I need a regex to make the validation of an username field. I need to accept just alphanumeric chars and max one space.
What I tried is:
^[a-z0-9]+[ ][a-z0-9]+$
It's working but it doesn't seems the right solution for this problem. Can someone guide me on how to optimize this regex? Thanks.
The following regular expression will do it: ^[a-zA-Z0-9]+ ?[a-zA-Z0-9]+$
You missed only the ? character after the space, what matches between zero and one times.
See full explanation at regex101...
Here you can visualize your regular expressions...
Related
I am new to using regex. I am trying to use the regex find and replace option in Notepad++.
I have used the following regex:
((?:)|(\+)|(-))(\d)((?:)|(\+)|(-))(/)((?:)|(\+)|(-))(\d)((?:)|(\+)|(-))
For the following text:
2/2
+2/+2
-2/-2
2+/2+
2-/2-
But I am able to get matches only for the first three. The last two, it only gives partial matches, excluding the last "+" and the "-". I am wondering if there is any upper limit for the number of groups (which i doubt is unlikely) that can be used or any upper limit for the maximum length of the regex. I am not sure why my regex is failing. Or if there is anything wrong with my regex, please correct it.
This is not an issue with Notepad++'s regex engine. The problem is that when you have alternations like (?:)|(\+)|(-), the regex engine will attempt to match the different options in the order they are specified. Since you specified an empty group first, it will attempt to match an empty string first, only matching the + or - if it needs to backtrack. This essentially makes the alternation lazy—it will never match any character unless it has to.
vks's answer works perfectly well, but just in case you actually needed those capturing groups separated out, you can do the same thing just by rewriting your alternations like this:
((\+)|(-)|(?:))(\d)((\+)|(-)|(?:))(/)((\+)|(-)|(?:))(\d)((\+)|(-)|(?:))
or even more simply, like this:
((\+)|(-)|)(\d)((\+)|(-)|)(/)((\+)|(-)|)(\d)((\+)|(-)|)
([-+]?)(\d)([-+]?)(/)([-+]?)(\d)([-+]?)
You can use this simple regex to match all cases.See here.
https://www.regex101.com/r/fG5pZ8/19
I need to parse an array-like text with regular expression and get the match groups.
One example of then text I want to parse is this:
['red','green', 'blue']
I want to use match groups, because I want to extract them.
I am using this regular expression, but the groups found by it are not like what I expected:
\[ *('.+?')( *, *('.+?'))* *\]
The idea is to parse in this order:
A square bracket
Any number of spaces
A group with:
Single quote
Any character
Single quote
Zero or more groups of:
Any number of spaces
A comma
Any number of spaces
A group with
Single quote
Any character
Single quote
Any number of spaces
A square bracket
And get one group with each parsed array element.
Can you help me?
Hint: a easy way to test regexp is the site http://rubular.com
This isn't going to be a totalitarian answer, but I'm fairly certain you can't whitespace check by doing " *", at least it may depend on the language you're using.
Here's a C# regex example that shows some of the language requirements to check for whitespace: regex check for white space in middle of string
Edit: I see you added Ruby as your language, unfortunately I'm not verbose in Ruby so specifics I cannot help you with, sorry.
Edit2: Seeing as you're forcing yourself into Ruby to debug your regex statement, might I suggest: http://www.debuggex.com/ which tries to stay language independent?
Try this regex: '([^']+)', it should give you the following match groups red, green, blue according to rubular.com
You can match an arbitrary number of groups with one regex:
^\[\s*|(?:\G'([^']+)'\s*(?:,\s*|]$))+
or like this (should be more performant):
^\[\s*+|(?>\G'([^']++)'\s*+(?>,\s*+|]$))++
This work in ruby like asked before, in delphi I don't know.
So I wanted to limit a textbox which contains an apartment number which is optional.
Here is the regex in question:
([0-9]{1,4}[A-Z]?)|([A-Z])|(^$)
Simple enough eh?
