I want to remove the brackets with its content using preg_replace(), but i am unable to use a lazy(non-greedy) in the pattern since the end bracket is the end character, the text in between the brackets is always a random character length and can contain numbers, underscores, and hyphens.
code-
$array = array(
"Text i want to keep (txt to remove)",
"Random txt (some more random txt)",
"Keep this (remove)",
"I like bananas (txt)"
);
$pattern = "#pattern#";
foreach($array as $new_txt){
$new_outputs .= preg_replace($pattern, '', $new_txt)."\n";
}
echo $new_outputs;
Wanted output-
Text i want to keep
Random txt
Keep this
I like bananas
I do not use regular expressions much and couldn't find anything to solve my problem.
The following regular expression should do it:
$pattern = '#\(.*?\)#';
.*? is a non-greedy match of anything.
$new_outputs .= preg_replace('#\([^\)]*\)$#','',$new_txt);
This might help you:
$pattern = "/\([^)]*\)+/";
foreach($array as $new_txt){
$new_outputs .= preg_replace($pattern, '', $new_txt)."\n";
}
Related
$text = 'Hello #demo here!';
$pattern = '/#(.*?)[ ]/';
$replacement = '<strong>${1}</strong> ';
echo preg_replace($pattern, $replacement, $text);
This works, I get HTML like this: Hello <strong>demo</strong> here!. But this not works, when that #demo is at the end of string, example: $text = 'Hello #demo';. How can I change my pattern, so it will return same output whenever it is end of the string or not.
Question 2:
What if the string is like $text = 'Hello #demo!';, so it will not put ! as bolded text? Just catch space, end of string or not real-word.
Sorry for bad English, hope you know what I need.
In order to select a word beginning with the # symbol, this regex will work:
$pattern = "/#(\w+)\b/"
`\w` is a short hand character class for `[a-zA-Z0-9_]`. `\b` is an anchor for the beginning or end of a word, in this case the end. So the regex is saying: select something starting with an '#' followed by one or more word characters until the end of the word is reached.
Reference: http://www.regular-expressions.info/tutorial.
You could use a word boundary, that's what they're for:
$pattern = '/#(.+?)\b/';
This will work for question 2 also
You can add an option to match the end of the string:
#(.*?)(?= |\p{P}?$)
Replace with <strong>$1</strong>.
You can also use \p{P} (any Unicode punctuation symbol) to prevent punctuation from bold formatting.
Here is a demo.
I want to convert certain patterns into links and it works fine as far as normal user ids are considered.But now i want to do the same for encrypted ids as well.
Below is my code:(works)
$text = "hi how are you guys???... ##[Sam Thomas:10181] ##[Jack Daniel:11074] ##[Paul Walker:11043] ";
$pattern = "/##\[([^:]*):(\d*)\]/";
$matches = array();
preg_match_all($pattern, $text, $matches);
$output = preg_replace($pattern, "$1", $text);
Now i need to do link the text like:
"hi how are you guys???... ##[Sam Thomas:ZGNjAmD9ac3K] ##[Jack Daniel:ZGNjAmD9ac3K] ##[Paul Walker:ZGNjAmD9ac3K] ";
But this encrypted is not identified by above regular expression...
##\[([^:]*):(.*?)\]
^^
Try this.See demo.Just change \d* to .*? to accept anything or \w* to accept only numbers and letters.or [^\]]* or [0-9a-zA-Z] as well.
https://regex101.com/r/vD5iH9/52
Change your regex to accept numbers and letters as well.
Something like this -
##\[([^:]*):([0-9a-zA-Z]*)\]
^^^^^^^^^^^ Replaced \d
Demo
I'm trying to remove all words of less than 3 characters from a string, specifically with RegEx.
The following doesn't work because it is looking for double spaces. I suppose I could convert all spaces to double spaces beforehand and then convert them back after, but that doesn't seem very efficient. Any ideas?
$text='an of and then some an ee halved or or whenever';
$text=preg_replace('# [a-z]{1,2} #',' ',' '.$text.' ');
echo trim($text);
Removing the Short Words
You can use this:
$replaced = preg_replace('~\b[a-z]{1,2}\b\~', '', $yourstring);
In the demo, see the substitutions at the bottom.
Explanation
\b is a word boundary that matches a position where one side is a letter, and the other side is not a letter (for instance a space character, or the beginning of the string)
[a-z]{1,2} matches one or two letters
\b another word boundary
Replace with the empty string.
