http://regexr.com/3ars8
^(?=.*[0-9])(?=.*[A-z])[0-9A-z-]{17}$
Should match "17 alphanumeric chars, hyphens allowed too, must include at least one letter and at least one number"
It'll correctly match:
ABCDF31U100027743
and correctly decline to match:
AB$DF31U100027743
(and almost any other non-alphanumeric char)
but will apparently allow:
AB^DF31U100027743
Because your character class [A-z] matches this symbol.
[A-z] matches [, \, ], ^, _, `, and the English letters.
Actually, it is a common mistake. You should use [a-zA-Z] instead to only allow English letters.
Here is a visualization from Expresso, showing what the range [A-z] actually covers:
So, this regex (with i option) won't capture your string.
^(?=.*[0-9])(?=.*[a-z])[0-9a-z-]{17}$
In my opinion, it is always safer to use Ignorecase option to avoid such an issue and shorten the regex.
regex uses ASCII printable characters from the space to the tilde range.
Whenever we use [A-z] token it matches the following table highlighted characters. If we use [ -~] token it matches starting from SPACE to tilde.
You're allowing A-z (capital 'A' through lower 'z'). You don't say what regex package you're using, but it's not necessarily clear that A-Z and a-z are contiguous; there could be other characters in between. Try this instead:
^(?=.*[0-9])(?=.*[A-Za-z])[0-9A-Za-z-]{17}$
It seems to meet your criteria for me in regexpal.
I need to identify matching course number that have xx.3xxxxxx.
These are some examples of the course numbers.
26.3730004
27.0210000
26.3730009
26.7114001
23.9610071
26.0A34430
23.3670005
26.0B05430
I tried many patterns one example I used is the pattern below. It did not get any match.
"[^0-9]{2}\Q.\E3[^0-9]+$"
I tried using grep and grepl. I actually need the code to return indexes.
This code shows my attempt to tag the rows that have matches.
Teacher$virtual[
which(
grepl("[^0-9]{2}\\Q.\\E3[^0-9]+$",Teacher$CourseNumber))]
<- "1"
I need to remove any row from my dataframe that have the course number with that pattern. XX.3XXXXXX
But, my code did not find any match. Can you please help me?
You should use
grepl("^[0-9]{2}\\.3", Teacher$CourseNumber)
See the regex graph:
Details:
^ - start of a string
[0-9]{2} - two digits
\\. - a dot (note that a regex escape is a literal backslash, but inside a string literal, "...", a single backslash is used to form string escape sequences, hence the backslash must be double to obtain a literal backslash char necessary for a regex escape)
3 - a 3 char.
NOTE: If you want to use in-pattern quoting with \Q and \E (in between which all chars are treated literally) you need to use PCRE regex, add perl=TRUE and use
grepl("^[0-9]{2}\\Q.\\E3", Teacher$CourseNumber, perl=TRUE)
Now, the dot is treated as a literal dot, not a . metacharacter that matches any char but a line break char (in a PCRE regex, . does not match line break chars by default).
Here, this simple expression would likely cover that:
^[0-9]{2}\.[3].+$
which has a [3] boundary right after the .. It would probably work without start and end anchors:
[0-9]{2}\.[3].+
Demo
We can add or reduce the boundaries, if it'd be necessary.
Simple regex question. I have a string on the following format:
this is a [sample] string with [some] special words. [another one]
What is the regular expression to extract the words within the square brackets, ie.
sample
some
another one
Note: In my use case, brackets cannot be nested.
You can use the following regex globally:
\[(.*?)\]
Explanation:
\[ : [ is a meta char and needs to be escaped if you want to match it literally.
(.*?) : match everything in a non-greedy way and capture it.
\] : ] is a meta char and needs to be escaped if you want to match it literally.
(?<=\[).+?(?=\])
Will capture content without brackets
(?<=\[) - positive lookbehind for [
.*? - non greedy match for the content
(?=\]) - positive lookahead for ]
EDIT: for nested brackets the below regex should work:
(\[(?:\[??[^\[]*?\]))
This should work out ok:
\[([^]]+)\]
Can brackets be nested?
If not: \[([^]]+)\] matches one item, including square brackets. Backreference \1 will contain the item to be match. If your regex flavor supports lookaround, use
(?<=\[)[^]]+(?=\])
This will only match the item inside brackets.
