So I hadn't done any regexps for a while, so I thought I'd brush up on my memory. I'm trying to convert a string like a*b*c into a<b>b</b>c. I've already gotten that working, but now I want to keep a string like a\*b\*c from turning into a\<b>b\</b>c, but rather, into a*b*c. Here's the code I'm using now:
$string = preg_replace("/\*([\s\S]*?)\*/", "<b>$1</b>", $input);
I've tried putting this \\\\{0} in before the asterisks, and that didn't work. Neither did [^\\\\].
Try negative lookbehind:
"/(?<!\\\\)\*([\s\S]*?)(?<!\\\\)\*/"
This only matches a * if it's not preceded by a \.
This is brittle, though; it would also fail if the string is escaped backslash \\*bold* text.
Related
I have a substitution code that replaces all instances of XD with an XD smiley face... thing is, should a link include the string 'XD', it then breaks the link.
I want it to only replace the XD, if it is followed by a whitespace, as in 'XD ', except I can't seem to get it to work (tried  , /\s/ and as in 'XD ')
Chances are I'm getting something really obvious wrong, but I can't find any help (all of it seems to be about removing whitespace, not requiring it), so I'm hoping someone can help me.
Here's the code for reference:
function BB_CODE($content) {
$content = str_replace("XD", "<img src=\"images/smilies/icon_xd.gif\" alt=\"XD\">", $content);
}
The content is user input. Thanks for any help!
You should surround "XD" with %
$content= str_replace("%XD%", "<img src=\"images/smilies/icon_xd.gif\" alt=\"XD\">", $content);
EDIT :
Or using preg_replace
preg_replace("/XD/", "<img src=\"images/smilies/icon_xd.gif\" alt=\"XD\">", $content);
So you want to replace XD only if it's alone?
preg_replace('/\bXD\b/', '(ಠ‿ಠ)', "Then I was like XD")
Use \b to watch for word boundaries instead of \s. This means it works at the beginning and the end of string too, like in my example.
With preg_replace() there are two common gotchas:
The separator char, I used / here by convention but in my own code I prefer %. You could write the regex as '%\bXD\b%' with the same meaning.
Escaping the backslashes, I used a single quoted string so I don't have to escape the backslash in \b. If you use double qoutes, you have to escape it, like so: "/\\bXD\\b/"
I have a string that looks like this
../Clean_Smarty_Projekt/tpl/templates_c\.
../Clean_Smarty_Projekt/tpl/templates_c\..
I want to replace ../, \. and \.. with a regulare expression.
Before, I did this like this:
$result = str_replace(array("../","\..","\."),"",$str);
And there it (pattern) has to be in this order because changing it makes the output a little buggy. So I decided to use a regular expression.
Now I came up with this pattern
$result = preg_replace('/(\.\.\/)|(\\[\.]{1,2})/',"",$str);
What actually returns only empty strings...
Reason: (\\[\.]{1,2})
In Regex101 its all ok. (Took me a couple of minutes to realize that I don't need the /g in preg_replace)
If I use this pattern in preg_replace I have to do (\\\\[\.]{1,2}) to get it to work. But that's obviously wrong because im not searching for two slashes.
Of course I know the escaping rulse (escaping slashes).
Why doesn't this match correctly ?
I suggest you to use a different php delimiter. Within the / delimiter, you need to use three \\\ or four \\\\ backslashes to match a single backslash.
$string = '../Clean_Smarty_Projekt/tpl/templates_c\.'."\n".'../Clean_Smarty_Projekt/tpl/templates_c\..';
echo preg_replace('~\.\./|\\\.{1,2}~', '', $string)
Output:
Clean_Smarty_Projekt/tpl/templates_c
Clean_Smarty_Projekt/tpl/templates_c
This is the text sample:
$text = "asd dasjfd fdsfsd http://11111.com/asdasd/?s=423%423%2F gfsdf http://22222.com/asdasd/?s=423%423%2F
asdfggasd http://3333333.com/asdasd/?s=423%423%2F";
This is my regex pattern:
preg_match_all( "#http:\/\/(.*?)[\s|\n]#is", $text, $m );
That match the first two urls, but how do I match the last one? I tried adding [\s|\n|$] but that will also only match the first two urls.
Don't try to match \n (there's no line break after all!) and instead use $ (which will match to the end of the string).
