RE: [PHP] Displaying Results

I used this workaround - if you see something inherently wrong with it, let me know: printf("<a href='http://www.mydomain.net/index.asp?type=%s&%s=%s'><b>%s</b> (%s)</a><br>",$by,$by,$$by,$$by,$total) So that the output goes like <a href="http://www.mydomain.net/index.asp?type=type1&type1=value">type 1 (total#)</a> And that gets passed to this: $type = $HTTP_GET_VARS["type"]; $value = $HTTP_GET_VARS[$type]; $result = mysql_query("SELECT * FROM cars WHERE $type ='$value'"); Little more work, but is this safe to use? It's been working great! -----Original Message----- From: Jason Wong [mailto:php-general@gremlins.com.hk] Sent: Tuesday, April 16, 2002 9:58 PM To: php-general@lists.php.net Subject: Re: [PHP] Displaying Results On Wednesday 17 April 2002 08:21, Jason Soza wrote: > Sorry, I just noticed that the count() function will do at least the > first part of my question, i.e. SELECT year, COUNT(*) FROM cars GROUP > BY year > > But the second part still has me a bit stumped. I know that you can > pass a variable using something like script.php?year=1991, but doesn't > that assume that in your script you have something like: > $query = ("SELECT * FROM mytable WHERE year=$year") > > So the $year is the only variable that gets replaced by the script.php? > year=1991 you called it with. How could I make it so that the > entire 'year=$year' part of the query gets replaced by what comes after > the ? in script.php?year=1991 ? You can do something like: script.php?qry=year%3D1991 %3D is the url encoding for '='. Then in your script: $query = ("SELECT * FROM mytable WHERE $qry") -- Jason Wong -> Gremlins Associates -> www.gremlins.com.hk Open Source Software Systems Integrators * Web Design & Hosting * Internet & Intranet Applications Development * /* The chief danger in life is that you may take too many precautions. -- Alfred Adler */ -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php