I'm trying to search a string in PHP using preg_match(), to return a string that takes the format of 'xdr', where x is a single digit.
The string is otherwise made of characters, numbers, '.' full stops and spaces.
I'm hopeless with regular expressions, can somebody help me out?
I've tried ^([a-zA-Z0-9]+)\s2dr$ but it doesn't work
For example, the string might look like A really big 2.0 was 3dr once
As far as I understand your needs, this regex will work to match a string that contains some number followed by dr:
/\b\d+dr\b/
in action in preg_match:
preg_match('/\b\d+dr\b/', $string);
explanation:
/ : regex delim
\b : word boundary
\d+ : one or more digit
dr : literaly dr
\b : word boundary
/ : regex delim
If you want the other characters to be only characters, numbers, '.' full stops and spaces use this:
preg_match('/^[\w\d. ]*\b\d+dr\b[\w\d. ]*$/', $string);
And, to be unicode compatible:
preg_match('/^[\pL\pN. ]*\b\pN+dr\b[\pL\pN. ]*$/', $string);
You're close.
The following regex will match a single digit followed by one or more alpha characters, numbers, full stops and spaces, and ends in dr.
^\d[a-zA-Z0-9. ]+dr$
Note: Not sure if you want the literal dr, if not, drop it from the end.
Related
I need (in PHP) to split a sententse by the word that cannot be the first or the last one in the sentence. Say the word is "pression" and here is my regex
/^.+?(?=[\s\.\,\:\;])pression(?=[\s\.\,\:\;]).+$/i
Live here: https://regex101.com/r/CHAhKj/1/
First, it doesn't match.
Next, I think - it is at all possible to split that way? I tryed simplified example
print_r(preg_split('/^.+pizza.+$/', 'my pizza is cool'));
live here http://sandbox.onlinephpfunctions.com/code/10b674900fc1ef44ec79bfaf80e83fe1f4248d02
and it prints an array of 2 empty strings, when I expect
['my ', ' is cool']
I need (in PHP) to split a sentence by the word that cannot be the first or the last one in the sentence
You may use this regex:
(?<=[^\s.?]\h)pression(?=\h[^\s.?])
RegEx Demo
RegEx Details:
(?<=[^\s.?]\h): Lookbehind to assert that ahead of current position we have a space and a character that not a whitespace, not a dot and not a ?.
pression: Match word pression
(?=\h[^\s.?]): Lookahead to assert that before current position we have a space and a character that not a whitespace, not a dot and not a ?
First, ^.+?(?=[\s\.\,\:\;])pression(?=[\s\.\,\:\;]).+$ can't match any string at all because the (?=[\s\.\,\:\;])p part requires p to be also either a whitespace char, or a ., ,, : or ;, which invalidates the whole match at once.
Second, ^.+pizza.+$ pattern does not ensure the pizza matched is not the first or last word in a sentence as . matches whitespace, too. It does not return anything meaningful, because preg_split uses the match to break string into chunks, and the two empty values are 1) start of string and 2) empty string positions.
That said, all you need is:
preg_match('~^(.*?\w\W+)pression(\W+\w.*)$~is', $text, $m)
See the regex demo. Details:
^ - start of string
(.*?\w\W+) - Capturing group 1: any zero or more chars, as few as possible, then a word char and then one or more non-word chars
pression - a word
(\W+\w.*) - Capturing group 2: one or more non-word chars, a word char, and then any zero or more chars as many as possible
$ - end of string.
s makes the . match across lines and i flag makes the pattern match in a case insensitive way.
See the PHP demo:
$text = "You can use any regular expression pression inside the lookahead ";
if (preg_match('~^(.*?\w\W+)pression(\W+\w.*)$~is', $text, $m)) {
echo $m[1] . " << | >> " . $m[2];
}
// => You can use any regular expression << | >> inside the lookahead
I have a regular expression to escape all special characters in a search string. This works great, however I can't seem to get it to work with word boundaries. For example, with the haystack
add +
or
add (+)
and the needle
+
the regular expression /\+/gi matches the "+". However the regular expression /\b\+/gi doesn't. Any ideas on how to make this work?
Using
add (plus)
as the haystack and /\bplus/gi as the regex, it matches fine. I just can't figure out why the escaped characters are having problems.
\b is a zero-width assertion: it doesn't consume any characters, it just asserts that a certain condition holds at a given position. A word boundary asserts that the position is either preceded by a word character and not followed by one, or followed by a word character and not preceded by one. (A "word character" is a letter, a digit, or an underscore.) In your string:
add +
...there's a word boundary at the beginning because the a is not preceded by a word character, and there's one after the second d because it's not followed by a word character. The \b in your regex (/\b\+/) is trying to match between the space and the +, which doesn't work because neither of those is a word character.
Try changing it to:
/\b\s?+/gi
Edit:
Extend this concept as far as you want. If you want the first + after any word boundary:
/\b[^+]*+/gi
Boundaries are very conditional assertions; what they anchor depends on what they touch. See this answer for a detailed explanation, along with what else you can do to deal with it.
So basically, I have a big string with some other information, and somewhere at the end, I have the following structure of a string:
62AC979D-5277D720
It is numbers and uppercase letters. I would like to extract this substring from many lines of the bigger strings which all contain it at different places. I have tried:
preg_match('/^[\w]+$/', $string);
But I really don't have much experience with regular expressions. Can someone provide the regex necessary or at least tell me where I am mistaken? Thank you for your time!
