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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/
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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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const item = "Dell Model 23506A Laptop M2";
const item2 = item.replace(/\d{2,}/g, "");
The above code will remove any words with "more than 2 digits found in a row, but will only remove the numbers from the Word.
Example the end result of the above will be "Dell Model A Laptop M2", leaving the A from 23506A.
How do you write the logic, that if more than 2 digits found in a row in a Word, to remove the entire word as a result.
Example the end result of the above should be "Dell Model Laptop M2"
Thus removing 23056A entirely, because more than 2 digits we're found in a row (right after another).
This should work:
item.replace(/[^\s]*\d{2,}[^\s]*/g, "");
You may also wish to get rid of the adjacent space:
item.replace(/[^\s]*\d{2,}[^\s]*\s?/g, "");
If you don't want to remove all other strings that could possibly contain 2 digits like for example 1,50, you can assert at least a char a-zA-Z
\b(?=\w*[A-Za-z])\w*\d{2}\w*\b
Explanation
\b A word boundary
(?=\w*[A-Za-z]) Positive lookahead, assert a char a-zA-Z
\w*\d{2}\w* Match 2 digits between optional word characters
\b A word boundary
Regex demo
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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 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)
")"