I am trying to formulate a regex for the following scenario :
The String to match : mName87.com
So, the string may consist of any number of alpha numeric characters , but can contain only a single dot anywhere in the string .
I formulated this regex : [a-zA-Z0-9.], but it matches even multiple dots(.)
What am i doing wrong here ?
The regex you provided matches only a single character in the whole string you're trying to validate. There are a few things to take care of in your scenario
You want to match over the whole string, so your regex must start with ^ (beginning of the string) and end with $ (end of the string).
Then you want to accept any number of alpha-numeric characters, this is done with [a-zA-Z0-9]+, here the + means one or more characters.
Then match the point: \. (you must escape it here)
Finally accept more characters again.
All together the regex would then be:
^[a-zA-Z0-9]+\.[a-zA-Z0-9]+$
You can use this regex:
\\w*\\.\\w*
You can try here
Try with:
^([a-zA-Z0-9]+\.)+[a-zA-Z]$
use this regular expression ^[a-zA-Z0-9]*\.[a-zA-Z0-9.]*$
EDITED:
Try
([a-zA-Z0-9]+\.[a-zA-Z0-9]+)|(\.[a-zA-Z0-9]+)|([a-zA-Z0-9]+\.)
That is: [a word that ends with a dot] OR [two words and the dot in the middle] OR [a word that starts with a dot]
Related
Need to allow user to enter only Numbers or alphabets or spaces or hyphens OR combination of any of the above.
and i tried the following
String regex = "/^[0-9A-Za-z\s\-]+$/";
sampleString.matches(regex);
but it is not working properly. would somebody help me to fix please.
Issue : your regex is trying to match / symbol at the beginning and at the end
In java there is no need of / before and after regex so use, java!=javascript
"^[0-9A-Za-z\\s-]+$"
^[0-9A-Za-z\\s-]+$ : ^ beginning of match
[0-9A-Za-z\\s-]+ : one or more alphabets, numbers , spaces and -
$ : end of match
You are close but need to make two changes.
The first is to double-escape (i.e., use \\ instead of \). This is due to the weirdness of Java (see the section "Backslashes, escapes, and quoting" in Javadoc for the Pattern class). The second thing is to drop the explicit reference to the start and end of the string. That's going to be implied when using matches(). So the correct Java code is
String regex = "[0-9A-Za-z\\s\\-]+";
sampleString.matches(regex);
While that will work, you can also replace the "0-9" reference with \d and drop the escaping of the "-". That gives you
String regex = "[\\dA-Za-z\\s-]+";
For example I have text like below :
case1:
(1) Hello, how are you?
case2:
Hi. (1) How're you doing?
Now I want to match the text which starts with (\d+).
I have tried the following regex but nothing is working.
^[\(\d+\)], ^\(\d+\).
[] are used to match any of the things you specify inside the brackets, and are to be followed by a quantifier.
The second regexp will work: ^\(\d+\), so check your code.
Check also so there's no space in front of the first parenthesis, or add \s* in front.
EDIT: Also, java can be tricky with escapes depending on if the regexp you type is directly translated to a regexp or is first a string literal. You may need to double escape your escapes.
In Java you have to escape parenthesis, so "\\(\\d+\\)" should match (1) in case one and two. Adding ^ as you did "^\\(\\d+\\)" will match only case1.
You have to use double back slashes within java string. Consider this
"\n" give you [line break]
"\\n" give you [backslash][n]
If you are going to downvote my post, at least comment to tell me WHY it's not useful.
I believe Java's Regex Engine supports Positive Lookbehind, in which case you can use the following regex:
(?<=[(][0-9]{1,9999}[)]\s?)\b.*$
Which matches:
The literal text (
Any digit [0-9], between 1 and 9999 times {1,9999}
The literal text )
A space, between 0 and 1 times \s?
