To validate String in Java I can use String.matches(). I would like to validate a simple string "*.txt" where "*" means anything. Input e.g. test.txt is correct, but test.tt is not correct, because of ".tt". I tried to use matches("[*].txt"), but it doesn't work. How can I improve this matches? Thanks.
Do not use code, you don't understand!
For your simple problem you could totally avoid using a regular expression and just use
yourString.endsWith(".txt")
and if you want to perform this comparison case insensitive (i.e. allow ".TXT" or ".tXt") use
yourString.toLowerCase().endsWith(".txt")
If you want to learn more about regular expressions in java, I'd recomment a tutorial. For example this one.
You may try this for txt files:
"file.txt".matches("^.*[.]txt$")
Basically ^ means the start of your string. .* means match anything greedy, hence as much as you can get to make the expression match. And [.] means match the dot character. The suffix txt is just the txt text itself. And finally $ is the anchor for the end of the string, which ensures that the string does not contain anything more.
Use .+, it means any character having one or unlimited lengths. It will ensure to avoid the inputs like only .txt
matches(".+[.]txt")
FYI: [.] simply matches with the dot character.
Related
I have a YAML file with multiple lines and I know there's one line that looks like this:
...
schemas: core,ext,plugin
...
Note that there is unknown number of whitespaces at the beginning of this line (because YAML). The line can be identified uniquely by the schemas: expression. The number of existing values for the schemas property is unknown, but greater than zero. And I do not know what these values are, except that one of them might be foo.
I would like to use a regex match-and-replace to append the word ,foo to this line if foo is not already contained in the list of values at any position. foo might appear on any other line but I want to ignore these instances. I don't want the other lines to be modified.
I've tried different regular expressions with lookarounds and capture groups, but none did the job. My latest attempt that looked promising at first was:
(?s)(?!.*foo)(.*schemas:.*)
But this does not match if foo is contained on any other line, which is not what I want.
Any assistance would be very much appreciated. Thanks.
(I use the Java regex engine, btw.)
Would this work?
^(?!.*foo)(\s*schemas:.*)$
If you want to make sure stuff like
food, fool, etc.
matches you can use this:
^(?!.*(?:foo\s*$|foo,))(\s*schemas:.*)$
Replacement:
$1,foo
If I understood your question correctly, you want to make sure only one line is checked for the negative lookahead. This should accomplish that. I tested it on https://regex101.com/ using the Java 8 engine. You can also check what each operator does there.
Explanation:
wrapping the expression with
^$
makes sure that only one line is considered at a time.
The negative lookahead
(?!.*(?:foo\s*$|foo,))
looks for any "foo" followed by either (whitespaces and a newline) or a comma within this line. If you want to make the expression faster you could probably turn the lookahead into a lookbehind, so that the simpler check for "schemas:" comes first. However, I don't know if this actually improves performance.
^(\s*schemas:.*)(?<!(?:foo\s?$|foo,))$
With lookbehinds you can't use the * quantifier, so the regex would match if foo is followed by more than one whitespace.
I am trying to write a regular expression to verify the presence of a specific number in a fixed position in a String.
String: 109300300330066611111111100000000017000656052086116020170111Name 1
Number to find: 111111111 (Staring from position 17)
I have written the following regular expression:
^.{16}(?<Ones>111111111)(.*)
My understanding is:
Let first 16 characters be whatever they are
Use the Named Capturing Group to grab the specific word
Let the rest of the characters be whatever they are
I am new to regex, is there any issue with the above approach?
Can it be done in other/better way?
I am using Java 8.
Without more details of why you're doing what you're doing, there's just one possible improvement I can see. You repeated any character 16 times at the beginning of the string rather than writing out 16 .s, which is nice and readable, but then, it would be nice to do the same for the repeated 1s:
^.{16}(?<Ones>1{9})(.*)
Otherwise, the string of 1s is hard to understand without the coder manually counting how many there are in the regex.
If you want to hard-code the ones and you know the starting position and you just wnat to know if it is there, using a regex seems unnecessary. you can use this:
String s = "109300300330066611111111100000000017000656052086116020170111Name 1";
if (s.indexOf("111111111").equals(16) doSomething();
Another possible solution without regex:
if(s.substring(16,25).equals("111111111") doSomething();
Otherwise your regex looks good.
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])?
Good morning. I realize there are a ton of questions out there regarding replace and replaceAll() but i havnt seen this.
What im looking to do is parse a string (which contains valid html to a point) then after I see the second instance of <p> in the string i want to remove everything that starts with & and ends with ; until i see the next </p>
To do the second part I was hoping to use something along the lines of s.replaceAll("&*;","")
That doesnt work but hopefully it gets my point across that I am looking to replace anything that starts with & and ends with ;
You should probably leave the parsing to a DOM parser (see this question). I can almost guarantee you'll have to do this to find text within the <p> tags.
For the replacement logic, String.replaceAll uses regular expressions, which can do the matching you want.
The "wildcard" in regular expressions that you want is the .* expression. Using your example:
String ampStr = "This &escape;String";
String removed = ampStr.replaceAll("&.*;", "");
System.out.println(removed);
This outputs This String. This is because the . represents any character, and the * means "this character 0 or more times." So .* basically means "any number of characters." However, feeding it:
"This &escape;String &anotherescape;Extended"
will probably not do what you want, and it will output This Extended. To fix this, you specify exactly what you want to look for instead of the . character. This is done using [^;], which means "any character that's not a semicolon:
String removed = ampStr.replaceAll("&[^;]*;", "");
This has performance benefits over &.*?; for non-matching strings, so I highly recommend using this version, especially since not all HTML files will contain a &abc; token and the &.*?; version can have huge performance bottle-necks as a result.
The expression you want is:
s.replaceAll("&.*?;","");
But do you really want to be parsing HTML this way? You may be better off using an XML parser.
I need a simple way to implement the contains function using matches. I believe this is my starting point:
xxx.matches("'.*yyy.*'");
But I need to make it a universal method and pre-process whatever I search for to be accepted by matches! This must be done using only the escape '\' character!
Imagine a string SEARCH_FOR that can contain some special characters that must be "regex escaped"...
String SEARCH_FOR="*.\\"
xxx.matches("'.*" + SEARCH_FOR + ".*'");
Are there any catches? Special situations? Any other "special chars should be taken into account?
Are you looking for Pattern.quote(String) ?
This escapes special characters for you.
EDIT:
After reading the comments, I really hope you try Pattern.quote(yourString.toLowerCase()) as it sounds like you've been using Pattern.quote(yourString).toLowerCase(). If DataNucleus is applying the regex then there should be no problems with using the \Q and \E escape sequence.
Since you have really asked for it, ".\\".replaceAll("(\\.|\\$|\\+|\\*|\\\\)", "\\\\\$1") outputs \.\\
This will escape .'s, $'s, + 's, *'s and \'s. Note that the security of this is now all upon you. If you don't escape something you needed to, or you escape it incorrectly, you will either allow people to use regex inside the search term when you weren't expecting to or it won't returns results that you were expecting.