I am trying to get a string in between single quotes from the following document.write('mystring') I have tried the following pattern \'[A-Za-z0-9]+\' but I want the string without the single quote marks. How can I do that?
preg_match("/'([a-z0-9]+)'/i", $str, $matches);
//string contents are now in `$matches[1]`
You may also want to use \w or even something else to do the capturing unless you are absolutely sure that the string you want to acquire is purely alphanumeric. This also assumes that only apostrophes are used and not quotes.
Capture the word in a specific group:
\'([A-Za-z0-9]+)\'
Related
I have the right function, just not finding the right regex pattern to remove (ID:999999) from the string. This ID value varies but is all numeric. I like to remove everything including the brackets.
$string = "This is the value I would like removed. (ID:17937)";
$string = preg_replace('#(ID:['0-9']?)#si', "", $string);
Regex is not more forte! And need help with this one.
Try this:
$string = preg_replace('# \(ID:[0-9]+\)#si', "", $string);
You need to escape the parenthesis using backslashes \.
You shouldn't use quotes around the number range.
You should use + (one or more) instead of ? (zero or one).
You can add a space at the start, to avoid having a space at the end of the resulting string.
In PHP regex is in / and not #, after that, parentheses are for capture group so you must escape them to match them.
Also to use preg_replace replacement you will need to use capture group so in your case /(\(ID:[0-9]+\))/si will be the a nice regular expression.
Here are two options:
Code: (Demo)
$string = "This is the value I would like removed. (ID:17937)";
var_export(preg_replace('/ \(ID:\d+\)/',"",$string));
echo "\n\n";
var_export(strstr($string,' (ID:',true));
Output: (I used var_export() to show that the technique is "clean" and gives no trailing whitespaces)
'This is the value I would like removed.'
'This is the value I would like removed.'
Some points:
Regex is a better / more flexible solution if your ID substring can exist anywhere in the string.
Your regex pattern doesn't need a character class if you use the shorthand range character \d.
Regex generally speaking should only be used when standard string function will not suffice or when it is proven to be more efficient for a specific case.
If your ID substring always occurs at the end of the string, strstr() is an elegant/perfect function.
Both of my methods write a (space) before ID to make the output clean.
You don't need either s or i modifiers on your pattern, because s only matters if you use a . (dot) and your ID is probably always uppercase so you don't need a case-insensitive search.
I have the list of strings (in PHP):
a2c
bdR
dDv
"ddv
aaa
"aaa
What's the RegEx expression to match only the strings that are not starting with quotation mark? In this example there are four such strings. So, I need to match four strings only (to count them). For this list I'am using a loop, but I just need RegEx now. Thanks!
I tried with
[^"]([a-zA-Z0-9]*)*
but it still matching all strings even those that starting with quotation mark.
You are missing the start of string anchor ^, which means that your expression will match a string if it appears at any place inside it. Obviously the sequence "non-quote followed by anything, including end of string" appears inside all of your sample inputs.
This expression will match what you want:
^[^"]
It simply matches any input whose first character is not a double quote. There is no need to bother with the rest of the characters.
Try it:
^[^"](.*)/gm
(If more than one string is in the variable.)
$strigs_arr = array();
foreach($strings as $str){
//$str = '"aaa';
preg_match('/^(?P<string>[^"].*)/',$str,$match);
$strigs_arr[] = $match['string'];
}
echo "<pre>";
print_r($strigs_arr);
How can i extract https://domain.com/gamer?hid=.115f12756a8641 from the below string ,i.e from url
rrth:'http://www.google.co',cctp:'323',url:'https://domain.com/gamer?hid=.115f12756a8641',rrth:'https://another.com'
P.s :I am new to regular expression, I am learning .But above string seems to be formatted..so some sort of shortcut must be there.
If your input string is called $str:
preg_match('/url:\'(.*?)\'/', $str, $matches);
$url = $matches[1];
(.*?) captures everything between url:' and ' and can later be retrieved with $matches[1].
The ? is particularly important. It makes the repetition ungreedy, otherwise it would consume everything until the very last '.
If your actual input string contains multiple url:'...' section, use preg_match_all instead. $matches[1] will then be an array of all required values.
