Suppose I have a string that looks like:
"lets refer to [[merp] [that entry called merp]] and maybe also to that entry called [[blue] [blue]]"
The idea here is to replace a block of [[name][some text]] with some text.
So I'm trying to use regular expressions to find blocks that look like [[name][some text]], but I'm having tremendous difficulty.
Here's what I thought should work (in PHP):
preg_match_all('/\[\[.*\]\[.*\]/', $my_big_string, $matches)
But this just returns a single match, the string from '[[merp' to 'blue]]'. How can I get it to return the two matches [[merp][that entry called merp]] and [[blue][blue]]?
The regex you're looking for is \[\[(.+?)\]\s\[(.+?)\]\] and replace it with $2
The regex pattern matched inside the () braces are captured and can be back-referenced using $1, $2,...
Example on regex101.com
Quantifiers like the * are by default greedy,
which means, that as much as possible is matched to meet conditions. E.g. in your sample a regex like \[.*\] would match everything from the first [ to the last ] in the string. To change the default behaviour and make quantifiers lazy (ungreedy, reluctant):
Use the U (PCRE_UNGREEDY) modifier to make all quantifiers lazy
Put a ? after a specific quantifier. E.g. .*? as few of any characters as possible
1.) Using the U-modifier a pattern could look like:
/\[\[(.*)]\s*\[(.*)]]/Us
Additional used the s (PCRE_DOTALL) modifier to make the . dot also match newlines. And added some \s whitespaces in between ][ which are in your sample string. \s is a shorthand for [ \t\r\n\f].
There are two capturing groups (.*) to be replaced then. Test on regex101.com
2.) Instead using the ? to making each quantifier lazy:
/\[\[(.*?)]\s*\[(.*?)]]/s
Test on regex101.com
3.) Alternative without modifiers, if no square brackets are expected to be inside [...].
/\[\[([^]]*)]\s*\[([^]]*)]]/
Using a ^ negated character class to allow [^]]* any amount of characters, that are NOT ] in between [ and ]. This wouldn't require to rely on greediness. Also no . is used, so no s-modifier is needed.
Test on regex101.com
Replacement for all 3 examples according to your sample: \2 where \1 correspond matches of the first parenthesized group,...
Related
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
Given a text string (a markdown document) I need to achieve one of this two options:
to replace all the matches of a particular expression ((\W)(theWord)(\W)) all across the document EXCEPT the matches that are inside a markdown image syntax .
to replace all the matches of a particular expression ({{([^}}]+)}}\[\[[^\]\]]+\]\]) ONLY inside the markdown images, ie.: ![Blah {{theWord}}[[1234]] blah](url).
Both expressions are currently matching everything, no matter if inside the markdown image syntax or not, and I've already tried everything I could think.
Here is an example of the first option
And here is an example of the second option
Any help and/or clue will be highly appreciated.
Thanks in advance!
Well I modified first expression a little bit as I thought there are some extra capturing groups then made them by adding a lookahead trick:
-First one (Live demo):
\b(vitae)\b(?![^[]*]\s*\()
-Second one (Live demo):
{{([^}}]+)}}\[\[[^\]\]]+\]\](?=[^[]*]\s*\()
Lookahead part explanations:
(?! # Starting a negative lookahead
[^[]*] # Everything that's between brackets
\s* # Any whitespace
\( # Check if it's followed by an opening parentheses
) # End of lookahead which confirms the whole expression doesn't match between brackets
(?= means a positive lookahead
You can leverage the discard technique that it really useful for this cases. It consists of having below pattern:
patternToSkip1 (*SKIP)(*FAIL)|patternToSkip2 (*SKIP)(*FAIL)| MATCH THIS PATTERN
So, according you needs:
to replace all the matches of a particular expression ((\W)(theWord)(\W)) all across the document EXCEPT the matches that are inside a markdown image syntax
You can easily achieve this in pcre through (*SKIP)(*FAIL) flags, so for you case you can use a regex like this:
\[.*?\](*SKIP)(*FAIL)|\bTheWord\b
Or using your pattern:
\[.*?\](*SKIP)(*FAIL)|(\W)(theWord)(\W)
The idea behind this regex is tell regex engine to skip the content within [...]
