM codes is very well defined and also I would be very wary of any input which could be non valid, yet pass the validation.
Here is my solution:
regex
([A-TV-Z][0-9][A-Z0-9](\.?[A-Z0-9]{0,4})?)
... however I would avoid that.
Since your domain is (moste likely) medical software, people's lives (or at least well being) is at stake. Not to mention astronomical damages and the lawyers ever-chasing ambulances. Therefore avoid the easy solution, and implement the bomb proof one.
You could use the regex to establish that given code is definitely not valid. However if a code passes your regex it does not mean that it is valid.
bomb proof method
See this example: O09.7, O09.70, O09.71, O09.72, O09.73 are valid entries, but O09.1 is not valid.
Therefore just get all possible codes. According to this gist there are 42784 different codes. Just load them to memory and any code which is not in the set, is not valid. You could compress said list and be clever about the encoding in memory, to occupy less space, but verbatim all codes are under 300kB on disk, so few MBs max in memory, therefore not a massive cost to pay for a price of people not having left instead of right kidney removed.
I am developing a java app, running on android. I am trying to pick all words which do not contain any embedded digits or symbols.
The best I have come up with is:
\b[a-zA-Z]+[a-zA-Z]*+\b
Test Data:
this is a test , an0ther gr8 WW##ee one, w1n 1test test1 end
This results in picking the following: this, is, a, test, WW##ee, one, end
I need to eliminate the WW##ee from the results.
You shouldn't use a word boundary meta-character \b since it matches the position right after WW which sees a hash # character. This position is a word boundary itself. So you should pick up a different way:
(?<![\S&&[^,]])[a-zA-Z]+(?![\S&&[^,]])
Using character class intersection feature of Java's regex you are able to define punctuation characters that are allowed to follow or precede a word character. Here it is a comma ,.
You could use look behind and look ahead to check there is no #.
\b(?<!\#)[a-zA-Z]+(?!\#)\b
My solution has evolved a bit as I have gotten additional help with this. So, this is now my best solution but still a bit lacking. I have not been able to accept "as-is" while rejecting "-this-" and a similar case of accept "and/or" while rejecting "/slash/". Also for simplicity I have made the input data single word per line.
^(?:[\p{P}\p{S}])?((?:[\p{L}\p{Pd}'])+)(?:[\p{P}\p{S}])$
as-is is picked valid
-this- is valid but I wish it weren't
and/or is not valid but I wish it would be picked
/slash/ "slash" is picked valid
(test) "test" is picked valid
[test] "test" is picked valid
<test> "test" is picked valid
I am currently trying to add a few errormessages to my application. For that I am using JOptionPane.showMessageDialog(...);
Basically everything is working as I'd expect it to. But one thing is a bit of a pain.
I am using e.getMessage() to receive the description of the occuring error.
In the case of a sql connection error this is such a long message, that it can't possibly fit to the screen. So I thouht to split it after every sentence, using split([\\.]).
This is working as well, BUT: the message includes a part like this
Error: "java.net.SocketTimeoutException: Receive timed out"., which, of course ends up in:
Error: "java
net
SocketTimeoutException: Receive timed out"
How could I avoid this behaviour? Or is there possibly a better way to achieve the result of a splitted error message?
Why not just split on every space that has dot before it?
Try maybe split("(?<=[.])\\s+")
(?<=[.]) is positive-look-behind. It is used to make sure that group of spaces \\s+ have dot before it, but will not include this dot in match, so it will stay untouched after split, while white-spaces will be removed.
Not sure until your input and expected result are posted in full, but you could use "lookarounds" for that purpose.
For instance:
String input = "Error: \"java.net.SocketTimeoutException: Receive timed out\".";
System.out.println(Arrays.toString(input.split("(?<!\\w)\\.(?!\\w)")));
Output
[Error: "java.net.SocketTimeoutException: Receive timed out"]
Explanation
It splits the String based on (escaped) dot Patterns neither preceded nor followed by any word character
It prints the split Array (here, only 1 element since the package-delimiting dots do not match the Pattern as expected)
An alternative to use regexes is using WordUtils.wrap from the apache.commons.lang package. Using a regex has the advantage of not using an additional library, but makes the code a wee bit more unreadable. In your case not really a big issue, but as an added benefit, commons.lang contains a good deal of useful stuff which might come in handy in your project.
It is one of the libraries which is pretty much a constant in my tool-belt.
I cannot get a regexp that checks if password has at least one digit to work.
This has been answered everywhere but all the answers stop working if split up.
For example in this Working Password Validation if I remove:
(?=.*[a-z])(?=.*[A-Z])(?=.*[##$%^&+=])
from
^(?=.*[0-9])(?=.*[a-z])(?=.*[A-Z])(?=.*[##$%^&+=])(?=\S+$).{8,}$
in order to check for the presence of a single digit, the whole thing stops working
I'm new with regular expressions, this seems to make sense but it doesn't, show me the light if you can.
