$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.
Related
Hi I'm new to php and I need a little help
I need to change the text that is between ** in php string and put it between html tag
$text = "this is an *example*";
But I really don't know how and i need help
personally I would use explode, you can then piece the sentence back together if the example appears in the middle of a sentence
<?php
$text = "this is an *example*";
$pieces = explode("*", $text);
echo $pieces[0];
?>
Edit:
Since you're looking for what basically amounts to custom BB Code use this
$text = "this is an *example*";
$find = '~[\*](.*?)[\*]~s';
$replace = '<span style="color: green">$1</span>';
echo preg_replace($find,$replace,$text);
You can add this to a function and have it parse any text that gets passed to it, you can also make the find and replace variables into arrays and add more codes to it
You really should use a DOM parser for things like this, but if you can guaratee it will always be the * character you can use some regex:
$text = "this is an *example*";
$regex = '/(?<=\*)(.*?)(?=\*)/';
$replacement = 'ostrich';
$new_text = preg_replace($regex, $replacement, $text);
echo $new_text;
Returns
this is an *ostrich*
Here is how the regex works:
Positive Lookbehind (?<=\*)
\* matches the character * literally (case sensitive)
1st Capturing Group (.*?)
.*? matches any character (except for line terminators)
*? Quantifier — Matches between zero and unlimited times, as few times as possible, expanding as needed (lazy)
Positive Lookahead (?=\*)
\* matches the character * literally (case sensitive)
This regex essentially starts and ends by looking at what is ahead of and behind the search character you specified and leaves those characters intact during the replacement with preg_replace().
i'm not very firm with regular Expressions, so i have to ask you:
How to find out with PHP if a string contains a word starting with # ??
e.g. i have a string like "This is for #codeworxx" ???
I'm so sorry, but i have NO starting point for that :(
Hope you can help.
Thanks,
Sascha
okay thanks for the results - but i did a mistake - how to implement in eregi_replace ???
$text = eregi_replace('/\B#[^\B]+/','\\1', $text);
does not work??!?
why? do i not have to enter the same expression as pattern?
Match anything with has some whitespace in front of a # followed by something else than whitespace:
$ cat 1812901.php
<?php
echo preg_match("/\B#[^\B]+/", "This should #match it");
echo preg_match("/\B#[^\B]+/", "This should not# match");
echo preg_match("/\B#[^\B]+/", "This should match nothing and return 0");
echo "\n";
?>
$ php 1812901.php
100
break your string up like this:
$string = 'simple sentence with five words';
$words = explode(' ', $string );
Then you can loop trough the array and check if the first character of each word equals "#":
if ($stringInTheArray[0] == "#")
Assuming you define a word a sequence of letters with no white spaces between them, then this should be a good starting point for you:
$subject = "This is for #codeworxx";
$pattern = '/\s*#(.+?)\s/';
preg_match($pattern, $subject, $matches);
print_r($matches);
Explanation:
\s*#(.+?)\s - look for anything starting with #, group all the following letters, numbers, and anything which is not a whitespace (space, tab, newline), till the closest whitespace.
See the output of the $matches array for accessing the inner groups and the regex results.
#OP, no need regex. Just PHP string methods
$mystr='This is for #codeworxx';
$str = explode(" ",$mystr);
foreach($str as $k=>$word){
if(substr($word,0,1)=="#"){
print $word;
}
}
Just incase this is helpful to someone in the future
/((?<!\S)#\w+(?!\S))/
This will match any word containing alphanumeric characters, starting with "#." It will not match words with "#" anywhere but the start of the word.
Matching cases:
#username
foo #username bar
foo #username1 bar #username2
Failing cases:
foo#username
#username$
##username
I have a string as
This is a sample text. This text will be used as a dummy for "various" RegEx "operations" using PHP.
I want to select and replace all the first alphabet of each word (in the example : T,i,a,s,t,T,t,w,b,u,a,d,f,",R,",u,P). How do I do it?
I tried /\b.{1}\w+\b/. I read the expression as "select any character that has length of 1 followed by word of any length" but didn't work.
You may try this regex as well:
(?<=\s|^)([a-zA-Z"])
Demo
Your regex - /\b.{1}\w+\b/ - matches any string that is not enclosed in word characters, starts with any symbol that is in a position after a word boundary (thus, it can even be whitespace if there is a letter/digit/underscore in front of it), followed with 1 or more alphanumeric symbols (\w) up to the word boundary.
That \b. is the culprit here.
