I need Regular Expression for a string [closed] - javascript

Closed. This question needs to be more focused. It is not currently accepting answers.
Want to improve this question? Update the question so it focuses on one problem only by editing this post.
Closed 8 years ago.
Improve this question
I need a regular expression for a string that
starts with Alphabets (no number)
max Length 8
No special characters or space.
string can have number or _ except for starting character.

This would work:
/^[a-z][a-z0-9_]{0,7}$/i
For example,
/^[a-z][a-z0-9_]{0,7}$/i.test('a1234567'); // true
/^[a-z][a-z0-9_]{0,7}$/i.test('01234567'); // false

The \w shorthand is for all letters, numbers and underscores. [A-Za-z] is overkill, the /i flag will get you all letters, case insensitive.
Therefore, a super simple regex for what you need is:
/^[a-z]\w{0,7}$/i
/^[a-z]\w{0,7}$/i.test("a1234567");
> true
/^[a-z]\w{0,7}$/i.test("a12345697");
> false
/^[a-z]\w{0,7}$/i.test("01234567");
> false

Try this out:
/^[A-Za-z]{1}[a-zA-Z0-9_]{0,7}$/

Try this one:
/^[a-zA-Z][0-9a-zA-Z_]{0,7}$/
This requires an alpha start character, and optionally allows up to 7 more characters which are either alphanumeric or underscore.
EDIT: Thanks, Jesse for the correction.

And another version with lookaheads :)
if (subject.match(/^(?=[a-z]\w{0,7}$)/i)) {
// Successful match
}
Explanation :
"^" + // Assert position at the beginning of the string
"(?=" + // Assert that the regex below can be matched, starting at this position (positive lookahead)
"[a-z]" + // Match a single character in the range between “a” and “z”
"\\w" + // Match a single character that is a “word character” (letters, digits, etc.)
"{0,7}" + // Between zero and 7 times, as many times as possible, giving back as needed (greedy)
"$" + // Assert position at the end of the string (or before the line break at the end of the string, if any)
")"

Related

complex regular expression for angular [closed]

Closed. This question needs to be more focused. It is not currently accepting answers.
Want to improve this question? Update the question so it focuses on one problem only by editing this post.
Closed 4 years ago.
Improve this question
I want a regular expression which validates my input e.g "S19/999999/090". my input field should accept this kind of input. means the first character should be an uppercase alphabet from A-Z after that two digits from 0-9. after that one backslash. after that 6 digits from 0-9 after that one backslash. after that 3 digits from 0-9.
E.g "S19/999999/090"
I don't understand how to create this kind of regular expression.
Thank you in advance
check the attached fiddle. [https://regex101.com/r/IQsyv2/1/]
and regex will be like -
^[A-Z][\d]{2}/[\d]{6}/[\d]{3}$
here initially to notice I have specified ^ and $ these char are used to define the full length match, mean ^ - says starting and $ says ending.
[A-Z] - mean check uppercase only once.
[\d]{2} - mean [\d] any digit, {2} - 2 times
\/ -- mean match backslash
[\d]{6} - only digit 6 times exactly.
\/ -- mean match backslash
[\d]{3} - only digit 3 times exactly.
Try this
[A-Z]\d{2}\/\d{6}\/\d{3}
[A-Z] - one character A-Z uppercase
\d - digital
\d{n} - n digital
/\ - /

Constructing a regex with three subpatterns and a maximum length of 28 characters [closed]

Closed. This question needs to be more focused. It is not currently accepting answers.
Want to improve this question? Update the question so it focuses on one problem only by editing this post.
Closed 4 years ago.
Improve this question
I need to create an expression that fit this requirements :
The string must be composed by 3 substring
The first substring accept 0-9a-zA-Z, the minimum length is 1 and there is notte max length
The second substring must be " - "
The last have The first's one same condition
Total maximum string length must be 28 chars
It is possible to accomplish this requirement with Regex?
Following regex should work fine:
/(?=^.{3,28}$)[^\W_]+\-[^\W_]+/
var array = [
"123456790-123456789012345678",
"123456790-1234567890123456789",
"adsd-dsds"
];
var re = /(?=^.{3,28}$)[^\W_]+\-[^\W_]+/;
array.forEach(e => console.log(re.test(e) ? e.match(re)[0]: "match failed"));
Breakdown shamelessly copied from regex101.com:
Positive Lookahead (?=^.{3,28}$)
^ asserts position at start of a line
.{3,28} matches any character (except for line terminators)
{3,28} Quantifier — Matches between 3 and 28 times, as many times as possible, giving back as needed (greedy)
$ asserts position at the end of a line
Match a single character not present in the list below [^\W_]+
+ Quantifier — Matches between one and unlimited times, as many times as possible, giving back as needed (greedy)
\W matches any non-word character (equal to [^a-zA-Z0-9_])
_ matches the character _ literally (case sensitive)
\- matches the character - literally (case sensitive)
Match a single character not present in the list below [^\W_]+
+ Quantifier — Matches between one and unlimited times, as many times as possible, giving back as needed (greedy)
\W matches any non-word character (equal to [^a-zA-Z0-9_])
_ matches the character _ literally (case sensitive)

