How do I get "Lrumipsm1" from "Lörum ipsäm 1!"?
So what I need is to only get a-z and 0-9 from a string, using php.
E.g. by using a regular expression (pcre) and replacing all characters that are not within the class of "acceptable" characters by ''.
$in = "Lörum ipsäm 1!";
$result = preg_replace('/[^a-z0-9]+/i', '', $in);
echo $result;
see also: http://docs.php.net/preg_replace
edit:
[a-z0-9] is the class of all characters a....z and 0...9
[^...] negates a class, i.e. [^a-z0-9] contains all characters that are not within a...z0...9
+ is a quantifier with the meaning "1 or more times", [^a-z0-9]+ matches one or more (consecutive) characters that are not within a...z0..9.
The option i makes the pattern case-insensitive, i.e. [a-z] also matches A...Z
you can do this also
$in = "Lörum ipsäm 1!";
$result = preg_replace('/[^[:alnum:]]/i', '', $in);
echo $result;
Related
I Have one string like below.
$string = "2346#$ABSC$%###234567";
Now I want last character from this string that is not numeric or special character, It should be only A-a to Z-z.
Means, I need only "C" from this string.
I have try this formula:
substr($string, -1);
You should look into regular expressions using something like preg_match()
An expression like this would match:
/([a-z])[^a-z]*$/i
It means:
([a-z]) Capture an a-z character (the i at the end makes it case-insensitive)
[^a-z]*$ followed by 0 or more non a-z characters until the end of the string
See an example.
This should work for you:
(Here I just replace everything expect a-zA-Z with an empty string. After this I just access the last character)
<?php
$string = '2346#$ABSC$%###234567';
$string = preg_replace("/[^a-zA-Z]/", "", $string);
echo $string[strlen($string)-1];
?>
output:
C
The proper regex is: ([a-z])[^a-z]*$
I am looking to find and replace words in a long string. I want to find words that start looks like this: $test$ and replace it with nothing.
I have tried a lot of things and can't figure out the regular expression. This is the last one I tried:
preg_replace("/\b\\$(.*)\\$\b/im", '', $text);
No matter what I do, I can't get it to replace words that begin and end with a dollar sign.
Use single quotes instead of double quotes and remove the double escape.
$text = preg_replace('/\$(.*?)\$/', '', $text);
Also a word boundary \b does not consume any characters, it asserts that on one side there is a word character, and on the other side there is not. You need to remove the word boundary for this to work and you have nothing containing word characters in your regular expression, so the i modifier is useless here and you have no anchors so remove the m (multi-line) modifier as well.
As well * is a greedy operator. Therefore, .* will match as much as it can and still allow the remainder of the regular expression to match. To be clear on this, it will replace the entire string:
$text = '$fooo$ bar $baz$ quz $foobar$';
var_dump(preg_replace('/\$(.*)\$/', '', $text));
# => string(0) ""
I recommend using a non-greedy operator *? here. Once you specify the question mark, you're stating (don't be greedy.. as soon as you find a ending $... stop, you're done.)
$text = '$fooo$ bar $baz$ quz $foobar$';
var_dump(preg_replace('/\$(.*?)\$/', '', $text));
# => string(10) " bar quz "
Edit
To fix your problem, you can use \S which matches any non-white space character.
$text = '$20.00 is the $total$';
var_dump(preg_replace('/\$\S+\$/', '', $text));
# string(14) "$20.00 is the "
There are three different positions that qualify as word boundaries \b:
Before the first character in the string, if the first character is a word character.
After the last character in the string, if the last character is a word character.
Between two characters in the string, where one is a word character and the other is not a word character.
$ is not a word character, so don't use \b or it won't work. Also, there is no need for the double escaping and no need for the im modifiers:
preg_replace('/\$(.*)\$/', '', $text);
I would use:
preg_replace('/\$[^$]+\$/', '', $text);
You can use preg_quote to help you out on 'quoting':
$t = preg_replace('/' . preg_quote('$', '/') . '.*?' . preg_quote('$', '/') . '/', '', $text);
echo $t;
From the docs:
This is useful if you have a run-time string that you need to match in some text and the string may contain special regex characters.
The special regular expression characters are: . \ + * ? [ ^ ] $ ( ) { } = ! < > | : -
Contrary to your use of word boundary markers (\b), you actually want the inverse effect (\B)-- you want to make sure that there ISN'T a word character next to the non-word character $.
You also don't need to use capturing parentheses because you are not using a backreference in your replacement string.
\S+ means one or more non-whitespace characters -- with greedy/possessive matching.