I'm not really sure what you mean by the whole things stops working.
(?=.*[a-z])(?=.*[A-Z])(?=.*[##$%^&+=])
All the above does is mandate that:
A lower case letter must appear at least once
An upper-case letter must appear at least once
Any of ##$%^&+= must appear at least once
So, there is no reason that taking them out should break anything--they are essentially independent components.
There are myriad ways to check if a String contains a number. How you want to check really depends on your specific requirements. The method used in the presented regex does this through a positive-look ahead: ^(?=.*[0-9])
^ : begins with
.* : matches 0 or more non-newline characters
[0-9]: is a character class that matches the numbers [0,1,2,...,9]
?= is the positive look ahead, which in this case says to match iff there exists at least one number
Hope that helped. You can start off with Oracle's Tutorial on Regular Expressions. After you digest that, I'm sure you'll be able to find more advanced resources via Google.
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Using a regular expression to validate an email address
This is homework, I've been working on it for a while, I've done lots of reading and feel I have gotten pretty familiar with regex for a beginner.
I am trying to find a regular expression for validating/invalidating a list of emails. There are two addresses which are giving me problems, I can't get them both to validate the correct way at the same time. I've gone through a dozen different expressions that work for all the other emails on the list but I can't get those two at the same time.
First, the addresses.
me#example..com - invalid
someone.nothere#1.0.0.127 - valid
The part of my expression which validates the suffix
I originally started with
#.+\\.[[a-z]0-9]+
And had a second pattern for checking some more invalid addresses and checked the email against both patterns, one checked for validity the other invalidity but my professor said he wanted it all in on expression.
#[[\\w]+\\.[\\w]+]+
or
#[\\w]+\\.[\\w]+
I've tried it written many, many different ways but I'm pretty sure I was just using different syntax to express these two expressions.
I know what I want it to do, I want it to match a character class of "character+"."character+"+
The plus sign being at least one. It works for the invalid class when I only allow the character class to repeat one time(and obviously the ip doesn't get matched), but when I allow the character class to repeat itself it matches the second period even thought it isn't preceded by a character. I don't understand why.
I've even tried grouping everything with () and putting {1} after the escaped . and changing the \w to a-z and replacing + with {1,}; nothing seems to require the period to surrounded by characters.
You need a negative look-ahead :
#\w+\.(?!\.)
See http://www.regular-expressions.info/lookaround.html
test in Perl :
Perl> $_ = 'someone.nothere#1.0.0.127'
someone.nothere#1.0.0.127
Perl> print "OK\n" if /\#\w+\.(?!\.)/
OK
1
Perl> $_ = 'me#example..com'
me#example..com
Perl> print "OK\n" if /\#\w+\.(?!\.)/
Perl>
#([\\w]+\\.)+[\\w]+
Matches at least one word character, followed by a '.'. This is repeated at least once, and is then followed by at least on more word character.
I think you want this:
#[\\w]+(\\.[\\w]+)+
This matches a "word" followed by one or more "." "word" sequences. (You can also do the grouping the other way around; e.g. see Dailin's answer.)
The problem with what you are doing before was that you were trying to embed a repeat inside a character class. That doesn't make sense, and there is no syntax that would support it. A character class defines a set of characters and matches against one character. Nothing more.
The official standard RFC 2822 describes the syntax that valid email addresses with this regular expression:
(?:[a-z0-9!#$%&'*+/=?^_`{|}~-]+(?:\.[a-z0-9!#$%&'*+/=?^_`{|}~-]+)*|"(?:[\x01-\x08\x0b\x0c\x0e-\x1f\x21\x23-\x5b\x5d-\x7f]|\\[\x01-\x09\x0b\x0c\x0e-\x7f])*")#(?:(?:[a-z0-9](?:[a-z0-9-]*[a-z0-9])?\.)+[a-z0-9](?:[a-z0-9-]*[a-z0-9])?|\[(?:(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.){3}(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?|[a-z0-9-]*[a-z0-9]:(?:[\x01-\x08\x0b\x0c\x0e-\x1f\x21-\x5a\x53-\x7f]|\\[\x01-\x09\x0b\x0c\x0e-\x7f])+)\])
More practical implementation of RFC 2822 (if we omit the syntax using double quotes and square brackets), which will still match 99.99% of all email addresses in actual use today, is:
[a-z0-9!#$%&'*+/=?^_`{|}~-]+(?:\.[a-z0-9!#$%&'*+/=?^_`{|}~-]+)*#(?:[a-z0-9](?:[a-z0-9-]*[a-z0-9])?\.)+[a-z0-9](?:[a-z0-9-]*[a-z0-9])?