Working demo
The first regex is easily fixed with a SKIP-FAIL trick:
\!\[.*?\]\(http[^)]*\)(*SKIP)(*FAIL)|\bvitae\b
To replace with the word of your choice. It is a totally valid way in PHP (PCRE) regex to match something outside some markers.
See Demo 1
As for the second one, it is harder, but acheivable with \G that ensures we match consecutively inside some markers:
(\!\[.*?|(?<!^)\G)((?>(?!\]\(http).)*?){{([^}]+?)}}\[{2}[^]]+?\]{2}(?=.*?\]\(http[^)]*?\))
To replace with $1$2{{NEW_REPLACED_TEXT}}[[NEW_DIGITS]]
See Demo 2
PHP:
$re1 = "#\!\[.*?\]\(http[^)]*\)(*SKIP)(*FAIL)|\bvitae\b#i";
$re2 = "#(\!\[.*?|(?<!^)\G)((?>(?!\]\(http).)*?){{([^}]+?)}}\[{2}[^]]+?\]{2}(?=.*?\]\(http[^)]*?\))#i";
I have a string. An example might be "Contact /u/someone on reddit, or visit /r/subreddit or /r/subreddit2"
I want to replace any instance of "/r/x" and "/u/x" with "[/r/x](http://reddit.com/r/x)" and "[/u/x](http://reddit.com/u/x)" basically.
So I'm not sure how to 1) find "/r/" and then expand that to the rest of the word (until there's a space), then 2) take that full "/r/x" and replace with my pattern, and most importantly 3) do this for all "/r/" and "/u/" matches in a single go...
The only way I know to do this would be to write a function to walk the string, character by character, until I found "/", then look for "r" and "/" to follow; then keep going until I found a space. That would give me the beginning and ending characters, so I could do a string replacement; then calculate the new end point, and continue walking the string.
This feels... dumb. I have a feeling there's a relatively simple way to do this, and I just don't know how to google to get all the relevant parts.
A simple preg_replace will do what you want.
Try:
$string = preg_replace('#(/(?:u|r)/[a-zA-Z0-9_-]+)#', '[\1](http://reddit.com\1)', $string);
Here is an example: http://ideone.com/dvz2zB
You should see if you can discover what characters are valid in a Reddit name or in a Reddit username and modify the [a-zA-Z0-9_-] charset accordingly.
You are looking for a regular expression.
A basic pattern starts out as a fixed string. /u/ or /r/ which would match those exactly. This can be simplified to match one or another with /(?:u|r)/ which would match the same as those two patterns. Next you would want to match everything from that point up to a space. You would use a negative character group [^ ] which will match any character that is not a space, and apply a modifier, *, to match as many characters as possible that match that group. /(?:u|r)/[^ ]*
You can take that pattern further and add a lookbehind, (?<= ) to ensure your match is preceded by a space so you're not matching a partial which results in (?<= )/(?:u|r)/[^ ]*. You wrap all of that to make a capturing group ((?<= )/(?:u|r)/[^ ]*). This will capture the contents within the parenthesis to allow for a replacement pattern. You can express your chosen replacement using the \1 reference to the first captured group as [\1](http://reddit.com\1).
In php you would pass the matching pattern, replacement pattern, and subject string to the preg_replace function.
In my opinion regex would be an overkill for such a simple operation. If you just want to replace instance of "/r/x" with "[r/x](http://reddit.com/r/x)" and "/u/x" with "[/u/x](http://reddit.com/u/x)" you should use str_replace although with preg_replace it'll lessen the code.
str_replace("/r/x","[/r/x](http://reddit.com/r/x)","whatever_string");
use regex for intricate search string and replace. you can also use http://www.jslab.dk/tools.regex.php regular expression generator if you have something complex to capture in the string.
I'm currently building a chat system with reply function.
