php validate a pattern with regex - php

I try to validate a pattern like 123*5000.
The pattern has 3 parts.
First, the [123] must contain three digits unique number.
The second is *, and the third is 5000, must be an integer, not unique is fine.
If the pattern is 223*443 it will return false (first part 223, number 2 it not unique).
If the pattern is 908*22 it will return true (first part contain unique number, second part is *, third part is integer).
If the pattern is 34*5000 it will return false (first part only contain 2 digits), and etc.
I try regex \^[0-9][0-9][0-9][*][0-9]+$\ but it did not solve the first part pattern that must contains unique number.
Can anyone help me?

You might write the pattern as:
^(?!\d*(\d)\d*\1)\d{3}\*\d+$
The pattern matches:
^ Start of string
(?!\d*(\d)\d*\1) Negative lookahead, assert not 2 of the same digits
\d{3}\*\d+ Match 3 digits, then * and 1 or more digits
$ End of string
See a regex demo.

Related

My pattern in preg_match not working as i want

I want to get:
false for A3312+A192389+B2323+B948348
false for A6712+A1922389
false for A4512
true for A4552+B948348
(only one Aelement and one or more Belement)
I tried:
print_r(preg_match('/^A((?!\+A).)\+B(.*)$/', $string));
So it looks as if the basic pattern you're after is "A(digits)+B(digits)"
Your expression seems a bit over-complicated for that purpose, I'd simply use:
preg_match('/^A\d+(\+B\d+)+$/', $input, $match);
If the input can be alphanumeric (A(alnum)+B(alnum), just use
preg_match('/^A[:alnum:]+(\+B[:alnum:]+)+$/', $input, $match);
instead.
Basically, the 2 absolute hard requirements are: the input string should start with an upper-case A, and there should be one + sign, followed by an upper-case B. Whatever the characters in between should be, you just have to add a character group that best fits your requirements. From the examples you gave \d+ (one or more digits) seems to fit the bill. If "A00FF33+B123ABC" should be valid, I'd either use [:alnum:] or [0-9A-F] (for hex values) instead.
The trick for the one-or-more requirement is to create a group for the +Belement part of the match, and repeat that group one or more times:
\+B\d+ //matches once
(\+B\d+)+ //matches once or more
This regex will work for you:
^A\d+\+B\d+(?:\+B\d+)*$
Use it as:
preg_match('/^A\d+\+B\d+(?:\+B\d+)*$/', $string);
This matches A followed by digits the +B followed by digits repeated 1 or more times.
A4552+B948348 is matched,
A4552+B948348+B948348+B948348 is matched,
A3312+A192389+B2323+B948348 is not matched
And with not-only decimals after A/B
^A[^\\+]+(?:\\+B[^\\+]+)+$

Match string that doesn't have number after letter

I've got a scenario as follows. Our systems needs to pull filters from a string passed in as a query parameter, but also throw a 404 error if the string isn't correctly formatted. So let's take the following three strings as an exmple:
pf0pt1000r
pfasdfadf
pf2000pt2100
By the application requirements, only #3 is supposed to match as a "valid" string. So my current regex to match that is /([a-z]+)(\d+)/. But this also matches #1, if not entirely, but it still matches.
My problem thus is twofold - I need 2 patterns, 1 that will match only the 3rd string in this list, and another that will match the "not-acceptable" strings 1 and 2. I believe there must be some way to "negate" a pattern (so then I'd technically only need one pattern, I'm assuming), but I'm not sure how exactly to do that.
Thanks for any help!
EDIT
For clarity's sake, let me explain. The "filter parameters" present here take the following structure - 1 or 2 letters, followed by a number of, well, numbers. That structure can repeat itself however many times. So for example, valid filter strings could be
pf100pt2000
pf100pt2000r2wp0to1
etc.
Invalid strings could be
pf10000pt2000r
pf3000pt2123wpno
... anything not following the structure above.
After clarifying the question:
^([a-zA-Z]{1,2}\d+)*$
Explanation:
[a-zA-Z] - a lower or upper case letter
{1,2} - one or two of those
\d+ - one or more digits
()* - the whole thing repeated any number of times
^$ - match the entire string from start(^) to end($)
You can use this regex for valid input:
^([a-zA-Z]+\d+)+$
RegEx Demo 1
To find invalid inputs use:
^(?!([a-zA-Z]+\d+)+$).+$
RegEx Demo 2
/^(?:(?:[a-z]+)(?:\d+))*$/
You were hella close, man. Just need to repeat that pattern over and over again till the end.
Change the * to a + to reject the empty string.
Oh, you had more specific requirements, try this:
/^(?:[a-z]{1,2}\d+)*$/
Broken down:
^ - Matches the start of the string an anchor
(?: - start a non-capturing group
[a-z] - A to Z. This you already had.
{1,2} - Repeat 1 or 2 times
\d+ - a digit or more You had this, too.
)* - Repeat that group ad nauseum
$ - Match the end of the string
If you only want digits at the end of the string, then
/\d$/
would do. \d = digit, $ = end of string.