A word boundary \b
Any character, between 0 and unlimited times .*
The end of a string $
I'm trying to compare following strings with regex:
#[xyz="1","2"'"4"] ------- valid
#[xyz] ------------- valid
#[xyz="a5","4r"'"8dsa"] -- valid
#[xyz="asd"] -- invalid
#[xyz"asd"] --- invalid
#[xyz="8s"'"4"] - invalid
The valid pattern should be:
#[xyz then = sign then some chars then , then some chars then ' then some chars and finally ]. This means if there is characters after xyz then they must be in format ="XXX","XXX"'"XXX".
Or only #[xyz]. No character after xyz.
I have tried following regex, but it did not worked:
String regex = "#[xyz=\"[a-zA-z][0-9]\",\"[a-zA-z][0-9]\"'\"[a-zA-z][0-9]\"]";
Here the quotations (in part after xyz) are optional and number of characters between quotes are also not fixed and there could also be some characters before and after this pattern like asdadad #[xyz] adadad.
You can use the regex:
#\[xyz(?:="[a-zA-z0-9]+","[a-zA-z0-9]+"'"[a-zA-z0-9]+")?\]
See it
Expressed as Java string it'll be:
String regex = "#\\[xyz=\"[a-zA-z0-9]+\",\"[a-zA-z0-9]+\"'\"[a-zA-z0-9]+\"\\]";
What was wrong with your regex?
[...] defines a character class. When you want to match literal [ and ] you need to escape it by preceding with a \.
[a-zA-z][0-9] match a single letter followed by a single digit. But you want one or more alphanumeric characters. So you need [a-zA-Z0-9]+
Use this:
String regex = "#\\[xyz(=\"[a-zA-z0-9]+\",\"[a-zA-z0-9]+\"'\"[a-zA-z0-9]+\")?\\]";
When you write [a-zA-z][0-9] it expects a letter character and a digit after it. And you also have to escape first and last square braces because square braces have special meaning in regexes.
Explanation:
[a-zA-z0-9]+ means alphanumeric character (but not an underline) one or more times.
(=\"[a-zA-z0-9]+\",\"[a-zA-z0-9]+\"'\"[a-zA-z0-9]+\")? means that expression in parentheses can be one time or not at all.
Since square brackets have a special meaning in regex, you used it by yourself, they define character classes, you need to escape them if you want to match them literally.
String regex = "#\\[xyz=\"[a-zA-z][0-9]\",\"[a-zA-z][0-9]\"'\"[a-zA-z][0-9]\"\\]";
The next problem is with '"[a-zA-z][0-9]' you define "first a letter, second a digit", you need to join those classes and add a quantifier:
String regex = "#\\[xyz=\"[a-zA-z0-9]+\",\"[a-zA-z0-9]+\"'\"[a-zA-z0-9]+\"\\]";
See it here on Regexr
there could also be some characters before and after this pattern like
asdadad #[xyz] adadad.
Regex should be:
String regex = "(.)*#\\[xyz(=\"[a-zA-z0-9]+\",\"[a-zA-z0-9]+\"'\"[a-zA-z0-9]+\")?\\](.)*";
The First and last (.)* will allow any string before the pattern as you have mentioned in your edit. As said by #ademiban this (=\"[a-zA-z0-9]+\",\"[a-zA-z0-9]+\"'\"[a-zA-z0-9]+\")? will come one time or not at all. Other mistakes are also very well explained by Others +1 to all other.
I am running into this problem in Java.
I have data strings that contain entities enclosed between & and ; For e.g.
&Text.ABC;, &Links.InsertSomething;
These entities can be anything from the ini file we have.
I need to find these string in the input string and remove them. There can be none, one or more occurrences of these entities in the input string.
I am trying to use regex to pattern match and failing.
Can anyone suggest the regex for this problem?
Thanks!
Here is the regex:
"&[A-Za-z]+(\\.[A-Za-z]+)*;"
It starts by matching the character &, followed by one or more letters (both uppercase and lower case) ([A-Za-z]+). Then it matches a dot followed by one or more letters (\\.[A-Za-z]+). There can be any number of this, including zero. Finally, it matches the ; character.
