What is the Regex for Only One # Character? - php

Question

I am looking for a Regex to use in PHP in order to match one character; the # symbol.
For example, if I typed: P#ssword into an input, the Regex will match. If I typed P##ssword into an input, the regex will not match.
Here is my PHP Code that I am using:
<?php
session_start();
if($_SERVER['REQUEST_METHOD'] == "POST") {
$conn=mssql_connect('d','dd','d');
mssql_select_db('d',$conn);
if(! $conn )
{
die('Could not connect: ' . mssql_get_last_message());
}
$username = ($_POST['username']);
$password = ($_POST['password']);
if (preg_match("[\W]",$_POST["password"]))
{
if (!preg_match("^[^#]*#[^#]*$",$_POST["password"]))
{
header("location:logingbm.php");
} else {
}
}
if(!filter_var($_POST['username'], FILTER_VALIDATE_EMAIL))
{
if ($_POST["username"])
{
if ($_POST["password"])
{
$result = mssql_query("SELECT * FROM staffportal WHERE email='".$username."' AND
password='".$password."'");
if(mssql_num_rows($result) > 0) {
$_SESSION['staff_logged_in'] = 1;
$_SESSION['username'] = $username;
}}}} else {
if ($_POST["password"])
{
$result = mssql_query("SELECT * FROM staffportal WHERE email='".$username."' AND
password='".$password."'");
if(mssql_num_rows($result) > 0) {
$_SESSION['staff_logged_in'] = 1;
$_SESSION['username'] = $username;
}}}}
if(!isset($_SESSION['staff_logged_in'])) {
header("location:logingbm.php");
echo "<script>alert('Incorrect log-in information!');</script>";
} else {
header("location:staffportal.php");
}
?>

Answer 1

Other lightweight approaches...
Without regex
Just use substr_count (see demo)
<?php
$str1 = "pa#s#s";
$str2 = "pa#ss";
echo (substr_count($str1,"#")==1)?"beauty\n":"abject\n"; // abject
echo (substr_count($str2,"#")==1)?"beauty\n":"abject\n"; // beauty
With regex
EDIT: just saw that Sam wrote something equivalent.
If you want to use regex, you could use this fairly simple regex:
#
How? This code (see demo)
<?php
$str1 = "pa#s#s";
$str2 = "pa#ss";
$regex = "~#~";
echo (preg_match_all($regex,$str1,$m)==1)?"beauty\n":"abject\n"; // abject
echo (preg_match_all($regex,$str2,$m)==1)?"beauty\n":"abject\n"; // beauty

Answer 2

The easiest way would be to use the return value of preg_match_all().
Returns the number of full pattern matches (which might be zero), or FALSE if an error occurred.
Example:
$count = preg_match_all('/#/', $password, $matches);
Non regex solution (based off of #cdhowie's comment):
$string = 'P#ssword';
$length = strlen($string);
$count = 0;
for($i = 0; $i < $length; $i++) {
if($string[$i] === '#') {
$count++;
}
}
This works because you can access characters of Strings as you would with normal arrays ($var = 'foo'; $var[0] = 'f';).

Answer 3

As I said in my comment, your pattern needs delimiters /, #, ~ or whatever you want (see the PHP doc for that and test yourself).
To be quickly sure that a string contains only one #, you can do that:
if (preg_match('~\A[^#]*#[^#]*\z~', $yourstr))
echo 'There is one #';
else
echo 'There is more than one # or zero #';

Answer 4

This regexp will do what you want:
^[^#]*#[^#]*$
This matches any line that contains one and only one #.
Explanation
^ matches the beginning of the line
[^#]* matches everything before the #
# matches the # character
[^#]* matches everything after the #
$ matches the end of the line
Use
preg_match("#^[^#]*#[^#]*$#", $passwd); //Matches $passwd if it contains only one character

Answer 5

Here's what your regex code means:
If there is at least one non-word character in the string ([\W]), there must be exactly one at-sign (#). There may be any number of any other characters before and after the at-sign: letters, digits, control characters, punctuation, anything. Anything but #.
What I'm wondering is, are you trying to say there can be not more than one at-sign (i.e. zero or one?) That's pretty simple, conceptually; just get rid of the first regex check ("[\W]") and change the second regex to this:
"^[^#]*(?:#[^#]*)?$"
In other words:
Start by consuming not at-signs you see. If you see a #, go ahead and consume it, then resume matching whatever not at-signs remain. If that doesn't leave you at the end of the string, it can only mean there were more than one #. Abandon the attempt immediately and report a failed match.
Of course, this still leaves you with the problem of which other characters you want to allow. I'm pretty sure [^#]* is not what you want.
Also, "[\W]" may be working as you intended, but it's only by accident. You could have written it "/\W/" or "~\W~" "(\W)" and it would work just the same. You may have meant those square brackets to form a character class, but they're not even part of the regex; they're the regex delimiters.
So why did it work, you ask? \W is a predefined character class, equivalent to [^\w]. You can use it inside a regular character class, but it works fine on its own.