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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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I need validate a input using regex (i think it's better) but in this input can be one or more "sentences" and be [A-Z] size 1. How can i do that?
E.g.:
A,B,D,G,J,X no repeat letters but this validate i do in code. I think regex is better 'cause validate a entire sentence instead letter by letter using a loop and split. My english is rusty, appreciate some help to improve =)
Note, Assumption is you want a single letter
If you just want to validate:
if (/([A-Z]*)?,([A-Z]*),?/.test(subject)) {
// Successful match
} else {
// Match attempt failed
}
If you are using to get extract values:
result = subject.match(/([A-Z]*)?,([A-Z]*),?/g);
Maybe this can help you ([A-Z],)+[A-Z] it will match a serie of uppercase letter followed by comma, and end with uppercase letter :
regex demo
A,B,D,G,J,X -> matches
A,B,DE,G,J,X -> not matches
A,B,D,G,J,XY -> not matches
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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])
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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
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I need to test whether a string contains only punctuation symbols except one character symbol say hypen (-)
'??...' -> true
'.!sf' -> false
'..-??' -> false
I use library XRegEx library that allows to match for example punctuation with p{P} but I need to exclude some characters from this match.
I use following pattern:
new XRegExp("^\\p{P}+$")
How can I except hypen symbol from "-" this match?
N.B. Original question was about "leters":
I need to test whether a string contains letters except one character say letter "m"
/^[a-ln-z]+$/.test('abcfx')
// true
/^[a-ln-z]+$/.test('abcfx12.!')
// false
/^[a-ln-z]+$/.test('abcmfx')
// false
Using positive lookahead:
/^(?=[^m]+$)[a-z]+$/.test('abcfx')
// true
/^(?=[^m]+$)[a-z]+$/.test('abcfx12.!')
// false
/^(?=[^m]+$)[a-z]+$/.test('abcmfx')
// false
What you are looking for is probably this regex, which accepts ranges from a to l and from n to z
^[a-ln-z]+$
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I want to test if a string can match a pattern in javascript
problem is there is a random string in the middle
pattern: "account/os/some_random_string/call_back"
so a string look like below will match
var mystring = "account/os/1234567/call_back"
thanks
You want a regex for starts with account/os/ and ends with /call_back, here's one:
/^account\/os\/.*\/call_back$/
.* will match any random string (including the empty string.). If you want a minimum length on the random string you change the *:
.* : 0 or more characters
.+ : 1 or more characters
.{n,} : n or more characters (replace n with an actual number)
Well, it depends. If you want every single character between account/os/ and /call_back, use this:
var randomStr = mystring.match(/account\/os\/(.*)\/call_back/)[1];
The match will return an array with 2 elements, the first one with the entire match, the 2nd one with the group (.*). If you are completely sure that you have at least one character there, replace * with +.
If you know something more specific about the text you have to collect, here are some replacements for the .(dot is matching almost everything):
[A-z] for any of A, B, .. , Y, Z, a, b, .. , y, z
[0-9] for any digit
You can mix then and go fancy, like this:
[A-Ea-e0-36-8]
So, your pattern may look like this one:
/account\/os\/([A-Ea-e0-36-8]*)\/call_back/
Your example have a number there, so you are probably looking for:
/account\/os\/([0-9]*)\/call_back/
or
/account\/os\/(\d*)\/call_back/
.. it's the same thing.
Hope that helps.
Edit: What JS answer doesn't have a jsfiddle? http://jsfiddle.net/U2Jhw/