Option 2: Also Remove Trailing Spaces
If you also want to remove the spaces after the words, we can add \s* at the end of the regex:
$replaced = preg_replace('~\b[a-z]{1,2}\b\s*~', '', $yourstring);
Reference
Word Boundaries
You can use the word boundary tag: \b:
Replace: \b[a-z]{1,2}\b with ''
Use this
preg_replace('/(\b.{1,2}\s)/','',$your_string);
As some solutions worked here, they had a problem with my language's "multichar characters", such as "ch". A simple explode and implode worked for me.
$maxWordLength = 3;
$string = "my super string";
$exploded = explode(" ", $string);
foreach($exploded as $key => $word) {
if(mb_strlen($word) < $maxWordLength) unset($exploded[$key]);
}
$string = implode(" ", $exploded);
echo $string;
// outputs "super string"
To me, it seems that this hack works fine with most PHP versions:
$string2 = preg_replace("/~\b[a-zA-Z0-9]{1,2}\b\~/i", "", trim($string1));
Where [a-zA-Z0-9] are the accepted Char/Number range.
I'm trying to match a certain word and replace part of the word with certain text but leave the rest of the word intact. It is my understanding that adding parentheses to part of the regex pattern means that the pattern match within the parentheses gets replaced when you use preg_replace()
for testing purposes I used:
$text = 'batman';
echo $new_text = preg_replace('#(bat)man#', 'aqua', $text);
I only want 'bat' to be replaced by 'aqua' to get 'aquaman'. Instead, $new_text echoes 'aqua', leaving out the 'man' part.
preg_replace replaces all the string matched by regular expression
$text = 'batman';
echo $new_text = preg_replace('#bat(man)#', 'aqua\\1', $text);
Capture man instead and append it to your aqua prefix
Another way of doing that is to use assertions:
$text = 'batman';
echo $new_text = preg_replace('#bat(?=man)#', 'aqua', $text);
I would not use preg_* functions for this and just do str_replace() DOCs:
echo str_replace('batman', 'aquaman', $text);
This is simpler as a regex is not really needed in this case. Otherwise it would be with a regular expression:
echo $new_text = preg_replace('#bat(man)#', 'aqua\\1', $text);
This will substitute your man in after aqua when replacing the entire search phrase. preg_replace DOCs replaces the entire matching portion of the pattern.
The way you're trying to do it, it would be more like:
preg_replace('#bat(man)#', 'aqua$1', $text);
I'd using positive lookahead:
preg_replace('/bat(?=man)/', 'aqua', $text)
Demo here: http://ideone.com/G9F4q
The brackets are creating a capturing group, that means you can access the part matched by this group using \1.
you can do either what zerkms suggested or use a lookahead that does just check but not match.
$text = 'batman';
echo $new_text = preg_replace('#bat(?=man)#', 'aqua', $text);
This will match "bat" but only if it is followed by "man", and only "bat" is replaced.
I'm trying to use regular expressions (preg_match and preg_replace) to do the following:
Find a string like this:
{%title=append me to the title%}
Then extract out the title part and the append me to the title part. Which I can then use to perform a str_replace(), etc.
Given that I'm terrible at regular expressions, my code is failing...
preg_match('/\{\%title\=(\w+.)\%\}/', $string, $matches);
What pattern do I need? :/
I think it's because the \w operator doesn't match spaces. Because everything after the equal sign is required to fit in before your closing %, it all has to match whatever is inside those brackets (or else the entire expression fails to match).
This bit of code worked for me:
$str = '{%title=append me to the title%}';
preg_match('/{%title=([\w ]+)%}/', $str, $matches);
print_r($matches);
//gives:
//Array ([0] => {%title=append me to the title%} [1] => append me to the title )
Note that the use of the + (one or more) means that an empty expression, ie. {%title=%} won't match. Depending on what you expect for white space, you might want to use the \s after the \w character class instead of an actual space character. \s will match tabs, newlines, etc.
You can try:
$str = '{%title=append me to the title%}';
// capture the thing between % and = as title
// and between = and % as the other part.
if(preg_match('#{%(\w+)\s*=\s*(.*?)%}#',$str,$matches)) {
$title = $matches[1]; // extract the title.
$append = $matches[2]; // extract the appending part.
}
// find these.
$find = array("/$append/","/$title/");
// replace the found things with these.
$replace = array('IS GOOD','TITLE');
// use preg_replace for replacement.
$str = preg_replace($find,$replace,$str);
var_dump($str);
Output:
string(17) "{%TITLE=IS GOOD%}"
Note:
In your regex: /\{\%title\=(\w+.)\%\}/
There is no need to escape % as its
not a meta char.
There is no need to escape { and }.
These are meta char but only when
used as a quantifier in the form of
{min,max} or {,max} or {min,}
or {num}. So in your case they are treated literally.
Try this:
preg_match('/(title)\=(.*?)([%}])/s', $string, $matches);
The match[1] has your title and match[2] has the other part.