I need to match a string to have words and not numbers, I need to match special characters such as $!:{}_ if they are part of the word but ignore otherwise.
I have a regex that matches for word and ignores numbers but cannot work out how to match special characters if they are part of the word but ignore otherwise.
Here is what I have correctly - /^d\s+/
Any help would be appreciated.
Allow words and letters only, and ignore any numbers. Special
characters such as (-_‘[]{}“£$&%!:;/) should either be ignored or
treated as part of the word they sit within.
Try using String.prototype.replace() with RegExp /\s\d+.*\d|\s+[^a-z]/ig to replace space character followed by digit followed by any character followed by digit , or space character followed by any character not a-z case insensitive
var str = "This is a test of 1,2,3 word-count - test.";
str = str.replace(/\s\d+.*\d|\s+[^a-z]/ig, "");
document.body.textContent = str;
Related
lets say i have this string that contain Non-Space Characters:
thisHُ is a long string I cant display
as you can see this "Hُ " contain Non-Space Characters now i want to short the string so i did this:
var text = "thisHُ is a long string I cant display"
text =text.replace(/^((.){0,9})(.*)/gm, "$1");
console.log(text);
this will give:
thisHُ is
but i don't want to count the Non-Space Characters i want to shorten the string with ignoring "counting" Non-Space Characters also i want to shorten string without cutting the words boundary.
You may use
s.match(/^(?:\s*\S){1,9}\S*/)[0]
s.match(/\S*(?:\S\s*){1,9}$/)[0]
See the regex demo #1 and regex demo #2.
The ^(?:\s*\S){1,9}\S* regex matches one to nine occurrences of 0+ whitespaces followed with a single non-whitespace char at the start of string and then any 0+ non-whitespace chars.
The \S*(?:\S\s*){1,9}$ regex will match 0+ non-whitespace chars and then one to nine occurrences of a single non-whitespace char followed with 0+ whitespaces at the end of string.
JavaScript demo:
const text = "thisHُ is a long string I cant display";
const startMatch = text.match(/^(?:\s*\S){1,9}\S*/);
const endMatch = text.match(/\S*(?:\S\s*){1,9}$/);
if (startMatch) console.log(`Match at start: ${startMatch[0]}`);
if (endMatch) console.log(`Match at end: ${endMatch[0]}`);
I use the function below to linkify usernames that:
start with a letter a-z or number 0-9
contains letters a-z, numbers 0-9 and "-"
have a length of 2-50 characters.
.
function linkifyUsernames(text){
return text.replace(/#\b([0-9a-z-]{2,49})\b/ig,
"<a href='/profile/$1' target='_blank'>#$1</a>");
}
The function above works OK but the only problem is that it breaks words. For example,
#abcdéíú
The function linkifies the first part of the word
#abcdéíú
but I need a function that does not convert to links any words that start with # but contains other characters than a-z0-9-. So, the word #abcdéíú must stay untauched.
Word bondaries for some reason don't help.
This doesn't fall between the [0-9a-z] range, try this range [a-z\u00E0-\u00FC], shown in the following example:
https://regex101.com/r/vW2mR9/1
function linkifyUsernames(text) {
var pattern = /^(#)(\w{2,12})/ig;
if(text.match(pattern).length>0 && text.match(pattern)[0]=== text)
return text.replace(pattern, "<a href='/profile/$2' target='_blank'>#$2</a>");
else return text;
}
$('div').append(linkifyUsernames('#abcd'))
$('div').append('<br/>')
$('div').append(linkifyUsernames('#abcdéáú'))
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.0/jquery.min.js"></script>
<div>
</div>
The a-z will not cover the special characters like à,é.
You can simplify using \w notation like:
/^(#)(\w{2,12})/ig
Here example:
https://regex101.com/r/nZ7uI7/1
As stated on: http://www.regular-expressions.info/wordboundaries.html
There are three different positions that qualify as word boundaries:
Before the first character in the string, if the first character is a word character.
