use a character class to split the values: /(-?[\d.]+)/ -? May start with a negative such as -123 [\d.]+ Has one or more numbers and decimals var string = “stuff 1.23! (456) 789 stuff -234”.split(/(-?[\d.]+)/); console.log(string)
Use a regular expression with a capture group and RegEx.exec() to extract any substrings between three apostrophes. The pattern uses lazy matching (.*?) to match any sequence of characters (including none) between the apostrophes. const str = ” My random text ”’tag1”’ ”’tag2”’ “; re = /”'(.*?)”’/g; matches = []; while (match = re.exec(str)) { … Read more
String areaCode = number.substring(0,3); String prefix = number.substring(3,6); String rest = number.substring(6); Hope that helps!
I think what you are trying to know is the generic regular expression which starts with any alphabets and has only alpha-numeric characters in it. Look, the generic format (in java, for string pattern matching) for your query is this: String pattern = “[a-zA-Z][a-zA-Z0-9]*”; or String pattern = “[a-zA-Z](\\w)*”; represents kleene star operator which denotes … Read more
$str = “COVId 1234 | SARS 4567 | EBOLA 2332”; preg_replace(‘([a-zA-Z]+ \d+ )’, ‘$0 |’, $str); echo $str; // COVId 1234 | SARS 4567 | EBOLA 2332 [a-ZA-Z]+ matches all alphabetic chars. is just to match a space. \d+ matches digits. Note the plural.
Code See regex in use here ^(?!0+$)\d{1,5}$ Results Results as described by the header for the section. ** VALID ** 1 12345 00001 01 10 ** INVALID ** 0 00000 123456 Explanation ^ Assert position at the start of the line (?!0+$) Negative lookahead ensuring what follows is not 0 one or more times, followed … Read more
(?:…) denotes a non capturing group. ?: is used when you want to group an expression, but you want to avoid to return it as a matched/captured portion of the string. The ^ does not mean negation when it is located outside square breackets. It means match from the beginning of the string.
Something like this? ^PREFIX(\.[a-z]+)+$ Example matches: PREFIX.test.test PREFIX.test.test.test // and so on
For PHP you can do something like this: $string = “one two three four five six seven eight nine One Two Three Four Five Six Seven Eight Nine oNE tWO tHREE fOuR fIVE sIX sEVEN eIGHT nINE ONE TWO THREE FOuR FIVE SIX SEVEN EIGHT NINE”; $replace = array(0,1,2,3,4,5,6,7,8,9); $search = array(‘zero’, ‘one’, ‘two’, ‘three’, … Read more
You can use the following regex: -?\d+(?:,\d{3})*(?:\.\d+)? It matches an optional hyphen (-?), then 1 or more digits (\d+), then optional 3-digit groups ((?:,\d{3})*), then an optional decimal part ((?:\.\d+)?). var s = “hello-sdvf-1,234.23 23 everybody 4”; var res = s.match(/-?\d+(?:,\d{3})*(?:\.\d+)?/); document.getElementById(“r”).innerHTML = res; <div id=”r”/> Or, to match multiple values: var re = /-?\d+(?:,\d{3})*(?:\.\d+)?/g; … Read more