I'm using these tools to test my regex:
Regex Analyzer
Regex Validator
Here are the expected results:
Valid
"1234A"
"Z"
"(Empty string)"
Invalid
"A1234"
"fhfdsahds527523832dvhsfdg"
Obviously if I'm here, the invalid ones are accepted by the regex. The goal of this regex is accept either 1 to 4 numbers with an optional letter, or a single letter or an empty string.
I just can't seem to figure out what's not working, I mean it is a simple enough regex we have here. I'm probably missing something as I'm not very good with regexes, but this syntax seems ok to my eyes. Hopefully someone here can point to my error.
Thanks for all help, it is greatly appreciated.
You need to use the ^ and $ anchors for your first two options as well. Also you can include the second option into the first one (which immediately matches the third variant as well):
^[0-9]{0,4}[A-Z]?$
Without the anchors your regular expression matches because it will just pick a single letter from anywhere within your string.
Depending on the language, you can also use a negative look ahead.
^[0-9]{0,4}[A-Za-z](?!.*[0-9])
Breakdown:
^[0-9]{0,4} = This look for any number 0 through 4 times at the beginning of the string
[A-Za-z] = This look for any characters (Both cases)
(?!.*[0-9]) = This will only allow the letters if there are no numbers anywhere after the letter.
I haven't quite figured out how to validate against a null character, but that might be easier done using tools from whatever language you are using. Something along this logic:
if String Doesn't equal $null Then check the Rexex
Something along those lines, just adjusted for however you would do it in your language.
I used RegEx Skinner to validate the answers.
Edit: Fixed error from comments
I am trying to learn regular expressions and was hoping someone could help me out. WOuld appreciate if someone can help me come up with a regular expression to validate that an input must be of the form
Graph: XY5, YZ4, ST7
Each part such as XY5 represents an edge in the graph and the number represents a the edge weight. There can be any number of such edges.
This is what I have till now. It's probably not correct
"^Graph:\\s{1}[A-ZA-Z\\d,\\s]+"
This might be what you're looking for:
/^Graph: (?:[A-Z]{2}\d(?:$|, ?))+/
See it here in action: http://regexr.com?309av
Here's an explanation of what the regex does (screenshot from RegexBuddy, which is probably the best tool for you if you're trying to learn Regular Expressions):
Try this
/^Graph:(\s+[A-Z][A-Z]\d+)+$/
You should explain your input format a little better. This might do it, from the single example I have and what you said. It does not allow a graph to be empty, which may or may not be part of your requirements.
"^Graph:(\s\w{2}\d+,?)+"
to explain:
^Graph: will cover the start of the line
(\s\w{2}\d+,?)+
\s is a space
\w{2} matches exactly 2 alphanumeric characters (hint: you could make this better!)
\d+ matches 1 or more digits, since I am assuming an edge can have a two digit length ( such as 10)
,? matches a comma optionally. (hint: you could make this better as well, as it will not necessitate a comma between each entry!, maybe by using an or and the end of string delimiter!)
I purposely left some room for improvement, because if you think of some of it on your own, you will accomplish your goal of becoming better with regular expressions.
i have following regular expression but it's not working properly it takes only three values after # sign but i want it to be any number length
"/^[a-zA-Z0-9_\.\-]+\#([a-zA-Z0-9\-]+\.)+[a-zA-Z0-9]{2,4}$/"
this#thi This is validated
this#this It is not validating this expression
Can you please tell me what's the problem with the expression...
Thanks
If you want your regex to match "any number length" then why are you using {2,4}?
I think a better example of the strings you're trying to match might give others a better idea of what you want, because based on your regex it is a bit confusing what you're looking for.
Try this:
^[a-zA-Z0-9_.-]+#([a-zA-Z0-9-]+\.)+[a-zA-Z0-9]{2,4}$
The main problem is that you didn't escape the dot: \.. In regular expression the dot matches everything (mostly), making your regex quite liberal.