PHP regular expression issue preg_match() on timestamps - php

Facing issue with regular expression
2013-05-29 15:15:12 string I am matching with /^(\d{4})-(\d{2})-(\d{2})({\s}+(\d{2}):(\d{2}):(\d{2}))?$/ with preg_match but not validating ... its giving false.
What should be regexp to match 2013-05-29 15:15:12 or 2013-05-29 pattern.

Let's take a look at your regex first. Between the date and the time you're matching {\s}+. This means "the character {, followed by a space/tab, followed by one or more }'s".
Replace {\s} with ?:\s+ (a non capturing group matching one or more spaces/tabs) so the full regex is
^(\d{4})-(\d{2})-(\d{2})(?:\s+(\d{2}):(\d{2}):(\d{2}))?$
DEMO

The {\s}+ is wrong. It should be \s+. The curly brackets are used as quantifiers or literals only.

Related

how can i using Regex for find string covered in [ ]? [duplicate]

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

Regex match section within string

I have a string foo-foo-AB1234-foo-AB12345678. The string can be in any format, is there a way of matching only the following pattern letter,letter,digits 3-5 ?
I have the following implementation:
preg_match_all('/[A-Za-z]{2}[0-9]{3,6}/', $string, $matches);
Unfortunately this finds a match on AB1234 AND AB12345678 which has more than 6 digits. I only wish to find a match on AB1234 in this instance.
I tried:
preg_match_all('/^[A-Za-z]{2}[0-9]{3,6}$/', $string, $matches);
You will notice ^ and $ to mark the beginning and end, but this only applies to the string, not the section, therefore no match is found.
I understand why the code is behaving like it is. It makes logical sense. I can't figure out the solution though.
You must be looking for word boundaries \b:
\b\p{L}{2}\p{N}{3,5}\b
See demo
Note that \p{L} matches a Unicode letter, and \p{N} matches a Unicode number.
You can as well use your modified regex \b[a-zA-Z]{2}[0-9]{3,5}\b. Note that using anchors makes your regex match only at the beginning of a string (with ^) or/and at the end of the string (with $).
In case you have underscored words (like foo-foo_AB1234_foo_AB12345678_string), you will need a slight modification:
(?<=\b|_)\p{L}{2}\p{N}{3,5}(?=\b|_)
You have to end your regular expression with a pattern for a non-digit. In Java this would be \D, this should be the same in PHP.

PHP Regex extract numbers inside brackets

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.

Character classes strange behavior in alternations in regular expressions

I'm trying to write a simple regular expression that recognizes a sequence of characters that are not columns or are escaped columns.
I.e:
foo:bar //Does not match
but
foo\:bar //Does match
By my knowledge of Regular Languages, such language can be described by the regular expression
/([^:]|\\[:])*/
You can see a graphical representation of this expression in the wonderful tool Regexper
Using php's preg_match (that is based on the PCRE engine), such expression does not match "foo\:bar".
However, if substitute the class with the single char:
/([^:]|\\:)*/
the expression matches.
Do you have an explanation for this? Is this a sort of limitation of the PCRE engine on character classes?
PS: Testing the first expression on RegExr, that is based on AS3 Regexp engine, does not offer a match, while changing the alternation order:
/(\\[:]|[^:])*/
it does match, while the same expression does not match in PCRE.
preg_match() accepts a regular expression pattern as a string, so you need to double escape everything.
^(?:[^:\\\\]|\\\\:)+$
This matches one or more characters that are not colons or escape characters [^:\\\\], or an escaped colon \\\\:.
Why your first regular expression didn't work: /([^:]|\\[:])*/.
This matches a non-colon [^:], or it matches \\[:] which matches a literal [ followed by a literal : and then a literal ].
Why this works : /([^:]|\\:)*/ ?
This matches a non-colon [^:], or it matches a literal \\: so it effectively matches everything.
Edit: Why /([^:]|E[:])*/ won't match fooE:bar ?
This is what happens: [^:] matches the f then it matches o then the other o then it matches the E, now it finds a colon : and it can't match it, but since by default the PCRE engine doesn't look for the longest possible match it is satisfied with what is has matched so far and stops right there and returns fooE as a match without trying the other alternative E[:] (which is equal by the way to E:) at all.
If you want to match the entire sequence then you will to use an expression like this one:
/([^:E]|E[:])*/
This prevents [^:] from consuming that E.
You can try this. This allow the secuence \\: to have a chance before the negated character class [^:].
^(?:\\:|[^:])+$
If you use the values in the alternation bar inverted as in ^((?:[^:]|\\:)+$ it will not match escaped colon \: because the first alternative will consume the slash (\) before the second expression have a chance to try.

Regex to detect the colon and sides of it?

First see my string please:
$a = "[ child : parent ]";
How can I detect that the pattern is:
[(optional space)word or character(optional space) : (optional space)word or character(optional space)]
You can catch this as follows in PHP:
Your regular expression is /\[ *\w+ *: *\w+ *]/
You would write code that would look like this to see if it matched.
if (preg_match('/regex/', $string)) {
// do things
}
Explanation of the Regular Expression
There is a backslash (\) before the open bracket because
[ has special meaning in regular expressions. The backslash
prevents its special meaning from being used.
The asterisk (*) matches 0 or more of the previous character expression. In this
case, it matches 0 or more spaces. If you instead used the
expression \s*, it would match 0 or more white-space characters
(space, tab, line break). Finally, if you wanted it to match 0 or 1
of the previous character, you would use ? instead of *.
The plus (+) matches 1 or more of the previous character expression. The \w character expression matches a letter, digit, or underscore. If you don't want underscores to match, you should instead use a character class. For example, you could use [A-Za-z0-9].
You can find more information on regular expressions at http://www.regular-expressions.info and http://www.regular-expressions.info/php.html
From your sample text I'd say you mean a human word and not \w regex word
preg_match('/\[ ?([a-z]+) ?: ?([a-z]+) ?\]/i', $a, $matches);
Explained demo: http://regex101.com/r/hB2oV9
$matches will save both values, test with var_dump($matches);
I'm not sure on the php-specific version of regex, but this should work:
\[ ?\w+ ? : ?\w+ ?\]
Here is a site that I've used in the past to find regular expressions for my needed patterns.
use this regex \[\s*\w+\s*:\s*\w+\s*\]
I would probably do it like this
preg_match('/^\[\s?\w+\s+:\s+\w+\s?\]$/', $string)

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