Re: SQL syntax?

Umm.. I meant you need to put $sql .= "FROM workorderform "; between these: $sql .= "AdSize AS Ad_Size, CPM AS CPM_Rate, "; $sql .= "ORDER BY StartDate DESC"; :) Terion Miller wrote: > On Fri, Dec 5, 2008 at 3:57 PM, Allan Arguelles <allan@arguelles.com.ph>wrote: > > >>> $sql = "SELECT WorkOrderID AS Work_Order_ID, DATE_FORMAT(StartDate, >>> >> '%b. >> >>> %e, %Y %l:%i %p') AS Start_Date, >>> DATE_FORMAT(EndDate, '%b. %e, %Y %l:%i %p') AS End_Date, "; >>> $sql .= "Advertiser AS Advertiser_Name,AccountNum AS Account_Number, >>> Impressions AS Ad_Impressions, "; >>> $sql .= "AdSize AS Ad_Size, CPM AS CPM_Rate, "; >>> $sql .= "ORDER BY StartDate DESC"; >>> >>> >> You forgot the tables, plus you have an extra comma after CPM_Rate. >> >> well I changed it to: >> >> $sql = "SELECT workorderform.WorkOrderID AS Work_Order_ID, >> DATE_FORMAT(workorderform.StartDate, '%b. %e, %Y %l:%i %p') AS Start_Date, >> DATE_FORMAT(workorderform.EndDate, '%b. %e, %Y %l:%i %p') AS End_Date, >> "; >> $sql .= "workorderform.Advertiser AS >> Advertiser_Name,workorderform.AccountNum AS Account_Number, >> workorderform.Impressions AS Ad_Impressions, "; >> $sql .= "workorderform.AdSize AS Ad_Size, workorderform.CPM AS >> CPM_Rate "; >> $sql .= "ORDER BY StartDate DESC"; >> >> and got the same error.... >> >> >> > >