Phone no contain this patteren AABBCC e.g 112233

I want to check if phone no contains this pattern AABBCC
Where A[0-9],B[0-9],C[0,9] They should be different e.g 112233,553322,887766
Let Us Suppose
I Have a phone no 03334112233
It will say yes pattern matched.
PHP Code but It Is For Exact String
$str = 'aabbaabbccaass'; //or whatever
if (preg_match('/(?!.*?aabbcc)^.*$/', $str))
echo "accepted\n";
else
echo "rejected\n";
Problem i don't know how to do if string is for numbers
Possible Duplicate
but it does not contain answer and exact detail.
Edited :
I want to match the last 6 characters of the string in this pattern AABBCC e.g 03329112233
To match number with AABBCC format, you can use this pattern:
(?:(\d)\1(?!\1)){2}(\d)\2
example of use:
if (preg_match('/(?:(\d)\1(?!\1)){2}(\d)\2/', $str)
echo "rejected\n";
else
echo "accepted\n";
But if you have other tests to do (for example that there is only digits), it can be more flexible to use it in this way:
if (preg_match('/(?!.*(?:(\d)\1(?!\1)){2}(\d)\2)^\d+$/', $str)
echo "accepted\n";
else
echo "rejected\n";
pattern details:
(?: # open a non capturing group that describes a repeated digit
(\d) # capture the first digit with group 1
\1 # a backreference to group 1 (the same digit thus)
(?!\1) # check with a negative lookahead that the same digit doesn't follow
){2} # repeat the group two times
(\d)\2 # same thing for digits 5 & 6 (the lookahead isn't needed here)
Note that the digit in the capture group change at each repetition of the non capturing group (because the negative lookahead forces it).
Notice: if you want to reject numbers that contains, for example, 111122 or 112222 or 111111, you only need to remove the negative lookahead.
if you want to reject numbers with the format 112211 or 448844, you must change the pattern like this: (\d)\1(?!\d{0,2}\1)(\d)\2(?!\2)(\d)\3
As I understand, you only want to match the last 6 characters of the string, if they are digits, and of 3 all different digit pairs. Would also use a lookahead and some pattern like this:
(?>((\d)\2)(?!.*\1)){3}$
\2 checks for an equivalent of 2nd capturing group, which is one digit (shorthand \d)
using a negative lookahead to check, if not followed by .* any amount of any characters, followed by equivalent of 1st capturing group (which contains 2 equal digits).
{3} 3 repitions at $ end of string.
Test on regex101.com, Regex FAQ
Your regex should be like this:
^((\d)\2){3}$
It is simpler and also works.
You can use capturing groups and backreferences like this:
if (preg_match('/(?!.*(.)\1(.)\2(.)\3)^.*$/', $str))
The (.) will match any single character and assign it to a group. The first instance is assigned to group 1, the second to group 2 and so on. Later in the pattern, the backreference \1 will match exactly what was previously captured in group the first group, \2 will match what was captured in the second group, etc.
You probably will also want to use \d to match any single digit (it's only necessary to use this outside of the lookahead) and a {n,m} quantifier to match between n and m digits. For example, the following will match any sequence of 7 to 10 digits that does not contain a subsequence like AABBCC:
if (preg_match('/(?!.*(.)\1(.)\2(.)\3)^\d{7,10}$/', $str))

