i'm trying to create a regex pattern that returns true if a certain word is not found. Ive tried using [^word] but that doesn't match up against a word just the individual characters as they appear.
I need preg_match(using php) to return true cause there are other words that I to match and return true.
if you are looking for a string within a string (no pattern needed) then use strstr() or stristr()
if (!preg_match("/word/", $string)) { // <-- true, if not match
// code here
}
Related
I need regular expression which matches a URL and return the desired value
Example (if the URL matches to)
1. http://example.com/amp
2. http://example.com/amp/
3. http://example.com/amp~
THEN
it should return: ?amp=1
ELSE
it should return: false
You should be able to use preg_replace to append ?amp= to the end of a matching string. Its functionality already does the if/else functional you require,
If matches are found, the new subject will be returned, otherwise subject will be returned unchanged or NULL if an error occurred.
(or I misread the it should return noting)
-http://php.net/manual/en/function.preg-replace.php
Something like
amp\K( |\/|~)$
Should do it
$string = 'http://example.com/amp~';
echo preg_replace('/amp\K( |\/|~)$/', '$1?amp=1', $string);
The $1 is optional, not sure if you wanted the found character included or not.
PHP Demo: https://eval.in/780432
Regex demo: https://regex101.com/r/JgcrLu/1/
$ is the end of the string. () is a capturing and alteration group. |s are alterations. \K skips the previously matched regex part.
You didn't specify the programming language you're using but you probably need something like:
php:
$new = preg_replace('%/amp\b(?:/|~)?%si', '/?amp=1', $old);
python:
new_string = re.sub(r"/amp\b(?:/|~)?", "/?amp=1", old_string, 0, re.IGNORECASE)
Regex Demo
I have made this regex:
(?<=span class="ope">)?[a-z0-9]+?\.(pl|com|net\.pl|tk|org|org\.pl|eu)|$(?=<\/span>)$
It does match the strings like: example.pl, example12.com, something.eu but it will also match the dontwantthis.com.
My question is how to don't match a string in case if it contains the dontwantthis string?
You're probably following your regex with a loop to cycle through matches. In this case, it's probably easiest to just check for the presence of the dontwantthis substring and continue if it's there. Trying to implement it in regex is just asking for trouble.
It seems that you are extracting content from span elements using a regular expression. Now, despite all the reasons why this is not such a good idea...
... just keep the expression you have. Then, if you have a match, filter out the matched entries that should be rejected.
var $match = extractContentFromHtml($html); // use regex here, return false if no match
if ($match && validMatch($match)) {
// do something
}
where validMatch(string) should check if the value exists in some array, for example.
Hi i have a problem with preg_match function.
i want to check whole text against pattern and return true if whole text matched with pattern and false for not matched or partially matched with pattern but i can't do this with php preg_match function.
for example i use bellow func to check if input is digit or no.
if(preg_match("/[\d]/","99ab") return true; else return false;
above code return true but i want to this return false.
and other regex that i want to check is users first name and last name with preg_match.
and another regex for phone numbers with this format (000) 000-0000.
please help me to solve this problem.
Just change the pattern like this:
"/^[\d]+$/"
Remember that:
^: means the start of the string
$: means the end of the string
You need to use ^ to indicate the start of the string and $ for the end
if (preg_match("/^[\d]+$/","99ab")) return true;
else return false;
I can't get the preg_match to find a word anywhere a the string.
I have this:
$bad_words = "/(\bsuck\b)|(\bsucks\b)|(\bporn\b)|";
$text = "sucky";
if(preg_match($bad_term_filter, trim($feedback_review_comment)) != 0 )
I need to return true but it only returns true if its an exact match, for example if
$text = "suck";
that returns true
\b is the word boundary anchor. It looks like you're trying to find if some word occurs anywhere regardless of the word boundaries, so I think the pattern you want is simply:
suck|porn
You also do not want the last empty alternate, because that will match everything (all string contains an empty string). There is no need to explicitly look for sucks, because it already contains suck.
References
regular-expressions.info/Anchors and Character Classes, and Optional
In PHP, how do I check if a String contains only letters? I want to write an if statement that will return false if there is (white space, number, symbol) or anything else other than a-z and A-Z.
My string must contain ONLY letters.
I thought I could do it this way, but I'm doing it wrong:
if( ereg("[a-zA-Z]+", $myString))
return true;
else
return false;
How do I find out if myString contains only letters?
Yeah this works fine. Thanks
if(myString.matches("^[a-zA-Z]+$"))
Never heard of ereg, but I'd guess that it will match on substrings.
In that case, you want to include anchors on either end of your regexp so as to force a match on the whole string:
"^[a-zA-Z]+$"
Also, you could simplify your function to read
return ereg("^[a-zA-Z]+$", $myString);
because the if to return true or false from what's already a boolean is redundant.
Alternatively, you could match on any character that's not a letter, and return the complement of the result:
return !ereg("[^a-zA-Z]", $myString);
Note the ^ at the beginning of the character set, which inverts it. Also note that you no longer need the + after it, as a single "bad" character will cause a match.
Finally... this advice is for Java because you have a Java tag on your question. But the $ in $myString makes it look like you're dealing with, maybe Perl or PHP? Some clarification might help.
Your code looks like PHP. It would return true if the string has a letter in it. To make sure the string has only letters you need to use the start and end anchors:
In Java you can make use of the matches method of the String class:
boolean hasOnlyLetters(String str) {
return str.matches("^[a-zA-Z]+$");
}
In PHP the function ereg is deprecated now. You need to use the preg_match as replacement. The PHP equivalent of the above function is:
function hasOnlyLetters($str) {
return preg_match('/^[a-z]+$/i',$str);
}
I'm going to be different and use Character.isLetter definition of what is a letter.
if (myString.matches("\\p{javaLetter}*"))
Note that this matches more than just [A-Za-z]*.
A character is considered to be a letter if its general category type, provided by Character.getType(ch), is any of the following: UPPERCASE_LETTER, LOWERCASE_LETTER, TITLECASE_LETTER, MODIFIER_LETTER, OTHER_LETTER
Not all letters have case. Many characters are letters but are neither uppercase nor lowercase nor titlecase.
The \p{javaXXX} character classes is defined in Pattern API.
Alternatively, try checking if it contains anything other than letters: [^A-Za-z]
The easiest way to do a "is ALL characters of a given type" is to check if ANY character is NOT of the type.
So if \W denotes a non-character, then just check for one of those.