luci2200 Posted March 20, 2008 Share Posted March 20, 2008 Hello, I have a problem retrieving and show data from a database. I have the following query: $host1="localhost"; // Host name $username1="user"; // Mysql username $password1="********"; // Mysql password $db_name1="dbname"; // Database name $link1 = mysql_connect($host1, $user1, $password1) or die("I can't connect!"); $tabela = mysql_select_db($db_name1, $link1) or die("I can't select table!"); $query="SELECT "; $query.='trafficSummaries.sitesID AS sitesID'; $query.=','; $query.='sites.site AS site'; $query.=','; $query.='SUM(trafficSummaries.inCache+trafficSummaries.outCache) AS bytes'; $query.=','; $query.='TRUNCATE((SUM(trafficSummaries.inCache)/SUM(trafficSummaries.inCache+trafficSummaries.outCache))*100,0) AS cachePercent'; $query.=' FROM trafficSummaries'; $query.=' JOIN sites ON '; $query.='trafficSummaries.sitesID=sites.id'; $query.=' AND '; $query.='trafficSummaries.date=sites.date'; $query.=' WHERE '; $query.="trafficSummaries.ip='".$pageVars['hostiplong']."'"; $query.=" AND"; $query.=" trafficSummaries.date='".$pageVars['date']."'"; $query.=" GROUP BY trafficSummaries.sitesID"; $query.=' ORDER BY '.$pageVars['OrderBy'].' '.$pageVars['OrderMethod']; $resultIP = mysql_query($query, $link1); Here I try to get the objects from the tables: echo "<center> <table> <tr> <th></th> <th>SITE</th> <th width=110>Traffic</th> </tr>"; while ($row = mysql_fetch_object($$resultIP)) { echo "<tr onMouseOver=bgColor=#C5D3E7; onMouseOut=bgColor=#DAE3F0> <td style=text-align: center;><b>$row->site</b></td> <td style=text-align: left;>$row->bytes</a></td> <td></td> </tr>"; } echo "<tr> <th>TOTALS</th> <th style=text-align: right;></th> <th style=text-align: right;></th> </tr> </table> </center>"; This is the first time I using such query to get informations from more tables. I used the output of of the $query into mysql and I get the informations, but when I try from php I get no error. Can anyone give me an idea? Thank you! Quote Link to comment Share on other sites More sharing options...
fenway Posted March 20, 2008 Share Posted March 20, 2008 First, echo $query. Second, add a mysql_error() call to make sure there are no syntax errors. Quote Link to comment Share on other sites More sharing options...
luci2200 Posted March 20, 2008 Author Share Posted March 20, 2008 I found the problem. On the while loop at resultIP were 2 $ at the beginning of that variable. Was my mistake! Thank you! Quote Link to comment Share on other sites More sharing options...
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