You may use the following regex pattern: (?:Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec) \d{1,2}(?: \d{4})? var dates = [“Nov 29 2019”, “abc 0 May 30 2020”, “ddd Apr 3 2021 efg”, “0 Jan 3 hellodewdde deded”, “Green eggs and ham”]; var months = “(?:Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec)”; var regex = new RegExp(months + ” \\d{1,2}(?: \\d{4})?”); var matches = []; for (var i=0; … Read more
You don’t necessarily have to create a Document in order to change an XML document. You can quite easily set up streaming XML parser to do this in an efficient manner. Have a look at the javax.xml.stream package. solved Find a xml element with Regex and replace its value
String.trim() method remove trailing and leading white spaces from the string. but since the string you are testing doesn’t have any trailing or leading white spaces, the below two will certainly return the same string. str.replaceFirst(“(.*)<?xml”,”<?xml”); str.trim().replaceFirst(“(.*)<?xml”,”<?xml”) However, your regular expression only removes leading white spaces, so if the testing string has trailing white spaces … Read more
Here’s how I’d do it: str = “8 560,90 cur.” str.gsub(/[^\d,]/, ”).to_i # => 8560 This removes every character that isn’t a digit or a comma, yielding “8560,90”, then calls to_i on it, which gives 8560. This will work for any string as long as you want every digit before the first comma to be part … Read more
$string=”1234-ABCD-EFGH”; preg_match(“~[0-9]{4}-[A-Z]{4}-[A-Z]{4}~”, $string, $matches); print_r($matches); This code will output the $matches array that contains the match information. preg_match will return true or false depending on whether the string matched. If you prefer to also match lower case characters, use this one: preg_match(“~[0-9]{4}-[A-Za-z]{4}-[A-Za-z]{4}~”, $string, $matches); 1 solved php – Regular expression to validate this string [closed]
For a simple mathematical operation like Number Operator Number you can use this regex: ^[0-9]+(\+|\-|\*|\/)[0-9]+$ For a complex expression like Number Operator Number Operator Number… try this: ^[0-9]+((\+|\-|\*|\/)[0-9]+)+$ 1 solved Construct regular expression that matches numeric expressions like 5+65 [closed]
I need to split it to get only 1 and 6 and I only want whole number I want to ignore the floating numbers You can use String.prototype.split() with RegExp /\D|\d+\.\d+/ to split characters that are not digits, or digits followed by . character followed by digits to handle for example 2.456, Array.prototype.filter() with parameter … Read more
This is normally solved with just a String#replaceAll, but since you have a custom, dynamic replacement string, you can to use a Matcher to efficiently and concisely do the string replacement. public static String parse(String command, String… args) { StringBuffer sb = new StringBuffer(); Matcher m = Pattern.compile(“\\$(\\d+)”).matcher(command); while (m.find()) { int num = Integer.parseInt(m.group(1)); … Read more
This is quite easy to o actually. Requirement #1 and #3 are the easiest. To test for whether a string contains two words and are separated by exactly one space, just use the split method. For requirement #2, test if the strings have a length that is larger than 2. Let’s say your string to … Read more
It can be achieved with the following regex: ^[^a-z'”*&<>\/]+$ See demo It allows any letter that is not ‘”*’&<>\/, and requires that there is no lowercase English letter. ^ – Start of string [^a-z'”*&<>\/]+ – 1 or more characters other than ‘, “, *, &, <, >, / or lowercase English character. $ – End … Read more
Try it out on regex101.com, looking on the right side: Hope you can see this ok and can grok it. solved strange regex experasion [closed]
If you insist on regex, you can use positive look ahead as preg_replace(‘/,(?=}$)/’, ”, “helloworld,}”) // helloworld} Regex Explanation , Matches , (?=}$) positive look ahead. Checks if the , is followed by } and then end of line $ 7 solved How can I check if the last 2 characters of a string are … Read more
Here’s a regex that just extracts the first number in the string: ^\D*(\d+).*$ If the number must be in front of min, then you can write it as: ^\D*(\d+)min.*$ solved Regex – Matching two numbers before a certain string [closed]
As others have posted, the regular expression you are looking for is: \d{4}-\d{6} The full code I would use: import re my_string = ‘Ticketing TSX – 2016-049172’ matches = re.findall(r”\d{4}-\d{6}”, my_string) print matches If, for example, the length of the second digit varies from 6 to 8 digits, you will need to update your regular … Read more
As you further explained that you actually do want to match any letter at the beginning, and any two digits at the end of the string, using a regular expression is indeed the shortest way to solve this. Regex re = new Regex(“^[a-z].*[0-9]{2}$”, RegexOptions.IgnoreCase); Console.WriteLine(re.IsMatch(“Apple02”)); // true Console.WriteLine(re.IsMatch(“Arrow”)); // false Console.WriteLine(re.IsMatch(“45Alty12”)); // false Console.WriteLine(re.IsMatch(“Basci98”)); // … Read more