Regex Rules for First and Second Character

I need help on following regular expression rules of javascript and php.
JS
var charFilter = new RegExp("^[A|B].+[^0123456789]$");
PHP
if (!preg_match('/^[A|B].+[^0123456789]$/', $data_array['sample_textfield'])) {
This regular expression is about
First character must be start with A or B and last character must not include 0 to 9.
I have another validation about, character must be min 3 character and max 6 number.
New rule I want to add is, second character cannot be C, if first letter is A.
Which means
ADA (is valid)
ACA (is not valid)
So I changed the regex code like this
JS
var charFilter = new RegExp("^(A[^C])|(B).+[^0123456789]$");
PHP
if (!preg_match('/^(A[^C])|(B).+[^0123456789]$/', $data_array['sample_textfield'])) {
It is worked for first and second character. If i type
ACA (it says invalid) , But if i type
AD3 (it says valid), it doesn't check the last character anymore. Last character must not contain 0 to 9 number, but it's show as valid.
Can anyone help me to fix that regex code for me ? Thank you so much.
Putting all of your requirements together, it seems that you want this pattern:
^(?=.{3,6}$)(?=A(?!C)|B).+\D$
That is:
From the beginning of the string ^
We can assert that there are between 3 to 6 of "any" characters to end of the string (?=.{3,6}$)
We can also assert that it starts with A not followed by C, or starts with B (?=A(?!C)|B)
And the whole thing doesn't end with a digit .+\D$
This will match (as seen on rubular.com):
= match = = no match =
ADA ACA
ABCD AD3
ABCDE ABCDEFG
ABCDEF
A123X
A X
Note that spaces are allowed by .+ and \D. If you insist on no spaces, you can use e.g. (?=\S{3,6}$) in the first part of the pattern.
(?=…) is positive lookahead; it asserts that a given pattern can be matched. (?!…) is negative lookahead; it asserts that a given pattern can NOT be matched.
References
regular-expressions.info
Lookarounds, Alternation, Anchors, Repetition, Dot, Character Class
Related questions
How does the regular expression (?<=#)[^#]+(?=#) work?
On alternation precedence
The problem with the original pattern is in misunderstanding the precedence of the alternation | specifier.
Consider the following pattern:
this|that-thing
This pattern consists of two alternates, one that matches "this", and another that matches "that-thing". Contrast this with the following pattern:
(this|that)-thing
Now this pattern matches "this-thing" or "that-thing", thanks to the grouping (…). Coincidentally it also creates a capturing group (which will capture either "this" or "that"). If you don't need the capturing feature, but you need the grouping aspect, use a non-capturing group ``(?:…)`.
Another example of where grouping is desired is with repetition: ha{3} matches "haaa", but (ha){3} matches "hahaha".
References
regular-expressions.info/Brackets for Grouping
Your OR is against the wrong grouping. Try:
^((A[^C])|(B)).+[^0123456789]$
In jasonbars solution the reason it doesn't match ABC is because it requires A followed by not C, which is two characters, followed by one or more of any character followed by a non number. Thus if the string begins with an A the minimum length is 4. You can solve this by using a look ahead assertion.
PHP
$pattern = '#^(A(?=[^C])|B).+\D$#';
i think it should be like
/^(A[^C]|B.).*[^0-9]$/
try this test code
$test = "
A
B
AB
AC
AAA
ABA
ACA
AA9
add more
";
$pat = '/^(A[^C]|B.).*[^0-9]$/';
foreach(preg_split('~\s+~', $test) as $p)
printf("%5s : %s\n<br>", $p, preg_match($pat, $p) ? "ok" : "not ok");

Can Someone explain this reg ex to me?