After the last character in the string, if the last character is a word character.
Between two characters in the string, where one is a word character and the other is not a word character.
If you try your sample string
#abcdéíú
with just a \w which is meant to match any word character you'll see that déíúare not word characters in regex word. So even if you use this [a-z\u00E0-\u00FC] from the prev answer the regex fail due to \b.
I need is the last match. In the case below the word test without the $ signs or any other special character:
Test String:
$this$ $is$ $a$ $test$
Regex:
\b(\w+)\b
The $ represents the end of the string, so...
\b(\w+)$
However, your test string seems to have dollar sign delimiters, so if those are always there, then you can use that instead of \b.
\$(\w+)\$$
var s = "$this$ $is$ $a$ $test$";
document.body.textContent = /\$(\w+)\$$/.exec(s)[1];
If there could be trailing spaces, then add \s* before the end.
\$(\w+)\$\s*$
And finally, if there could be other non-word stuff at the end, then use \W* instead.
\b(\w+)\W*$
In some cases a word may be proceeded by non-word characters, for example, take the following sentence:
Marvelous Marvin Hagler was a very talented boxer!
If we want to match the word boxer all previous answers will not suffice due the fact we have an exclamation mark character proceeding the word. In order for us to ensure a successful capture the following expression will suffice and in addition take into account extraneous whitespace, newlines and any non-word character.
[a-zA-Z]+?(?=\s*?[^\w]*?$)
https://regex101.com/r/D3bRHW/1
We are informing upon the following:
We are looking for letters only, either uppercase or lowercase.
We will expand only as necessary.
We leverage a positive lookahead.
We exclude any word boundary.
We expand that exclusion,
We assert end of line.
The benefit here are that we do not need to assert any flags or word boundaries, it will take into account non-word characters and we do not need to reach for negate.
var input = "$this$ $is$ $a$ $test$";
If you use var result = input.match("\b(\w+)\b") an array of all the matches will be returned next you can get it by using pop() on the result or by doing: result[result.length]
Your regex will find a word, and since regexes operate left to right it will find the first word.
A \w+ matches as many consecutive alphanumeric character as it can, but it must match at least 1.
A \b matches an alphanumeric character next to a non-alphanumeric character. In your case this matches the '$' characters.
What you need is to anchor your regex to the end of the input which is denoted in a regex by the $ character.
To support an input that may have more than just a '$' character at the end of the line, spaces or a period for instance, you can use \W+ which matches as many non-alphanumeric characters as it can:
\$(\w+)\W+$
Avoid regex - use .split and .pop the result. Use .replace to remove the special characters:
var match = str.split(' ').pop().replace(/[^\w\s]/gi, '');
DEMO
I have a password field in one form. Now I have to validate in such a way that the field value should be a 7 digits string along with a number. Otherwise it will return false.
Please help me.
Create regex first
Var regex = /\w{7}\d/i;
var yourvalue=$("#passwordid").value;
regex.test(yourvalue){
return true;
}
else{
return false
}
I’m sure there is a better way, but something like:
if ( /.{7}/.test(str) && /\d/.test(str) ) {
//OK
}
In your javascript you can use the RegExp object.
var regEx = new RegExp(pattern, modifiers);
or more simply:
var pattern = /pattern/modifiers;
E.g.
