Hello Dear Friends I am facing a issue as follows when I am setting up DB Parse error: syntax error, unexpected '=' ......projects/jobsblue/conn.php on line 11 And My conn.php is as follows ===========conn.php=============================== <? $db_host = "localhost"; $db_username = "db_user"; $db_password = "password"; $db_name = "db_name"; $site_name = "mydomain.com"; $site_title = "My Career Site"; $url = "http://www.yourdomain.com"; /////////////////////////// $connection = mysql_connect($db_host, $db_username, $db_password) or die(mysql_error()); $db = mysql_select_db($db_name, $connection); $qset = "select * from job_settings"; $rset = mysql_query($qset) or die(mysql_error()); $aset = mysql_fetch_array($rset); ?> Please help me this , Thanks a many in advance
I've the same issue here... Try to get this code working (i'm learning php just by luck) <?php $query = "SELECT * FROM #__ezrealty_locality WHERE id = $cityid"; $database->setQuery( $query ); $locrows = $database->loadObjectList(); if ($database->getErrorNum()) { echo $database->stderr(); return false; } echo" <table> <tr> <th>City List</th> </tr> while($locrows = mysql_fetch_array($query)){ extract($locrows); echo" <tr> <td>$cityid</td> </tr>"; } echo"</table>"; /*ERROR HERE*/ ?> PHP: and i get this error.. Parse error: syntax error, unexpected '>' in */testloc.php on line 27 pls need help I want it will look like this City List city 1 city 2 city 3 city 4
lochoe, look at the syntax highlighting on the forum, you're not closing an echo double quote properly
thanx iDemonix, i just write using textpad, so the syntax does not look really well with that.. i'm gonna use rapid php , since my friend suggest that application.. anyway, i'm really2 newbie in php.. very chicken.. thanx for ur reply