PHP: RegEx for string with 3-9 letters and 5-50 numbers

This is a classic "password validation"-type problem. For this, the "rough recipe" is to check each condition with a lookahead, then we match everything.

^(?=(?:[^A-Z]*[A-Z]){3,9}[^A-Z]*$)(?=(?:[^0-9]*[0-9]){5,50}[^0-9]*$)[A-Z0-9]*$

I'll explain this one below, but here's a variation that I'll leave for you to figure out.

^(?=(?:[^A-Z]*[A-Z]){3,9}[0-9]*$)(?=(?:[^0-9]*[0-9]){5,50}[A-Z]*$).*$

Let's look at the first regex piece by piece.

1. We anchor the regex between the head of string ^ and end of string $ assertions, ensuring that the match (if any) is the whole string.
2. We have two lookaheads: one for the capital letters, one for the digits.
3. After the lookaheads, [A-Z0-9]* matches the whole string (if it consists only of uppercase ASCII letters and digits). (Thanks to @TimPietzcker for pointing out that I was asleep at the wheel for starting out with a dot-star there.)

How do the lookaheads work?

The (?:[^A-Z]*[A-Z]){3,9}[^A-Z]*$) asserts that at the current position, i.e. the beginning of the string, we are able to match "any number of characters that are not capital letters, followed by a single capital letter", 3 to 9 times. This ensures we have enough capital letters. Note that the {3,9} is greedy, so we will match as many capital letters as possible. But we don't want to match more than we wish to allow, so after the expression quantifies by {3,9}, the lookahead checks that we can match "zero or any number" of characters that are not a capital letter, until the end of the string, marked by the anchor $.

The second lookahead works in similar fashion.

For a more in-depth explanation of this technique, you may want to peruse the password validation section of this page about regex lookarounds.

In case you are interested, here is a token-by-token explanation of the technique.

^                      the beginning of the string
(?=                    look ahead to see if there is:
 (?:                   group, but do not capture (between 3 and 9 times)
  [^A-Z]*              any character except: 'A' to 'Z' (0 or more times)
   [A-Z]               any character of: 'A' to 'Z'
 ){3,9}                end of grouping
  [^A-Z]*              any character except: 'A' to 'Z' (0 or more times)
$                      before an optional n, and the end of the string
)                      end of look-ahead
(?=                    look ahead to see if there is:
 (?:                   group, but do not capture (between 5 and 50 times)
  [^0-9]*              any character except: '0' to '9' (0 or more times)
   [0-9]               any character of: '0' to '9'
 ){5,50}               end of grouping
  [^0-9]*              any character except: '0' to '9' (0 or more times)
$                      before an optional n, and the end of the string
)                      end of look-ahead
[A-Z0-9]*              any character of: 'A' to 'Z', '0' to '9' (0 or more times)
$                      before an optional n, and the end of the string

The Arabic regex is:

[u0600-u06FF]

Actually, ?-? is a subset of this Arabic range, so I think you can remove them from the pattern.

So, in JS it will be

/^[a-z0-9+,()/'su0600-u06FF-]+$/i

See regex demo

For this PHP regex:

$str = preg_replace ( '{(.)1+}', '$1', $str );
$str = preg_replace ( '{[ '-_()]}', '', $str )

In Java:

str = str.replaceAll("(.)\1+", "$1");
str = str.replaceAll("[ '-_\(\)]", "");

I suggest you to provide your input and expected output then you will get better answers on how it can be done in PHP and/or Java.