I'm trying to make a regex that would allow input including at least one digit and at least one letter (no matter if upper or lower case) AND NOTHING ELSE. Here's what I've come up with:
<?php
if (preg_match('/(?=.*[a-z]+)(?=.*[0-9]+)([^\W])/i',$code)) {
echo "=)";
} else {
echo "=(";
}
?>
While it gives false if I use only digits or only letters, it gives true if I add $ or # or any other non-alphanumeric sign. Now, I tried putting ^\W into class brackets with both a-z and 0-9, tried to use something like ?=.*[^\W] or ?>! but I just can't get it work. Typing in non-alphanums still results in true. Halp meeee
You need to use anchors so that it matches against the entire string.
^(?=.*[a-z]+)(?=.*[0-9]+)(\w+)$
Since you are using php, why even use regex at all. You can use ctype_alnum()
http://php.net/manual/en/function.ctype-alnum.php
Related
I'm new to PHP coding and I made a mistake in the code.
I have like 400 occurrences of a method that I want to change but I don't know how.
I've heard about regular expressions, but I can't apply to this case because I dont know so much of RegExp.
I have this in my code, for example
<?php echo $lang['extension_not_allowed'] ?>
I want to change all the results with $lang for $this->lang(''), for example
<?php echo $this->lang('extension_not_allowed') ?>
There's any way to do it using Ctrl+Shift+H (Global Replace) in NetBeans?
Thanks.
You can use the following regex with replacement:
(<\?php.*?)\$lang\[([^]]*)]
And the replacement:
$1\$this->\$lang($2)
Or remove the (<\?php.*?) part if the $lang may appear on lines without <?php:
\$lang\[([^]]*)]
and replace with "\$this->\$lang($1)".
See demo
A couple of notes:
(<\?php.*?) - matches and captures the text <?php and 0 or more characters other than a newline, as few as possible (with .*?)
\$lang\[ - matches $lang[ literally (note that special regex characters must be escaped)
([^]]*) - matches and captures 0 or more characters other than a ] (we are using a character class [...])
] - a literal ].
PHP preg_match to accept new line
I want to pass every post/string through PHP preg_match function. I want to accept all the alpha-numerics and some special characters. Help me edit my syntax to allow newline. As the users fill textarea and press enter. Following syntax does not allow new line.
Please feedback whether following special characters are properly done or not
*/_:,.?#;-*
if (preg_match("/^[0-9a-zA-Z \/_:,.?#;-]+$/", $string)) {
echo 'good';
else {
echo 'bad';
}
You were almost there!
The DOTALL modifier mentioned by others is irrelevant to your regex.
To allow new lines, we just add \r\n to your character class. Your code becomes:
if (preg_match("/^[\r\n0-9a-zA-Z \/_:,.?#;-]+$/", $string)) {
echo 'good';
else {
echo 'bad';
}
Note that this test and the regex can be written in a tidier way:
echo (preg_match("~^[\r\n\w /:,.?#;-]+$~",$string))? "***Good!***" : "Bad!";
See the result of the online demo at the bottom.
\w matches letters, digits and underscores, so we can get rid of them in the character class
Changing the delimiter to a ~ allows you to use a / slash without escaping it (you need to escape delimiters)
it's always safe to add backslash to any non-alphanumeric characters so:
/^[0-9a-zA-Z \/\_\:\,\.\?\#\;\-]+$/
Also use character classes:
/^[[:alnum:] \/\_\:\,\.\?\#\;\-]+$/
oh about the new lines:
/^[[:alnum:] \r\n\/\_\:\,\.\?\#\;\-]+$/
to be able to do that string ^ (also, it'll be easier/safer to use single quotes)
'/^[[:alnum:] \\r\\n\/\_\:\,\.\?\#\;\-]+$/'
You can use an alternation to factor in the newlines:
/^(?:[0-9a-zA-Z \/_:,.?#;-]|\r?\n)+$/
Btw, you can shorten the expression a bit by replacing [A-Za-z0-9_] with [\w\d]:
/^(?:[\w\d \/:,.?#;-]|\r?\n)+$/
So:
if (preg_match('/^(?:[\w\d \/:,.?#;-]|\r?\n)+$/', $string)) {
echo "good";
} else {
echo "bad";
}
I've just made a few edits to a file and when testing it seemed to not work, I did a bit of debugging and found that preg_match was returning 0, I've looked into it and cannot see what the problem is, also since I haven't touched this part of the file, I'm confused as to what might have happened...
