php regular expression to check whether a number consists of 5 digits

Question

how to write a regular expression to check whether a number is consisting only of 5 digits?

Answer 1

This regular expression should work nicely:

/^\d{5}$/

This will check if a string consists of only 5 numbers.

• / is the delimiter. It is at the beginning and the end of a regular expression. (User-defined, you can use any character as delimiter).

• ^ is a start of string anchor.

• \d is a shorthand for [0-9], which is a character class matching only digits.

• {5} means repeat the last group or character 5 times.

• $ is the end of string anchor.

• / is the closing delimiter.

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If you want to make sure that the number doesn't start with 0, you can use the following variant:

/^[1-9]\d{4}$/

Where:

• / is the delimiter. It is at the beginning and the end of a regular expression. (User-defined, you can use any character as delimiter).

• ^ is a start of string anchor.

• [1-9] is a character class matching digits ranging from 1 to 9.

• \d is a shorthand for [0-9], which is a character class matching only digits.

• {4} means repeat the last group or character 4 times.

• $ is the end of string anchor.

• / is the closing delimiter.

---

Note that using regular expressions for this kind of validation is far from being ideal.

Answer 2

Try this:

^[0-9]{5}$

The ^ and $ mark the begin and end of the string where the characters described by [0-9] must be repeated 5 times. So this only matches numbers with exactly 5 digits (but it can still be 00000). If you want to allow only numbers from 10000 to 99999:

^[1-9][0-9]{4}$

And if you want to allow any number up to 5 digits (0 to 99999):

^(?:0|[1-9][0-9]{0,4})$

The (?:expr) is just a non-capturing grouping used for the alternation between zero and the other numbers with a non-zero leading digit.

Answer 3

This regex will make sure the number does not start with zeros:

if(preg_match('/^[1-9]\d{4}$/', $number))
    echo "Number is 5 digits\n";
else
    echo "Number is not five digits\n";

But why not use is_numeric() instead?

if(is_numeric($number) && $number >= 10000 && $number <= 99999)
    echo "Number is 5 digits\n";
else
    echo "Number is not five digits\n";

Or you can even just cast it to an integer to make sure it only has an integer value:

if(strval(intval($number)) === "$number" && $number >= 10000 && $number <= 99999)
    echo "Number is 5 digits\n";
else
    echo "Number is not five digits\n";

Answer 4

Although not using regexp, but hopefully faster:

$var = trim($var);

if(strlen($var) == 5 && ctype_digit($var))
{
//
}

EDIT: Added trim function. It's important! It removes spaces so that strlen() function works as expected. Also it makes sure $var is a STRING. You need to pass a string to ctype_digit. If you pass say, an integer it will return false!!!