I have been working on this problem for several days and it's starting to drive me crazy. I'm comfortable using regular expressions but this one thing seems to be escaping me.
I need to match a string between a set of characters if they exist otherwise it should match to the end of the line.
For example:
I'm just trying to get "content" out of the following example:
$str1="title:content #description"
$str2="title:content"
preg_match("/:(.*?)[(#)|(:)|(\*)]?$/",$str1,$content);
preg_match("/:(.*?)[(#)|(:)|(\*)]?$/",$str2,$content);
$str1 outputs:"content #description"
$str2 outputs:"content"
note: the strings may be in a different order or may not have a special character (#,:,or *) in it or they might have one so there's no "end of string" character that will be common besides "end of line".
I've tried every combination i can think of to make the entire "or" statement conditional and read a ton of posts with similar but not quite the same question.
You can write:
preg_match("/:(.*?)(?:[#:*]|$)/", $str1, $content);
(note that the match ends at one of #:* or end-of-string, using |; your version has the [#:*] optional, but makes the end-of-string mandatory.)
or simply:
preg_match("/:([^#:*]*)/", $str1, $content);
(meaning "a colon, followed by zero or more characters that aren't in #:*").
Related
This question already has answers here:
How do I replace certain parts of my string?
(5 answers)
Closed 2 years ago.
I'm creating a simple comment system connected by Steam API. Every Steam user connected in my website can automatically post things. But i'm changing some functions to replace things like the URLs.
My question is: When a user post something like,
"Hello I'm nice, have a look at http://www.cute.com"
Automatically replaces the http:// for the link without changing the http:// in the string.
Maybe something like this?
<?php
$str = "helloo im nice, have a look http://www.cute.com";
echo preg_replace("/http:\/\/(.+)\.(.+)\.(.+)/", "<a href='http://$1.$2.$3'>$1.$2.$3</a>", $str);
?>
This will convert any link into an anchor (or an a tag).
Alternative added
Alternatively, it might be a good idea to add support for https as well. In which case the following might be useful.
<?php
$str = "helloo im nice, have a look http://www.cute.com";
echo preg_replace("/http(s?):\/\/(.+)\.(.+)\.(.+)/", "<a href='http$1://$2.$3.$4'>http$1://$2.$3.$4</a>", $str);
?>
This takes advantage of the ? modifier which means "one or more of the preceding character". In this case it is the "s" character since it is "http" and "https" both match.
Explanation
This uses RegEx (or Regular Expressions) to create this.
The first parameter of the preg_replace function takes the RegEx (I like to test mine here: http://regexr.com/).
All RegExs must start and end with a forward slash. The bits inbetween are as follows.
http: is simply selecting a string that starts with "http:"
\/\/ is called "escaping" and that will select two forward slashes. Since forward slashes are special characters used in RegEx (start and end of a statement) they need to be escaped so that PHP doesn't think the RegEx has ended sooner.
(.+) The brackets are also special characters (though not escaped) and they are known as "capture groups". What this is used for is so that I can see what is between the "http://" and the ".com" (or whatever extension is used). The full stop (or period or ".") character selects anything.
\. Further on the escaping. Since full stop is used as a special character, we have to escape this one. What that means so far is that we are selecting "http://" then anything and then stopping at a full stop.
(.+) Last but not least is the final capture group. This, again selects anything from the string so that have our final capture group and RegEx complete.
Modifiers:
? means "one or more of the preceding character". This means that /tests?/ would match test and tests since s is the preceding character and in the first example we have 0 and in the second there is 1
+ means "one of more of the preceding character". In this case we are saying one of more of anything which means we expect at least one character to be provided.
The second parameter is our replace part.
In short, the $1 and $2 sections are to reference the two brackets from the above RegEx.
Some further reading
The PHP function I used
More information on Regular Expressions
RegEx capture groups
$string = 'helloo im nice, have a look http://www.cute.com';
$string = str_replace('http://', '', $string);
echo $string;
How would i use regular expressions to check for characters within the following string of text:
=== logo ===
I tried to use a regex tester but could come up with the correct expression for i've tried this:
/^[=]{3}$/
I want search within a string find where the text starts with 3 equal signs.
Find a string or any other characters within the equal signs.
Find 3 more equal signs.. ending the expression.
Thanks in advance.
Try using this regex:
/===[^=]+===/
If you want to capture the text, surround it in parentheses:
/===([^=]+)===/
Here's the fiddle: http://jsfiddle.net/jufXA/
If you might have equal signs in your text (but less than 3, obviously) you should instead match everything lazily (which is a tad slower):
/===(.+?)===/
Here's the fiddle: http://jsfiddle.net/jufXA/1/
How about as simple as...
/===(.+?)===/
For example:
$test = "here's ===something special===, like ===this=one===";
preg_match_all('/===(.+?)===/', $test, $matches);
var_dump($matches[1]);
Laziness is kinda virtue here: the regex engine won't advance past the first 'closing delimiter ==='. Without ?, however, you need to use negated character classes (but then again, what about ===something=like=this===?).
I prefer:
/([=]{3})\s*(.+?)\s*\1/.
This puts the text markup (three equal signs) in the beginning and then just uses a back reference for the end. It also trims your text of spaces, which is what you probably want.
I have created a Regular Expression (using php) below; which must match ALL terms within the given string that contains only a-z0-9, ., _ and -.
