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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
Related
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How could I get the specific substring between 'Y___(string i want to get)___N'？
For example：
"Y___INT_GET_ERROR_CONFIGS___N"
"INT_GET_ERROR_CONFIGS"
You can get everything inside by Y___(.*?)___N, you can use matchAll to get all instance that matches this case, and you can loop through and get the group value.
const str = `Y___INT_GET_ERROR_CONFIGS___N
INT_GET_ERROR_CONFIGS Y___INT_GET_ERROR_SOMETHING___N`
const result = str.matchAll(/Y___(.*?)___N/g);
for (match of result) {
console.log(match[1])
}
If it's just the first occurrence you wish you match, then:
'Y___(string you want to get)___N'.match(/(?<=Y___).+(?=___N)/)
Result:
"(string you want to get)"
If you want all such occurrences to be returned, then use the g flag:
`Y___(string you want to get)___N
.
.
.
Y___(second string you want)___N`.match(/(?<=Y___).+(?=___N)/g)
Result:
["(string you want to get)", "(second string you want)"]
Explanation:
(?<=Y___): A positive lookbehind stipulates that matches will be preceded by the contents of the lookbehind, namely "Y___". The contents of the lookbehind does not form part of the match result, and also does not consume characters during matching.
.+: Matches at least one instance of any character, but will match as many as possible.
(?=___N): A positive lookahead stipulates that matches will be proceeded by the contents of the lookahead, namely "___N". The contents of the lookahead does not form part of the match result, nor does it consume characters during matching.
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How to use regex to replace all the occurrences of space ( . ) '- with underscores.
I used chaining of replaceAll. Is there a better approach?
const str = "Certificate of Naturalization (From N-550 or N-570) or Certificate of U.S Citizenship (From N-560 or N-561)";
console.log(((str).toUpperCase()).replaceAll(' ', '_').replaceAll('\'', '_').replaceAll('(', '').replaceAll(')', '').replaceAll('-', '_').replaceAll('.', '_'));
You can put the characters you want to replace inside [] and you need to escape some special characters with \
there are 12 characters with special meanings: the backslash , the caret ^, the dollar sign $, the period or dot ., the vertical bar or pipe symbol |, the question mark ?, the asterisk or star *, the plus sign +, the opening parenthesis (, the closing parenthesis ), the opening square bracket [, and the opening curly brace {, These special characters are often called “metacharacters”. Most of them are errors when used alone.
If you want to use any of these characters as a literal in a regex, you need to escape them with a backslash.
const str = `Certificate of Naturalization (From N-550 or N-570) or Certificate of U.S Citizenship (From N-560 or N-561)`;
const result = str.replace(/[-'\s\(\)\.]/g, '_')
console.log(result)
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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 ($).
/^\w+ [A-Z]+( |$)/
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Given a string "translateX(-50%) scale(1.2)" with N transform functions
1) How can I match the names ["translateX", "scale"]?
2) How can I match the values ["-50%", "1.2"]?
If you absolutely must use regular expression to do this, you can use the exec() method in a loop, pushing the match result of the captured group(s) to the desired arrays of choice.
var str = 'translateX(-50%) scale(1.2)'
var re = /(\w+)\(([^)]*)\)/g,
names = [], vals = [];
while (m = re.exec(str)) {
names.push(m[1]), vals.push(m[2]);
}
console.log(names) //=> [ 'translateX', 'scale' ]
console.log(vals) //=> [ '-50%', '1.2' ]
The regular expression uses two capture groups, the first matches/captures word characters only, the second uses negation which will match any character except ) "zero or more" times.
Try something like (\w+)\((.+?)\):
(\w+): Match the regular expression below and capture its match into backreference number 1
\w+: Match a single character that is a “word character” (letters, digits, and underscores)
+: Between one and unlimited times, as many times as possible, giving back as needed (greedy)
\(: Match the character “(” literally
(.+?): Match the regular expression below and capture its match into backreference number 2
.+?: Match any single character that is not a line break character
+?: Between one and unlimited times, as few times as possible, expanding as needed (lazy)
\): Match the character “)” literally
var str = "translateX(-50%) scale(1.2)",
regex = /(\w+)\((.+?)\)/g,
match, names = [], values = [];
while(match = regex.exec(str)) {
names.push(match[1]);
values.push(match[2]);
}
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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)
")"