var password = "abcdefg1";
var pattern = /\w{7}\d/i;
var isMatch = pattern.test(password);
Here are some expressions:
[abc] Find any character between the brackets
[^abc] Find any character not between the brackets
[0-9] Find any digit from 0 to 9
[A-Z] Find any character from uppercase A to uppercase Z
[a-z] Find any character from lowercase a to lowercase z
[A-z] Find any character from uppercase A to lowercase z
[adgk] Find any character in the given set
[^adgk] Find any character outside the given set
(red|blue|green) Find any of the alternatives specified
Metacharacters:
. Find a single character, except newline or line terminator
\w Find a word character
\W Find a non-word character
\d Find a digit
\D Find a non-digit character
\s Find a whitespace character
\S Find a non-whitespace character
\b Find a match at the beginning/end of a word
\B Find a match not at the beginning/end of a word
\0 Find a NUL character
\n Find a new line character
\f Find a form feed character
\r Find a carriage return character
\t Find a tab character
\v Find a vertical tab character
\xxx Find the character specified by an octal number xxx
\xdd Find the character specified by a hexadecimal number dd
\uxxxx Find the Unicode character specified by a hexadecimal number xxxx
Quantifiers
n+ Matches any string that contains at least one n
n* Matches any string that contains zero or more occurrences of n
n? Matches any string that contains zero or one occurrences of n
n{X} Matches any string that contains a sequence of X n's
n{X,Y} Matches any string that contains a sequence of X to Y n's
n{X,} Matches any string that contains a sequence of at least X n's
n$ Matches any string with n at the end of it
^n Matches any string with n at the beginning of it
?=n Matches any string that is followed by a specific string n
?!n
I need help with regular expression. I need a expression which allows only alphabets with space for ex. college name.
I am using :
var regex = /^[a-zA-Z][a-zA-Z\\s]+$/;
but it's not working.
Just add the space to the [ ] :
var regex = /^[a-zA-Z ]*$/;
This is the better solution as it forces the input to start with an alphabetic character. The accepted answer is buggy as it does not force the input to start with an alphabetic character.
[a-zA-Z][a-zA-Z ]+
This will allow space between the characters and not allow numbers or special characters. It will also not allow the space at the start and end.
[a-zA-Z][a-zA-Z ]+[a-zA-Z]$
This will accept input with alphabets with spaces in between them but not only spaces. Also it works for taking single character inputs.
[a-zA-Z]+([\s][a-zA-Z]+)*
Special Characters & digits Are Not Allowed.
Spaces are only allowed between two words.
Only one space is allowed between two words.
Spaces at the start or at the end are consider to be invalid.
Single word name is also valid : ^[a-zA-z]+([\s][a-zA-Z]+)*$
Single word name is in-valid : ^[a-zA-z]+([\s][a-zA-Z]+)+$
Regular expression starting with lower case or upper case alphabets but not with space and can have space in between the alphabets is following.
/^[a-zA-Z][a-zA-Z ]*$/
This worked for me
/[^a-zA-Z, ]/
This will work too,
it will accept only the letters and space without any symbols and numbers.
^[a-zA-z\s]+$
^ asserts position at start of the string Match a single character
present in the list below [a-zA-z\s]
matches the previous token between one and unlimited times, as many times as possible, giving back as needed (greedy) a-z matches a single
character in the range between a (index 97) and z (index 122) (case
sensitive) A-z matches a single character in the range between A
(index 65) and z (index 122) (case sensitive) \s matches any
whitespace character (equivalent to [\r\n\t\f\v ]) $ asserts position
at the end of the string, or before the line terminator right at the
end of the string (if any)
This worked for me, simply type in javascript regex validation
/[A-Za-z ]/
This one "^[a-zA-Z ]*$" is wrong because it allows space as a first character and also allows only space as a name.
This will work perfectly. It will not allow space as a first character.
pattern = "^[A-Za-z]+[A-Za-z ]*$"
This works for me
function validate(text) {
let reg = /^[A-Za-z ]+$/; // valid alphabet with space
return reg.test(text);
}
console.log(validate('abcdef')); //true
console.log(validate('abcdef xyz')); //true
console.log(validate('abc def xyz')); //true
console.log(validate('abcdef123')); //false
console.log(validate('abcdef!.')); //false
console.log(validate('abcdef#12 3')); //false
This will restrict space as first character
FilteringTextInputFormatter.allow(RegExp('^[a-zA-Z][a-zA-Z ]*')),
This will work for not allowing spaces at beginning and accepts characters, numbers, and special characters
/(^\w+)\s?/