You can use this regex in java like this:
Pattern p = Pattern.compile("&[A-Za-z]+(\\.[A-Za-z]+)*;"); // java.util.regex.Pattern
String subject = "foo &Bar; baz\n";
String result = p.matcher(subject).replaceAll("");
Or just
"foo &Bar; baz\n".replaceAll("&[A-Za-z]+(\\.[A-Za-z]+)*;", "");
If you want to remove whitespaces after the matched tokens, you can use this re:
"&[A-Za-z]+(\\.[A-Za-z]+)*;\\s*" // the "\\s*" matches any number of whitespace
And there is a nice online regular expression tester which uses the java regexp library.
http://www.regexplanet.com/simple/index.html
You can try:
input=input.replaceAll("&[^.]+\\.[^;]+;(,\\s*&[^.]+\\.[^;]+;)*","");
See it
I'm no expert in regex but I need to parse some input I have no control over, and make sure I filter away any strings that don't have A-z and/or 0-9.
When I run this,
Pattern p = Pattern.compile("^[a-zA-Z0-9]*$"); //fixed typo
if(!p.matcher(gottenData).matches())
System.out.println(someData); //someData contains gottenData
certain spaces + an unknown symbol somehow slip through the filter (gottenData is the red rectangle):
In case you're wondering, it DOES also display Text, it's not all like that.
For now, I don't mind the [?] as long as it also contains some string along with it.
Please help.
[EDIT] as far as I can tell from the (very large) input, the [?]'s are either white spaces either nothing at all; maybe there's some sort of encoding issue, also perhaps something to do with #text nodes (input is xml)
The * quantifier matches "zero or more", which means it will match a string that does not contain any of the characters in your class. Try the + quantifier, which means "One or more": ^[a-zA-Z0-9]+$ will match strings made up of alphanumeric characters only. ^.*[a-zA-Z0-9]+.*$ will match any string containing one or more alphanumeric characters, although the leading .* will make it much slower. If you use Matcher.lookingAt() instead of Matcher.matches, it will not require a full string match and you can use the regex [a-zA-Z0-9]+.
You have an error in your regex: instead of [a-zA-z0-9]* it should be [a-zA-Z0-9]*.
You don't need ^ and $ around the regex.
Matcher.matches() always matches the complete string.
String gottenData = "a ";
Pattern p = Pattern.compile("[a-zA-z0-9]*");
if (!p.matcher(gottenData).matches())
System.out.println("doesn't match.");
this prints "doesn't match."
The correct answer is a combination of the above answers. First I imagine your intended character match is [a-zA-Z0-9]. Note that A-z isn't as bad as you might think it include all characters in the ASCII range between A and z, which is the letters plus a few extra (specifically [,\,],^,_,`).
A second potential problem as Martin mentioned is you may need to put in the start and end qualifiers, if you want the string to only consists of letters and numbers.
Finally you use the * operator which means 0 or more, therefore you can match 0 characters and matches will return true, so effectively your pattern will match any input. What you need is the + quantifier. So I will submit the pattern you are most likely looking for is:
^[a-zA-Z0-9]+$
You have to change the regexp to "^[a-zA-Z0-9]*$" to ensure that you are matching the entire string
Looks like it should be "a-zA-Z0-9", not "a-zA-z0-9", try correcting that...
Did anyone consider adding space to the regex [a-zA-Z0-9 ]*. this should match any normal text with chars, number and spaces. If you want quotes and other special chars add them to the regex too.
You can quickly test your regex at http://www.regexplanet.com/simple/
You can check input value is contained string and numbers? by using regex ^[a-zA-Z0-9]*$
if your value just contained numberString than its show match i.e, riz99, riz99z
else it will show not match i.e, 99z., riz99.z, riz99.9
Example code:
if(e.target.value.match('^[a-zA-Z0-9]*$')){
console.log('match')
}
else{
console.log('not match')
}
}
online working example