This might work, although (phrase)+ isn’t something I’ve tested (and would be happy to hear feedback on): /^((\/)|([a-z]+\/{0,1}[a-z]+)+)$/ I may have gotten carried away with the brackets… 1 solved regex – can contain slash, just not at the start or end
[…] matches any character defined in the character class, so [X{9,11}\\*{2,3}] actually means, a single character which is: X, or open brace, or 9, or comma, or 1, or 1 (yes you have it duplicated), or backslash, or asterisk…. So as your string have more than character in your string to-be-matched, such pattern will not … Read more
Depends on what you really want to do, for example if you only want to get the value of a given key, given that key and value are separated by colon and key/value are separated by comma, you can just split in this way string first by comma and then by colon : String[] couple … Read more
First of all, let’s see what it is going to be matched by the pattern you mentioned: ===================================================================== ^[a-zA-Z]+(([\’\ \][(^10$|^[0-9]{1,2}]))+(([\’\ \][(?:m|M|f|F|)$]))*$ ===================================================================== Assert position at the beginning of the string «^» Match a single character present in the list below «[a-zA-Z]+» Between one and unlimited times, as many times as possible, giving back as needed … Read more
If you’re a masochist you can do this old school VB3 style: string input = @”</script><div id=’PO_1WTXxKUTU98xDU1′><!–DO NOT REMOVE-CONTENTS PLACED HERE–></div>”; string startString = “div id='”; int startIndex = input.IndexOf(startString); if (startIndex != -1) { startIndex += startString.Length; int endIndex = input.IndexOf(“‘”, startIndex); string subString = input.Substring(startIndex, endIndex – startIndex); } solved Searching a String … Read more
Your perl backreferences are incorrect: perl -lne ‘$a{$1}++ if (/^(Unexpected exception) : (.*?)\s*$/) ^^^^^^^^^^^^^^^^^^^^^-$1 ^^^^^-$2 You want $a{$2}++ instead. Or convert that first () to a non-capturing group: (?:Un….) 1 solved A perl script to list all the exceptions and their number of occurrence
You could use this regex: ^(\w|\.)+$ Which is the same as: ^[a-zA-Z0-9_\.]+$ Here’s preview of the regex in action on regex101.com, and here’s a breakdown of it ^ matches the beginning of the string ( just groups the characters so a modifier can be applied \w matches any character that is a-z, A-Z, 0-9, and … Read more
the easiest way to parse a string to number is Number constructor (not only integer) Number(“0000.14”); // 0.14 Number(“000000″) // 0 or you can add plus sign before a strings +”00000″ // 0 +”000.12” // 0.12 1 solved How to handle consequence of zero’s in java script [closed]
As you guessed correctly, this has to do with capturing groups. According to the documentation for re.findall: If one or more groups are present in the pattern, return a list of groups Therefore, you need to make all your groups () non-capturing using the (?:) specifier. If there are no captured groups, it will return … Read more
Try this: ^[1-9][0-9]?$ The ? means that the previous token is optional. The ^ and $ are anchors for the start and end of the string respectively. solved Regular expression for 1 or more [closed]
^(?=.{8,10}$)\d+(?:[ _]\d+)*$ Explanation: ^ # Start of string (?=.{8,10}$) # Assert length 8-10 \d+ # One or more digits (?: # followed by [ _] # one space or underscore \d+ # one or more digits )* # zero or more times. $ # End of string 2 solved PHP regular expression for given condition … Read more
Resulting from the descussion, you need one regular expression for each check to perform. At least 8 characters: ^.{8,}$ No whitespace character: ^\S*$ Both combined: ^\S{8,}$ Explanation: ^ Start of the string $ End of the string . Any character (inclusive whitespace) \S Any characters that is not whitespace {8,} previous expression 8 or more … Read more
This will work ^\d{4}-\d{3,4}-[23]$ Regex Demo Regex Breakdown ^ #Start of string \d{4} #Match 4 digits – #Match – literally \d{3,4} #Match 3 or 4 digits – #Match – literally [23] #Match 2 or 3 $ #End of string 0 solved Regex pattern can’t figure out
Just try with: $input=”stackoverflow.com”; if (strpos($input, ‘www.’) !== 0) { $input=”www.” . $input; } 2 solved Add www. if it’s not already in the string [closed]
The user asked for something simple, so I gave him something simple. $input = “[link=http://www.google.com]test[/link]”; $replacement = preg_replace(‘/\[link=(.*?)\](.*?)\[\/link\]/’, ‘<a href=”https://stackoverflow.com/questions/10458103/”>$2</a>’, $input); Where /\[link=(.*?)\](.*?)\[\/link\]/ is the regex, <a href=”https://stackoverflow.com/questions/10458103/”>$2</a> is the format, $input is the input/data, and $replacement is the return. 1 solved How to make PHP BB Codes with RegExp?