I recently asked a question on formatting a telephone number and I got lots of responses. Most of the responses were great but one i really wanted to figure out what its doing because it worked great. If phone is the following how do the other lines work...what are they doing so i can learn
$phone = "(407)888-9999";
$phone = preg_replace("~[^0-9]~", "", $phone);
preg_match('~([0-9]{3})([0-9]{3})([0-9]{4})~', $phone, $matches);
Let's break the code into two lines.
preg_replace("~[^0-9]~", "", $phone);
First, we're going to replace matches to a regex with an empty string (in other words, delete matches from the string). The regex is [^0-9] (the ~ on each end is a delimiter). [...] in a regex defines a character class, which tells the regex engine to match one character within the class. Dashes are generally special characters inside a character class, and are used to specify a range (ie. 0-9 means all characters between 0 and 9, inclusive).
You can think of a character class like a shorthand for a big OR condition: ie. [0-9] is a shorthand for 1 or 2 or 3 or 4 or 5 or 6 or 7 or 8 or 9. Note that classes don't have to contain ranges, either -- [aeiou] is a character class that matches a or e or i or o or u (or in other words, any vowel).
When the first character in the class is ^, the class is negated, which means that the regex engine should match any character that isn't in the class. So when you put all that together, the first line removes anything that isn't a digit (a character between 0 and 9) from $phone.
preg_match('~([0-9]{3})([0-9]{3})([0-9]{4})~', $phone, $matches);
The second line tries to match $phone against a second expression, and puts the results into an array called $matches, if a match is made. You will note there are three sets of brackets; these define capturing groups -- ie. if there is a match of a pattern as a whole, you will end up with three submatches, which in this case will contain the area code, prefix and suffix of the phone number. In general, anything contained in brackets in a regular expression is capturing (while there are exceptions, they are beyond the scope of this explanation). Groups can be useful for other things too, without wanting the overhead of capturing, so a group can be made non-capturing by prefacing it with ?: (ie. (?:...)).
Each group does a similar thing: [0-9]{3} or [0-9]{4}. As we saw above, [0-9] defines a character class containing the digits between 0 and 9 (as the classes here don't start with ^, these are not negated groups). The {3} or {4} is a repetition operator, which says "match exactly 3 (or 4) of the previous token (or group)". So [0-9]{3} will match exactly three digits in a row, and [0-9]{4} will match exactly four digits in a row. Note that the digits don't have to be all the same (ie. 111), because the character class is evaluate for each repetition (so 123 will match because 1 matches [0-9], then 2 matches [0-9], and then 3 matches [0-9]).
In the preg_replace it looks for anything that is not, ^ inside of the [], 0-9 (basically not a number) and replaces / removes it from that string given the replacement is "".
For the first section, it pulls out the first 3 numbers ([0-9]{3}) the {3} is the number of characters to match the items inside the [] are what to match and since this is inside of paranthesis () it stores it as a match in the array $matches. The second part pulls out the next 3 numbers and the last part pulls out the last 4 numbers from $phone and stores the matches that were matched in $matches.
The ~ are delimeters for the regular expressions.
You know it's a regular expression from the regex tag.
So, you are pattern matching.
The pattern you are matching is: [^0-9] followed by the phone number.
[^0-9] is NOT '^' any one digit
So, the match after that is any 3 digits, followed by any 3 digits, followed by any 4 digits.
I don't think it will match because of the () around the area code and the dash are missing.
I'd do this:
~\(([0-9]{3})\)([0-9]{3})-([0-9]{4})~'
"[^0-9]" means everything but numbers from 0 to 9. So basically, first line replace everything but numbers with "" (nothing)
[0-9]{3} means number from 0 to 9, 3 times in a row.
So it check if you have 3 numbers then 3 numbers than 4 numbers and try to match it with $matches.
Check this tuts
Using Regular Expressions with PHP
http://www.webcheatsheet.com/php/regular_expressions.php
$phone = "(407)888-9999";
$phone = preg_replace("~[^0-9]~", "", $phone);
In php you have to delimit regex pattern in some non-alphanumeric character "~" is used here.
[^0-9] is regex pattern used to remove anything out of $phone that is not in 0-9 range remember [^...] will negate the pattern it precedes.
preg_match('~([0-9]{3})([0-9]{3})([0-9]{4})~', $phone, $matches);
Again in this line of code you have "~" as delimiter and
([0-9]{3}) this part of pattern will return 3 numbers from string (note: {} is used to specify range/number of characters to match) in a different output array dimension (check your $matches variable for result) using ( ) in a pattern results in groups/submatches

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