To match a substring between the first [ and last ], you may use
\[.*\] # Including open/close brackets
\[(.*)\] # Excluding open/close brackets (using a capturing group)
(?<=\[).*(?=\]) # Excluding open/close brackets (using lookarounds)
See a regex demo and a regex demo #2.
Use the following expressions to match strings between the closest square brackets:
Including the brackets:
\[[^][]*] - PCRE, Python re/regex, .NET, Golang, POSIX (grep, sed, bash)
\[[^\][]*] - ECMAScript (JavaScript, C++ std::regex, VBA RegExp)
\[[^\]\[]*] - Java, ICU regex
\[[^\]\[]*\] - Onigmo (Ruby, requires escaping of brackets everywhere)
Excluding the brackets:
(?<=\[)[^][]*(?=]) - PCRE, Python re/regex, .NET (C#, etc.), JGSoft Software
\[([^][]*)] - Bash, Golang - capture the contents between the square brackets with a pair of unescaped parentheses, also see below
\[([^\][]*)] - JavaScript, C++ std::regex, VBA RegExp
(?<=\[)[^\]\[]*(?=]) - Java regex, ICU (R stringr)
(?<=\[)[^\]\[]*(?=\]) - Onigmo (Ruby, requires escaping of brackets everywhere)
NOTE: * matches 0 or more characters, use + to match 1 or more to avoid empty string matches in the resulting list/array.
Whenever both lookaround support is available, the above solutions rely on them to exclude the leading/trailing open/close bracket. Otherwise, rely on capturing groups (links to most common solutions in some languages have been provided).
If you need to match nested parentheses, you may see the solutions in the Regular expression to match balanced parentheses thread and replace the round brackets with the square ones to get the necessary functionality. You should use capturing groups to access the contents with open/close bracket excluded:
\[((?:[^][]++|(?R))*)] - PHP PCRE
\[((?>[^][]+|(?<o>)\[|(?<-o>]))*)] - .NET demo
\[(?:[^\]\[]++|(\g<0>))*\] - Onigmo (Ruby) demo
If you do not want to include the brackets in the match, here's the regex: (?<=\[).*?(?=\])
Let's break it down
The . matches any character except for line terminators. The ?= is a positive lookahead. A positive lookahead finds a string when a certain string comes after it. The ?<= is a positive lookbehind. A positive lookbehind finds a string when a certain string precedes it. To quote this,
Look ahead positive (?=)
Find expression A where expression B follows:
A(?=B)
Look behind positive (?<=)
Find expression A where expression B
precedes:
(?<=B)A
The Alternative
If your regex engine does not support lookaheads and lookbehinds, then you can use the regex \[(.*?)\] to capture the innards of the brackets in a group and then you can manipulate the group as necessary.
How does this regex work?
The parentheses capture the characters in a group. The .*? gets all of the characters between the brackets (except for line terminators, unless you have the s flag enabled) in a way that is not greedy.
Just in case, you might have had unbalanced brackets, you can likely design some expression with recursion similar to,
\[(([^\]\[]+)|(?R))*+\]
which of course, it would relate to the language or RegEx engine that you might be using.
RegEx Demo 1
Other than that,
\[([^\]\[\r\n]*)\]
RegEx Demo 2
or,
(?<=\[)[^\]\[\r\n]*(?=\])
RegEx Demo 3
are good options to explore.
If you wish to simplify/modify/explore the expression, it's been explained on the top right panel of regex101.com. If you'd like, you can also watch in this link, how it would match against some sample inputs.
RegEx Circuit
jex.im visualizes regular expressions:
Test
const regex = /\[([^\]\[\r\n]*)\]/gm;
const str = `This is a [sample] string with [some] special words. [another one]
This is a [sample string with [some special words. [another one
This is a [sample[sample]] string with [[some][some]] special words. [[another one]]`;
let m;
while ((m = regex.exec(str)) !== null) {
// This is necessary to avoid infinite loops with zero-width matches
if (m.index === regex.lastIndex) {
regex.lastIndex++;
}
// The result can be accessed through the `m`-variable.
m.forEach((match, groupIndex) => {
console.log(`Found match, group ${groupIndex}: ${match}`);
});
}
Source
Regular expression to match balanced parentheses
(?<=\[).*?(?=\]) works good as per explanation given above. Here's a Python example:
import re
str = "Pagination.go('formPagination_bottom',2,'Page',true,'1',null,'2013')"
re.search('(?<=\[).*?(?=\])', str).group()
"'formPagination_bottom',2,'Page',true,'1',null,'2013'"
The #Tim Pietzcker's answer here
(?<=\[)[^]]+(?=\])
is almost the one I've been looking for. But there is one issue that some legacy browsers can fail on positive lookbehind.
