and please, next time paste the error or tell us at least, if it is a php error or a mysql error and the line on which it occured and mark that line in your sample code, so someone can look at it , understand it and help you.
Regards Michael "Michael Virnstein" <[EMAIL PROTECTED]> schrieb im Newsbeitrag [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... > $this usually is a self-reference inside a class. > Use it with care! > try to "echo $sql;", perhaps this tells you more. > > Regards Michael > > "James Opere" <[EMAIL PROTECTED]> schrieb im Newsbeitrag > FC788AB9771FD6118E6F0002A5AD7B8F7268AB@ICRAFNTTRAIN">news:FC788AB9771FD6118E6F0002A5AD7B8F7268AB@ICRAFNTTRAIN... > > Hi All, > > I'm trying to pass variables from one form to the other.I have a problem > > when i want to do the the following: > > 1.COUNT($variable) > > 2.DISTINCT($variable) > > ............. > > I realise i can not use the brackets in my query and the variable be > > recognised.When i add COUNT without the brackets i still get an error. > > Example. > > test.html > > <form action="me.php" method="post"> > > <input type="text" name="this"> > > .............. > > This is sent to : > > > > me.php > > <?php > > $db=mysql_connect('localhost','',''); > > mysql_select_db($database,$db); > > $sql="select COUNT($this) from $table group by $this"; > > ................ > > ?> > > This gives an error. > > Please help. > > > > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