This regex should do it for you,
([A-Z\d]{8}-[A-Z\d]{8})
in use
<?php
$string = 'This is 62AC979D-5277D720 the whole string.';
preg_match_all('~([A-Z\d]{8}-[A-Z\d]{8})~', $string, $value);
print_r($value[1]);
Your current regex fails I suspect because of the ^ and $. These mark the start and end of the string you are searching for (or line if the m modifier is used). The \w is also a-z, A-Z, 0-9 and _. I think you only care about capital letters and you want to allow only one dash. If the target will also always only be 8 characters you can add the {8} in place of the +. The () are to capture the value that is found. The first found value in $string will be $value[1][0].
Demo: http://sandbox.onlinephpfunctions.com/code/c6b2c391d95c5454a3c7ea81d5ac4a3bb8e49aef
preg_match_all('/\\b[0-9A-Z]+-[0-9A-Z]+\\b/')
This should do it for you.
preg_match('/\\b[0-9A-Z]{8}-[0-9A-Z]{8}\\b/', $string);
This works for the string you gave i.e 8 numbers or alphabets followed by - and then numbers and alphabets again
You try this.
preg_match('/^[0-9A-Z]{8}-[0-9A-Z]{8}$/', $string)
Could you help me with PHP function/regex that in given text finds all words starting with character ":" ?
..in other words all substrings that start with ":" and are separated with " " (a space)
Since :word should probably be valid, and I guess :word:another should be considered two words, then you cannot say that there is always a space.
Words in natural languages can be followed by dots and other characters.
In digital input, they can be followed by end of line.
I suggest using this regexp:
~:\w+~
It takes any : character followed by at least one alpha character and will end at any character that is not valid letter.
Example: on RegExr.com
You can also try ~:\w+\b~, where \b is word boundary (literally end of word), but I see it not necessary here.
Note: \w stands for [a-zA-Z0-9_] meaning it catches underscores _ and digits 0-9 as well. It works pretty much like variable/function naming in PHP
EDIT (some notes on usage):
You said that in given text (I understand that like input with random things) you want to extract all words prepended with :, for example :word. To do that easily, you should use preg_match_all() function with PREG_PATTERN_ORDER flag.
Example:
$regex = '~(:\w+)~';
if (preg_match_all($regex, $input, $matches, PREG_PATTERN_ORDER)) {
foreach ($matches[1] as $word) {
echo $word .'<br/>';
}
}
regex: /:\w+\s/g
\w Matchs any word character
\s Matchs whitespace character
This would work:
preg_match('/^:\w*\s$/g', $var);
Sorry, because I don't use PHP. But I suppose that your problem is that PHP would have reserved the character ":" for some reason in its regex implementation ?
Well, in that case, you still can catch any word beginning with ":" and ending with some space this way:
(...)
match('^\x3A[.]*[\s]');
("3A" is hexadecimal value for 58, which is the ASCII code for ":")
This should work, I think...
How would I go about removing numbers and a space from the start of a string?
For example, from '13 Adam Court, Cannock' remove '13 '
Because everyone else is going the \d+\s route I'll give you the brain-dead answer
$str = preg_replace("#([0-9]+ )#","",$str);
Word to the wise, don't use / as your delimiter in regex, you will experience the dreaded leaning-toothpick-problem when trying to do file paths or something like http://
:)
Use the same regex I gave in my JavaScript answer, but apply it using preg_replace():
preg_replace('/^\d+\s+/', '', $str);
Try this one :
^\d+ (.*)$
Like this :
preg_replace ("^\d+ (.*)$", "$1" , $string);
Resources :
preg_replace
regular-expressions.info
On the same topic :
Regular expression to remove number, then a space?
regular expression for matching number and spaces.
I'd use
/^\d+\s+/
It looks for a number of any size in the beginning of a string ^\d+
Then looks for a patch of whitespace after it \s+
When you use a backslash before certain letters it represents something...
\d represents a digit 0,1,2,3,4,5,6,7,8,9.
\s represents a space .
Add a plus sign (+) to the end and you can have...
\d+ a series of digits (number)
\s+ multiple spaces (typos etc.)
The same regex I gave you on your other question still applies. You just have to use preg_replace() instead.
Search for /^[\s\d]+/ and replace with the empty string. Eg:
$str = preg_replace(/^[\s\d]+/, '', $str);
This will remove digits and spaces in any order from the beginning of the string. For something that removes only a number followed by spaces, see BoltClock's answer.
If the input strings all have the same ecpected format and you will receive the same result from left trimming all numbers and spaces (no matter the order of their occurrence at the front of the string), then you don't actually need to fire up the regex engine.
I love regex, but know not to use it unless it provides a valuable advantage over a non-regex technique. Regex is often slower than non-regex techniques.
Use ltrim() with a character mask that includes spaces and digits.
Code: (Demo)
var_export(
ltrim('420 911 90210 666 keep this part', ' 0..9')
);
Output:
'keep this part'
It wouldn't matter if the string started with a space either. ltrim() will greedily remove all instances of spaces or numbers from the start of the string intil it can't anymore.