Edit:
I'd love to hear why my initial idea doesn't work, so in case you know it, let me know. I'd guess because [] tries to match one character, while end of line isn't one? :)
This one will work:
preg_match_all('#http://(\S+)#is', $text, $m);
Note that you don't have to escape the / due to them not being the delimiting character, but you'd have to escape the \ as you're using double quotes (so the string is parsed). Instead I used single quotes for this.
I'm not familar with PHP, so I don't have the exact syntax, but maybe this will give you something to try. the [] means a character class so |$ will literally look for a $. I think what you'll need is another look ahead so something like this:
#http:\/\/(.*)(?=(\s|$))
I apologize if this is way off, but maybe it will give you another angle to try.
See What is the best regular expression to check if a string is a valid URL?
It has some very long regular expressions that will match all urls.
Here is the input I am searching:
\u003cspan class=\"prs\">email_address#me.com\u003c\/span>
Trying to just return email_address#me.com.
My regex class=\\"prs\\">(.*?)\\ returns "class=\"prs\">email_address#me.com\" in RegExp which is OK, I can work with that result.
But I can't get it to work in PHP.
$regex = "/class=\\\"prs\\\">(.*?)\\/";
Gives me an error "No ending delimiter"
Can someone please help?
Your original code:
$regex = "/class=\\\"prs\\\">(.*?)\\/";
The reason you get No ending delimiter is that although you are escaping the backslash prior to the closing forward slash, what you have done is escaped it in the context of the PHP string, not in the context of the regex engine.
So the PHP string escaping mechanism does its thing, and by the time the regex engine gets it, it will look like this:
/class=\"prs\">(.*?)\/
This means that the regular expression engine will see the backslash at the end of the expression as escaping the forward slash that you are intending to use to close the expression.
The usual PHP solution to this kind of thing is to switch to using single-quoted string instead of a double-quoted one, but this still won't work, as \\ is an escaped backslash in both single and double quoted strings.
What you need to do is double up the number of backslash characters at the end of your string, so your code needs to look like this:
$regex = "/class=\\\"prs\\\">(.*?)\\\\/";
The way to prove what it's doing is to print the contents of the $regex variable, so you can see what the string will look like to the regex engine. These kinds of errors are actually very hard to spot, but looking at the actual content of the string will help you spot them.
Hope that helps.
If you change to single quotes it should fix it
$regex = '/class=\\\"prs\\\">(.*?)\\/';
Given a literal string such as:
Hello\n\n\n\n\n\n\n\n\n\n\n\nWorld
I would like to reduce the repeated \n's to a single \n.
I'm using PHP, and been playing around with a bunch of different regex patterns. So here's a simple example of the code:
$testRegex = '/(\\n){2,}/';
$test = 'Hello\n\n\n\n\n\n\n\n\nWorld';
$test2 = preg_replace($testRegex ,'\n',$test);
echo "<hr/>test regex<hr/>".$test2;
I'm new to PHP, not that new to regex, but it seems '\n' conforms to special rules. I'm still trying to nail those down.
Edit: I've placed the literal code I have in my php file here, if I do str_replace() I can get good things to happen, but that's not a complete solution obviously.
To match a literal \n with regex, your string literal needs four backslashes to produce a string with two backlashes that’s interpreted by the regex engine as an escape for one backslash.
$testRegex = '/(\\\\n){2,}/';
$test = 'Hello\n\n\n\n\n\n\n\n\n\n\n\nWorld';
$test2 = preg_replace($testRegex, '\n', $test);
Perhaps you need to double up the escape in the regular expression?
$pattern = "/\\n+/"
$awesome_string = preg_replace($pattern, "\n", $string);
Edit: Just read your comment on the accepted answer. Doesn't apply, but is still useful.
If you're intending on expanding this logic to include other forms of white-space too:
$output = echo preg_replace('%(\s)*%', '$1', $input);
Reduces all repeated white-space characters to single instances of the matched white-space character.
it indeed conforms to special rules, and you need to add the "multiline"-modifier, m. So your pattern would look like
$pattern = '/(\n)+/m'
which should provide you with the matches. See the doc for all modifiers and their detailed meaning.
Since you're trying to reduce all newlines to one, the pattern above should work with the rest of your code. Good luck!
Try this regular expression:
/[\n]*/