<?php
echo preg_match('/[A-Z]+[a-z]+[0-9]+/', 'testeR123');
?>
This is a snippet I'm using for debugging, I'm guessing my pattern is wrong, but I am probably wrong about that.
Thanks,
P110
According to your comment:
I'm just looking for it to check if there is an uppercase, lowercase and a number, but from the replies, my pattern checks for it in an order
have a try with:
preg_match('/^(?=.*[A-Z])(?=.*[a-z])(?=.*[0-9])[A-Za-z0-9]+$/', $input_string);
where
(?=.*[A-Z]) checks there are at least one uppercase
(?=.*[a-z]) checks there are at least one lowercase
(?=.*[0-9]) checks there are at least one digit
[A-Za-z0-9]+ checks there are only these characters.
(?=...) is called lookahead.
The problem is the order of the letters:
Try this:
echo preg_match('/[a-z]+[A-Z]+[0-9]+/', 'testeR123');
Or:
echo preg_match('/[A-Z]+[a-z]+[0-9]+/', 'Rtest123');
Or simpler
echo preg_match('/[A-Z]+[0-9]+/i', 'testeR123');
Your regex first test if there are Capital letters from A to Z then if there are lowercase letters from at to z and then if there are numbers. since your string starts with an lowercase it will not match.
i think you want to do this
[A-Za-z0-9]+
Or if you need that your string starts with a lowecase string then an uppercase string and then numbers you should change the regex to.
[a-z]+[A-Z]+[0-9]+
In that way your current string would fit the regex as well.
<?php
preg_match('/([A-Za-z0-9]+)/', 'testeR123', $match);
echo $match[1];
?>
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',$str);
}
I'm going to be different and use Character.isLetter definition of what is a letter.
if (myString.matches("\\p{javaLetter}*"))
Note that this matches more than just [A-Za-z]*.
A character is considered to be a letter if its general category type, provided by Character.getType(ch), is any of the following: UPPERCASE_LETTER, LOWERCASE_LETTER, TITLECASE_LETTER, MODIFIER_LETTER, OTHER_LETTER
Not all letters have case. Many characters are letters but are neither uppercase nor lowercase nor titlecase.
The \p{javaXXX} character classes is defined in Pattern API.
Alternatively, try checking if it contains anything other than letters: [^A-Za-z]
The easiest way to do a "is ALL characters of a given type" is to check if ANY character is NOT of the type.
So if \W denotes a non-character, then just check for one of those.
I want a regular expression to validate a nickname: 6 to 36 characters, it should contain at least one letter. Other allowed characters: 0-9 and underscores.
This is what I have now:
if(!preg_match('/^.*(?=\d{0,})(?=[a-zA-Z]{1,})(?=[a-zA-Z0-9_]{6,36}).*$/i', $value)){
echo 'bad';
}
else{
echo 'good';
}
This seems to work, but when a validate this strings for example:
11111111111a > is not valid, but it should
aaaaaaa!aaaa > is valid, but it shouldn't
Any ideas to make this regexp better?
I would actually split your task into two regex:
to find out whether it's a valid word: /^\w{6,36}$/i
to find out whether it contains a letter /[a-z]/i
I think it's much simpler this way.
Try this:
'/^(?=.*[a-z])\w{6,36}$/i'
Here are some of the problems with your original regex:
/^.*(?=\d{0,})(?=[a-zA-Z]{1,})(?=[a-zA-Z0-9_]{6,36}).*$/i
(?=\d{0,}): What is this for??? This is always true and doesn't do anything!
(?=[a-zA-Z]{1,}): You don't need the {1,} part, you just need to find one letter, and i flag also allows you to omit A-Z
/^.*: You're matching these outside of the lookaround; it should be inside
(?=[a-zA-Z0-9_]{6,36}).*$: this means that as long as there are between 6-36 \w characters, everything else in the rest of the string matches! The string can be 100 characters long mostly containing illegal characters and it will still match!
You can do it easily using two calls to preg_match as:
if( preg_match('/^[a-z0-9_]{6,36}$/i',$input) && preg_match('/[a-z]/i',$input)) {
// good
} else {
// bad
}