How to produce regex string to check to second word is uppercase [closed]

Closed. This question needs debugging details. It is not currently accepting answers.
Edit the question to include desired behavior, a specific problem or error, and the shortest code necessary to reproduce the problem. This will help others answer the question.
Closed 6 years ago.
Improve this question
I have two words within sentence.
EX: big BUTTON
here I need to check second word is in uppercase using regex expression.
Square brackets [] are your friend. They allow you to specify characters that will match. To match the first work, you need to check for any letter. This can be done with [a-zA-Z]. This will match any letter between a and z, as well as A and Z. for the second word, you only want to match uppercase, so only use [A-Z]. To get 1 or more matches, put a + after the closing bracket.
Putting this all together, with a space in between the words, you get [a-zA-Z]+ [A-Z]+.
The carat ^ is used to signify the start of the string, and the dollar sign $ is used to signify the end of the string. Your question somewhat vague, so here are a couple scenarios:
Each sentence is only two words with a space in between them: ^[a-zA-Z]+ [A-Z]+$
Each sentence has at least two words and may or may not end in a period: ^[a-zA-Z]+ [A-Z]+( |\.?$)
In the second example the parenthesis with a pipe (|) is used as an OR statement. The period is escaped since it is a special character (matches any single character). The question mark denotes 0 or 1 of the preceding character, which is a period. So ( |\.?$) will match a space or a sentence that ends with or without a period.
Here is a good site that has information on Regexes: http://www.regular-expressions.info/
This regexp looks for any sequence, starting at the beginning of the string (^), of alphanumeric characters (\w)--that's the first word--then a space, followed by a sequence of upper-case letters ([A-Z]+)--the second word--followed by either a space or the end of the string ($).
/^\w+ [A-Z]+( |$)/

Regex first letter not integer with jquery [closed]

Closed. This question needs details or clarity. It is not currently accepting answers.
Want to improve this question? Add details and clarify the problem by editing this post.
Closed 6 years ago.
Improve this question
I have textbox and I want get string value. But I want users to not be able to enter the string that has number on first letter. As matter of fact I want to replace number with '' null.
for example
1test =====convert=======> test
you can simply use ^[a-zA-Z]
^ starts with a-z or A-Z
or if you want special character too then use ^\D
^\D : Matches anything other than a decimal digit
Regex Demo
you can use $text.replace(/^[^0-9]+/, '')
/^ beginning of the line
[^0-9]+ match anything other than digits at-least once
thanks # Wiktor and Tushar
here is the solution: You can check on this live regex.
https://regex101.com/r/OJfyv4/1
$re = '/\b[a-z][a-z0-9]*/';
$str = '1test';
preg_match_all($re, $str, $matches);
// Print the entire match result
print_r($matches);
This works your case:
^\d+
https://regex101.com/r/daezA9/1
^ asserts position at start of the string
\d matches a digit (equal to [0-9])

(\d+,\d+) add so it is imuneto white spaces [closed]

Closed. This question needs details or clarity. It is not currently accepting answers.
Want to improve this question? Add details and clarify the problem by editing this post.
Closed 8 years ago.
Improve this question
How in javascript to make regex to recognize and extract integer numbers for coordinates which have format like
( number1, number2 )
between ( and number1 and , and number2 and ) can be arbitrary number of whitespaces (user are going to enter coordinates so I don't want to force strict format without whitespaces)
(\d+,\d+)
what to add to this so it works ?
There are a few choices, pending your actual input.
To match all "whitespace characters" (space, tab, carriage return, newline and form feed), you can use the \s shorthand approach. If you want a number, in this case \d+, to be surrounded by "0 or more" of these, you would use:
\s*\d+\s*
In your full pattern:
\( # opening parentheses
\s*\d+\s*, # first number followed by a comma
\s*\d+\s* # second number
\) # closing parentheses
Note: The parentheses are escaped here as they're special characters in a regular expression pattern.
Now, if you don't want to match "all whitespace" and were only interested in plain spaces, for example, you could use a matching character set of [ ] (i.e. a space between two brackets). In the pattern from above:
\(
[ ]*\d+[ ]*,
[ ]*\d+[ ]*
\)
Not really sure how you want to use the matches, I'm assuming you want the numbers returned individually so in that case, you can use the following:
var str = '(1, 2)';
var matches = str.match(/\(\s*(\d+)\s*,\s*(\d+)\s*\)/);
if (matches) {
var firstNumber = matches[1];
var secondNumber = matches[2];
// do stuffs
}
Note: In the pattern I used here, I've wrapped the \d+s in parentheses; this will "capture" those values in to groups which are then accessible by their "group index". So, the first (\d+) will be available in matches[1] and the second will be available in matches[2].
Try this regex: \(\s*\d+\s*,\s*\d+\s*\).
Fiddle

Categories