Simple regex:
preg_match('/url\s*\:\s*\'([^\']+)/i',$theString,$match);
echo $match[1];//should be the url
How it works:
/url\s*\:\s*: matches url + [any number of spaces] + : (colon)+ [any number of spaces]But we don't need this, that's where the second part comes in
\'([^\']+)/i: matches ', then the brackets (()) create a group, that will be stored separately in the $matches array. What will be matches is [^']+: Any character, except for the apostrophe (the [] create a character class, the ^ means: exclude these chars). So this class will match any character up to the point where it reaches the closing/delimiting apostrophe.
/i: in case the string might contain URL:'http://www.foo.bar', I've added that i, which is the case-insensitive flag.
That's about it.Perhaps you could sniff around here to get a better understanding of regex's
note: I've had to escape the single quotes, because the pattern string uses single quotes as delimiters: "/url\s*\:\s*'([^']+)/i" works just as well. If you don't know weather or not you'll be dealing with single or double quotes, you could replace the quotes with another char class:
preg_match('/url\s*\:\s*[\'"]([^\'"]+)/i',$string,$match);
Obviously, in that scenario, you'll have to escape the delimiters you've used for the pattern string...
I have user input where some tags are allowed inside square brackets. I've already wrote the regex pattern to find and validate what's inside the brackets.
In user input field opening-bracket could ([) be escaped with backslash, also backslash could be escaped with another backslash (\). I need look-behind sub-pattern to avoid odd number of consecutive backslashes before opening-bracket.
At the moment I must deal with something like this:
(?<!\\)(?:\\\\)*\[(?<inside brackets>.*?)]
It works fine, but problem is that this code still matches possible pairs of consecutive backslashes in front of brackets (even they are hidden) and look-behind just checks out if there's another single backslash appended to pairs (or directly to opening-bracket). I need to avoid them all inside look-behind group if possible.
Example:
my [test] string is ok
my \[test] string is wrong
my \\[test] string is ok
my \\\[test] string is wrong
my \\\\[test] string is ok
my \\\\\[test] string is wrong
...
etc
I work with PHP PCRE
Last time I checked, PHP did not support variable-length lookbehinds. That is why you cannot use the trivial solution (?<*\\).
The simplest workaround would be to simply match the entire thing, not just the brackets part:
(?<!\\)((?:\\\\)*)\[(?<inside_brackets>.*?)]
The difference is that now, if you're using that regex in a preg_replace, you gotta remember to prefix the replacement string by $1, to restore the backslashes being there.
You could do it without any look-behinds at all (the (\\\\|[^\\]) alternation eats anything but a single back-slash):
^(\\\\|[^\\])*\[(?<brackets>.*?)\]
Is there a way to back reference in the regular expression pattern?
Example input string:
Here is "some quoted" text.
Say I want to pull out the quoted text, I could create the following expression:
"([^"]+)"
This regular expression would match some quoted.
Say I want it to also support single quotes, I could change the expression to:
["']([^"']+)["']
But what if the input string has a mixture of quotes say Here is 'some quoted" text. I would not want the regex to match. Currently the regex in the second example would still match.
What I would like to be able to do is if the first quote is a double quote then the closing quote must be a double. And if the start quote is single quote then the closing quote must be single.
Can I use a back reference to achieve this?
My other related question: Getting text between quotes using regular expression
You can make use of the regex:
(["'])[^"']+\1
() : used for grouping
[..] : is the char class. so ["']
matches either " or ' equivalent
to "|'
[^..] : char class with negation.
It matches any char not listed after
the ^
+ : quantifier for one or more
\1 : backreferencing the first
group which is (["'])
In PHP you'd use this as:
preg_match('#(["\'])[^"\']+\1#',$str)
preg_match('/(["\'])([^"\']+)\1/', 'Here is \'quoted text" some quoted text.');
Explanation: (["'])([^"']+)\1/ I placed the first quote in parentheses. Because this is the first grouping, it's back reference number is 1. Then, where the closing quote would be, I placed \1 which means whichever character was matched in group 1.
/"\(.*?\)".*?\1/ should work, but it depends on the regular expression engine
This is old. But you need to provide the $matches variable in preg_match($pattern, $subject, &$matches)
Then you can use it var_dump($matches)
see https://www.php.net/manual/en/function.preg-match