My expression is: '~(?:\(|\s{0,},\s{0,})([a-z0-9._-]+)(?:\s{0,},\s{0,}|\))$~i'.
My target string is: ('word', word.2, a_word, another-word).
Expected terms in the results are: word.2, a_word, another-word.
I am currently getting: another-word.
My Goal
I am detecting a MySQL function from my target string, this works fine. I then want all of the fields from within that target string. It's for my own ORM.
I suppose there could be a situation where by further parenthesis are included inside this expression.
From what I can tell, you have a list of comma-separated terms and wish to find only the ones which satisfy [a-z0-9._\-]+. If so, this should be correct (it returns the correct results for your example at least):
'~(?<=[,(])\\s*([a-z0-9._-]+)\\s*(?=[,)])~i'
The main issues were:
$ at the end, which was anchoring the query to the end of the string
When matching all you continue from the end of the previous match - this means that if you match a comma/close parenthesis at the end of one match it's not there at match at the beginning of the next one. I've solved this with a lookbehind ((?<=...) and a lookahead ((?=...)
Your backslashes need to be double escaped since the first one may be stripped by PHP when parsing the string.
EDIT: Since you said in a comment that some of the terms may be strings that contain commas you will first want to run your input through this:
$input = preg_replace('~(\'([^\']+|(?<=\\\\)\')+\'|"([^"]+|(?<=\\\\)")+")~', '"STRING"', $input);
which should replace all strings with '"STRING"', which will work fine for matching the other regex.
Maybe using of regex is overkill. In this kind of text you can just remove parenthesis and explode string by comma.
I'm having trouble using preg_match to find and replace a string. The string of interest is:
<span style="font-size:0.6em">EXPIRATION DATE: 04/30/2011</span>
I need to target and replace the date, "04/30/2011" with a different date. Can someone throw me a bone a give me the regular expression to match this pattern using preg_match in PHP? I also need it to match in such a way that it only replaces up to the first closing span and not closing span tags later in the code, e.g.:
<span style="font-size:0.6em">EXPIRATION DATE: 04/30/2011</span><span class="hello"></span>
I'm not versed in regex, and although I've spent the last hour trying to learn enough to make this work, I'm utterly failing. Thanks so much!
EDIT: As you can see this has gotten me exhausted. I did mean preg_replace, not preg_match.
If you're after a replacement, consider using preg_replace(), something like
preg_replace('#(\d{2})/(\d{2})/(\d{4})#', '<new date>', $string);
How about this:
$toBeFoundPattern = '/([0-9][0-9])\/([0-9][0-9])\/([0-9][0-9][0-9][0-9])/';
$toBeReplacedPattern = '$2.$1.$3';
$inString = '<span style="font-size:0.6em">EXPIRATION DATE: 04/30/2011</span>';
// Will convert from US date format 04/30/2011 to european format 30.04.2011
echo preg_replace( $toBeFoundPattern, $toBeReplacedPattern, $inString );
and prints
EXPIRATION DATE: 30.04.2011
Patterns always begin and end with identical so called delimiter characters. Often the character / is used.
$1 references the string, which matched the first string matched by ([0-9][0-9]), $2 references be (...) and $3 the four letters matched by the last (...).
[...] matched a single character, which is one of those listed inside the brackets. E.g. [a-z] matches all lower case letters.
To use the special meaning character / inside of a pattern, you need to escape it by \ to make it be the literal slash character.
Update: Using {..} as pointed out below is shorthand for repeated patterns.
Regex should be:
(0[1-9]|1[012])[- /.](0[1-9]|[12][0-9]|3[01])[- /.](19|20)\d\d
If you want to only match one instance, this is OK. For multiple instances, use preg_match_all instead. Taken from http://www.regular-expressions.info/regexbuddy/datemmddyyyy.html.
Edit: are you looking to just search and replace inside a PHP script or do you want to do some javascript live replacement?
I'm looking for a simple replacement of [[wiki:Title]] into Title.
So far, I have:
$text = preg_replace("/\[\[wiki:(\w+)\]\]/","\\1", $text);
The above works for single words, but I'm trying to include spaces and on occasion special characters.
I get the \w+, but \w\s+ and/or \.+ aren't doing anything.
Could someone improve my understanding of basic regex? And I don't mean for anyone to simply point me to a webpage.
\w\s+ means "a word-character, followed by 1 or more spaces". You probably meant (\w|\s)+ ("1 or more of a word character or a space character").
\.+ means "one or more dots". You probably meant .+ (1 or more of any character - except newlines, unless in single-line mode).
The more robust way is to use
\[wiki:(.+?)\]
This means "1 or more of any character, but stop at first position where the rest matches", i.e. stop at first right bracket in this case. Without ? it would look for the longest available match - i.e. past the first bracket.
You need to use \[\[wiki:([\w\s]+)\]\]. Notice square brackets around \w\s.
If you are learning regular expressions, you will find this site useful for testing: http://rexv.org/
You're definitely getting there, but you've got a couple syntax errors.
When you're using multiple character classes like \w and \s, in order to match within that group, you have to put them in [square brackets] like so... ([\w\s]+) this basically means one or more of words or white space.
Putting a backslash in front of the period escapes it, meaning the regex is searching for a period.
As for matching special characters, that's more of a pain. I tried to come up with something quickly, but hopefully someone else can help you with that.
(Great cheat sheet here, I keep a copy on my desk at all times: http://www.addedbytes.com/cheat-sheets/regular-expressions-cheat-sheet/ )