Regular expression to compare two variables similar to SQL: field LIKE 'value%'

[bissatch]

I am trying to compare two variable but using regular expression:

$access = "glasgow"

$areaid = "glasgow-westend-byers_rd"

using the two variables, I would like the areaid variable to be
compared with the access variable. the regular expression function will
return true if $access matches the start of $areaid. So the two above
would return true

The following would also return true:

$access = "glasgow-westend"

$areaid = "glasgow-westend-byers_rd"

or

$areaid = "glasgow-westend-greatwestern_rd"

However, the following should return false though:

$access = "glasgow"

$areaid = "stirling-glasgow_rd"

Although the value for $access, 'glasgow', appears in the $areaid it
does not appear at the start.
If I were to do this in SQL it would be something like:

$select = "SELECT * FROM areas WHERE areaid LIKE '" . $access . '%"

Note the use of '%' at the end of the compare LIKE field

Unfortunetely I have never been able to properly get my head round
regular expressions.

Any help?

Cheers, burnsy

[Andy Jeffries]

bi******@yahoo.co.uk wrote:

I am trying to compare two variable but using regular expression:

$access = "glasgow"

$areaid = "glasgow-westend-byers_rd"

using the two variables, I would like the areaid variable to be
compared with the access variable. the regular expression function will
return true if $access matches the start of $areaid. So the two above
would return true

if (ereg("^$access", $areaid)) {
// ... do something
}

Cheers,
Andy

[John Dunlop]

bissatch wrote:

$areaid = "glasgow-westend-byers_rd"

on a sidenote, bissatch, the street between Dumbarton Road and Great
Western Road spells its name with an 're', does it not? it runs through
what was once called the Byres of Partick ('byre' = shed for keeping
cattle in). which begs the question, is there more heifers there now (on
a Friday night) than there was then?

--
Jock