So I had to made my day by myself :). I manged to write this:
/([^[]+(?=]))/g
Maybe it will help someone.
console.log("this is a [sample] string with [some] special words. [another one]".match(/([^[]+(?=]))/g));
if you want fillter only small alphabet letter between square bracket a-z
(\[[a-z]*\])
if you want small and caps letter a-zA-Z
(\[[a-zA-Z]*\])
if you want small caps and number letter a-zA-Z0-9
(\[[a-zA-Z0-9]*\])
if you want everything between square bracket
if you want text , number and symbols
(\[.*\])
This code will extract the content between square brackets and parentheses
(?:(?<=\().+?(?=\))|(?<=\[).+?(?=\]))
(?: non capturing group
(?<=\().+?(?=\)) positive lookbehind and lookahead to extract the text between parentheses
| or
(?<=\[).+?(?=\]) positive lookbehind and lookahead to extract the text between square brackets
In R, try:
x <- 'foo[bar]baz'
str_replace(x, ".*?\\[(.*?)\\].*", "\\1")
[1] "bar"
([[][a-z \s]+[]])
Above should work given the following explaination
characters within square brackets[] defines characte class which means pattern should match atleast one charcater mentioned within square brackets
\s specifies a space
+ means atleast one of the character mentioned previously to +.
I needed including newlines and including the brackets
\[[\s\S]+\]
If someone wants to match and select a string containing one or more dots inside square brackets like "[fu.bar]" use the following:
(?<=\[)(\w+\.\w+.*?)(?=\])
Regex Tester
What is the regex that matches these examples(6 characters, first is a letter, others are numbers):
u78945 - valid
s56123 - valid
456a12 - invalid
78561d - invalid
1234567 - invalid
i don't know if regular expressions are the same for every programming language. I need it for Regular Expression Validator control using VB ASP.NET.
Use this pattern:
^[a-z][0-9]{5}$
This will match any Latin letter (lower-case unless using case-insensitive matching) followed by 5 decimal digits.
Note: You could use \d instead of [0-9], but read this for an explanation about why they are different.
[a-zA-Z]\d{5}
If you are searching explicitly from the beginning of the line use ^
^[a-zA-Z]\d{5}
and append $ for the end of the line.
^[a(?i)-z(?i)]\d{5}$
The (?i) code enables the expression to accept any letter without case-sensitivity. The \d{5} looks for a sequence of numbers whose length is exactly 5.
I'm trying to complete a regular expression that will pull out matches based on their opening and closing characters, the closest I've gotten is
^(\[\[)[a-zA-Z.-_]+(\]\])
Which will match a string such as "[[word1]]" and bring me back all the matches if there is more than one, The problem is I want it to pick up matchs where there may be a space in so for example "[[word1 word2]]", now this will work if I add a space into my pattern above however this pops up a problem that it will only get one match for my entire string so for example if I have a string
"Hi [[Title]] [[Name]] [[surname]], How are you"
then the match will be [[Title]] [[Name]] [[surname]] rather than 3 matches [[Title]], [[Name]], [[surname]]. I'm sure I'm just a char or two away in the Regex but I'm stuck, How can I make it return the 3 matches.
Thanks
You just need to make you regex non-greedy by using a ? like:
^(\[\[)[a-zA-Z.-_ ]+?(\]\])
Also there is a bug in your regex. You've included - in the char class thinking of it as a literal hyphen. But - in a char class is a meta char. So it effectively will match all char between . (period) and _ (underscore). So you need to escape it as:
^(\[\[)[a-zA-Z.\-_ ]+?(\]\])
or you can put is in some other place in the regex so that it will not have things on both sides of it as:
^(\[\[)[a-zA-Z._ -]+?(\]\])
or
^(\[\[)[-a-zA-Z._ ]+?(\]\])
You need to turn off greedy matching. See these examples for different languages:
asp.net
java
javascript
You should use +? instead of +.
The one without the question mark will try to match as much as possible, while the one with the question mark as little as possible.
Another approach would be to use [^\]] as your characters instead of [a-zA-Z.-_]. That way, a match will never extend over your closing brackets.