Code: (Demo)
$text = '$foo$ boo hi$$ mon$k$ey $how thi$ $baz$ bar $foobar$';
var_export(
preg_replace(
'/\B\$\S+\$\B/',
'',
$text
)
);
Output:
' boo hi$$ mon$k$ey $how thi$ bar '
I need the regex to check if a string only contains numbers, letters, hyphens or underscore
$string1 = "This is a string*";
$string2 = "this_is-a-string";
if(preg_match('******', $string1){
echo "String 1 not acceptable acceptable";
// String2 acceptable
}
Code:
if(preg_match('/[^a-z_\-0-9]/i', $string))
{
echo "not valid string";
}
Explanation:
[] => character class definition
^ => negate the class
a-z => chars from 'a' to 'z'
_ => underscore
- => hyphen '-' (You need to escape it)
0-9 => numbers (from zero to nine)
The 'i' modifier at the end of the regex is for 'case-insensitive' if you don't put that you will need to add the upper case characters in the code before by doing A-Z
if(!preg_match('/^[\w-]+$/', $string1)) {
echo "String 1 not acceptable acceptable";
// String2 acceptable
}
Here is one equivalent of the accepted answer for the UTF-8 world.
if (!preg_match('/^[\p{L}\p{N}_-]+$/u', $string)){
//Disallowed Character In $string
}
Explanation:
[] => character class definition
p{L} => matches any kind of letter character from any language
p{N} => matches any kind of numeric character
_- => matches underscore and hyphen
+ => Quantifier — Matches between one to unlimited times (greedy)
/u => Unicode modifier. Pattern strings are treated as UTF-16. Also
causes escape sequences to match unicode characters
Note, that if the hyphen is the last character in the class definition it does not need to be escaped. If the dash appears elsewhere in the class definition it needs to be escaped, as it will be seen as a range character rather then a hyphen.
\w\- is probably the best but here just another alternative
Use [:alnum:]
if(!preg_match("/[^[:alnum:]\-_]/",$str)) echo "valid";
demo1 | demo2
Why to use regex? PHP has some built in functionality to do that
<?php
$valid_symbols = array('-', '_');
$string1 = "This is a string*";
$string2 = "this_is-a-string";
if(preg_match('/\s/',$string1) || !ctype_alnum(str_replace($valid_symbols, '', $string1))) {
echo "String 1 not acceptable acceptable";
}
?>
preg_match('/\s/',$username) will check for blank space
!ctype_alnum(str_replace($valid_symbols, '', $string1)) will check for valid_symbols
I want to change a specific character, only if it's previous and following character is of English characters. In other words, the target character is part of the word and not a start or end character.
For Example...
$string = "I am learn*ing *PHP today*";
I want this string to be converted as following.
$newString = "I am learn'ing *PHP today*";
$string = "I am learn*ing *PHP today*";
$newString = preg_replace('/(\w)\*(\w)/', '$1\'$2', $string);
// $newString = "I am learn'ing *PHP today* "
This will match an asterisk surrounded by word characters (letters, digits, underscores). If you only want to do alphabet characters you can do:
preg_replace('/([a-zA-Z])\*([a-zA-Z])/', '$1\'$2', 'I am learn*ing *PHP today*');
The most concise way would be to use "word boundary" characters in your pattern -- they represent a zero-width position between a "word" character and a "non-word" characters. Since * is a non-word character, the word boundaries require the both neighboring characters to be word characters.
No capture groups, no references.
Code: (Demo)
$string = "I am learn*ing *PHP today*";
echo preg_replace('~\b\*\b~', "'", $string);
Output:
I am learn'ing *PHP today*
To replace only alphabetical characters, you need to use a [a-z] as a character range, and use the i flag to make the regex case-insensitive. Since the character you want to replace is an asterisk, you also need to escape it with a backslash, because an asterisk means "match zero or more times" in a regular expression.
$newstring = preg_replace('/([a-z])\*([a-z])/i', "$1'$2", $string);
To replace all occurances of asteric surrounded by letter....
$string = preg_replace('/(\w)*(\w)/', '$1\'$2', $string);
AND
To replace all occurances of asteric where asteric is start and end character of the word....
$string = preg_replace('/*(\w+)*/','\'$1\'', $string);
I have this code:
$string = "123456ABcd9999";
$answer = ereg("([0-9]*)", $string, $digits);
echo $digits[0];
This outputs '123456'. I'd like it to output '1234569999' ie. all the digits. How can I achieve this. I've been trying lots of different regex things but can't figure it out.
First, don't use ereg (it's deprecated). Secondly, why not replace it out:
$answer = preg_replace('#\D#', '', $string);
Note that \D is the inverse of \d. So \d matches all decimal numeric characters (0-9), therefore \D matches anything that \d does not match...
You could use preg_replace for this, preg_replace("/[^0-9]/", "", $string) for example.