If you plan to match any non-whitespace preceded with a whitespace, you can just use
/(?<!\S)\S/
Or
/(?<=^|\s)\S/
See demo
Then, replace with any symbol you need.
You may try to use the following regex:
(.)[^\s]*\s?
Using the preg_match_all and implode the output result group 1
<?php
$string = 'This is a sample text. This text will be used as a dummy for'
. '"various" RegEx "operations" using PHP.';
$pattern = '/(.)[^\s]*\s?/';
$matches;
preg_match_all($pattern, $string, $matches);
$output = implode('', $matches[1]);
echo $output; //Output is TiastTtwbuaadf"R"uP
For replace use something like preg_replace_callback like:
$pattern = '/(.)([^\s]*\s?)/';
$output2 = preg_replace_callback($pattern,
function($match) { return '_' . $match[2]; }, $string);
//result: _his _s _ _ample _ext. _his _ext _ill _e _sed _s _ _ummy _or _various" _egEx _operations" _sing _HP.
I'm reformatting some text, and sometimes I have a string, where there is a sentence which is not ended by a dot.
I'm running various checks for this purpose, and one more I'd like is to "Add dot after last character before new line".
I'm not sure how to form the regular expression for this:]
$string = preg_replace("/???/", ".\n", $string);
Try this one:
$string = preg_replace("/(?<![.])(?=[\n\r]|$)/", ".", $string);
negative lookbehind (?<![.]) is checking previous character is not .
positive lookahead (?=[\n\r]|$) is checking next character is a newline or end of string.
like this I suppose:
<?php
$string = "Add dot after last character before new line\n";
$string = preg_replace("/(.)$/", "$1.\n", $string);
print $string;
?>
This way the dot will be added after the word line in the sentence and before the \n.
demo : http://ideone.com/J4g7tH
I'd do:
$string = "Add dot after last character before new line\n";
$string = preg_replace("/([^.\r\n])$/s", "$1.", $string);
Thanks for all the answers, but none of them really caught all scenarios right.
I fumbled my way to a good solution using the word boundary regex character class:
// Add dot after every word boundary that is followed by a new line.
$string = preg_replace("/[\b][\n]/", ".\n", $string);
I guess [\b][\n] could just as well be \b\n without square brackets.
This works for me:
$content = preg_replace("/(\w+)(\n)/", "$1.$2", $content);
It will match a word immediately followed by a new line, and add a dot in between.
Will match:
Hello\n
Will not match:
Hello \n
or
Hello.\n
i'm not very firm with regular Expressions, so i have to ask you:
How to find out with PHP if a string contains a word starting with # ??
e.g. i have a string like "This is for #codeworxx" ???
I'm so sorry, but i have NO starting point for that :(
Hope you can help.
Thanks,
Sascha
okay thanks for the results - but i did a mistake - how to implement in eregi_replace ???
$text = eregi_replace('/\B#[^\B]+/','\\1', $text);
does not work??!?
why? do i not have to enter the same expression as pattern?
Match anything with has some whitespace in front of a # followed by something else than whitespace:
$ cat 1812901.php
<?php
echo preg_match("/\B#[^\B]+/", "This should #match it");
echo preg_match("/\B#[^\B]+/", "This should not# match");
echo preg_match("/\B#[^\B]+/", "This should match nothing and return 0");
echo "\n";
?>
$ php 1812901.php
100
break your string up like this:
$string = 'simple sentence with five words';
$words = explode(' ', $string );
Then you can loop trough the array and check if the first character of each word equals "#":
if ($stringInTheArray[0] == "#")
Assuming you define a word a sequence of letters with no white spaces between them, then this should be a good starting point for you:
$subject = "This is for #codeworxx";
$pattern = '/\s*#(.+?)\s/';
preg_match($pattern, $subject, $matches);
print_r($matches);
Explanation:
\s*#(.+?)\s - look for anything starting with #, group all the following letters, numbers, and anything which is not a whitespace (space, tab, newline), till the closest whitespace.
See the output of the $matches array for accessing the inner groups and the regex results.
#OP, no need regex. Just PHP string methods
$mystr='This is for #codeworxx';
$str = explode(" ",$mystr);
foreach($str as $k=>$word){
if(substr($word,0,1)=="#"){
print $word;
}
}
Just incase this is helpful to someone in the future
/((?<!\S)#\w+(?!\S))/
This will match any word containing alphanumeric characters, starting with "#." It will not match words with "#" anywhere but the start of the word.
Matching cases:
#username
foo #username bar
foo #username1 bar #username2
Failing cases:
foo#username
#username$
##username