How can I match the numbers inside the '#' symbol and brackets, example: #[123456789]
This one works in JavaScript
/#\[(0-9_)+\]/g
But it doesn't work in PHP as it cannot recognize the /g modifier. So I tried this:
/\#\[[^0-9]\]/
I have the following example code:
$example_message = 'Hi #[123456789] :)';
$msg = preg_replace('/\#\[[^0-9]\]/', '$1', $example_message);
But it doesn't work, it won't capture those numbers inside #[ ]. Any suggestions? Thanks
You have some core problems in your regex, the main one being the ^ that negates your character class. So instead of [^0-9] matching any digit, it matches anything but a digit. Also, the g modifier doesn't exist in PHP (preg_replace() replaces globally and you can use preg_match_all() to match expressions globally).
You'll want to use a regex like /#\[(\d+)\]/ to match (with a group) all of the digits between #[ and ].
To do this globally on a string in PHP, use preg_match_all():
preg_match_all('/#\[(\d+)\]/', 'Hi #[123456789] :)', $matches);
var_dump($matches);
However, your code would be cleaner if you didn't rely on a match group (\d+). Instead you can use "lookarounds" like: (?<=#\[)\d+(?=\]). Also, if you will only have one digit per string, you should use preg_match() not preg_match_all().
Note: I left the example vague and linked to lots of documentation so you can read/learn better. If you have any questions, please ask. Also, if you want a better explanation on the regular expressions used (specifically the second one with lookarounds), let me know and I'll gladly elaborate.
Use the preg_match_all function in PHP if you’d like to produce the behaviour of the g modifier in Javascript. Use the preg_match function otherwise.
preg_match_all("/#\\[([0-9]+)\\]/", $example_message, $matches);
Explanation:
/ opening delimiter
# match the at sign
\\[ match the opening square bracket (metacharacter, so needs to be escaped)
( start capturing
[0-9] match a digit
+ match the previous once or more
) stop capturing
\\] match the closing square bracket (metacharacter, so needs to be escaped)
/ closing delimiter
Now $matches[1] contains all the numbers inside the square brackets.
I am tired of being frightened of regular expressions. The topic of this post is limited to PHP implementation of regular expressions, however, any generic regular expression advice would obviously be appreciated (i.e. don't confuse me with scope that is not applicable to PHP).
The following (I believe) will remove any whitespace between numbers. Maybe there is a better way to do so, but I still want to understand what is going on.
$pat="/\b(\d+)\s+(?=\d+\b)/";
$sub="123 345";
$string=preg_replace($pat, "$1", $sub);
Going through the pattern, my interpretation is:
\b A word boundary
\d+ A subpattern of 1 or more digits
\s+ One or more whitespaces
(?=\d+\b) Lookahead assertion of one or more digit followed by a word boundary?
Putting it all together, search for any word boundary followed by one or more digits and then some whitespace, and then do some sort of lookahead assertion on it, and save the results in $1 so it can replace the pattern?
Questions:
Is my above interpretation correct?
What is that lookahead assertion all about?
What is the purpose of the leading / and trailing /?
Is my above interpretation correct?
Yes, your interpretation is correct.
What is that lookahead assertion all about?
That lookahead assertion is a way for you to match characters that have a certain pattern in front of them, without actually having to match the pattern.
So basically, using the regex abcd(?=e) to match the string abcde will give you the match: abcd.
The reason that this matches is that the string abcde does in fact contain:
An a
Followed by a b
Followed by a c
Followed by a d that has an e after it (this is a single character!)
It is important to note that after the 4th item it also contains an actual "e" character, which we didn't match.
On the other hand, trying to match the string against the regex abcd(?=f) will fail, since the sequence:
"a", followed by "b", followed by "c", followed by "d that has an f in front of it"
is not found.
What is the purpose of the leading / and trailing /
Those are delimiters, and are used in PHP to distinguish the pattern part of your string from the modifier part of your string. A delimiter can be any character, although I prefer # signs myself. Remember that the character you are using as a delimiter needs to be escaped if it is used in your pattern.
It would be a good idea to watch this video, and the 4 that follow this:
http://blog.themeforest.net/screencasts/regular-expressions-for-dummies/
The rest of the series is found here:
http://blog.themeforest.net/?s=regex+for+dummies
A colleague sent me the series and after watching them all I was much more comfortable using Regular Expressions.
Another good idea would be installing RegexBuddy or Regexr. Especially RegexBuddy is very useful for understanding the workings of a regular expression.