Todd88 Posted March 5, 2008 Share Posted March 5, 2008 Fatal error: Call to a member function on a non-object in /mysite/proof.php on line 14 I am using a mysql database to get the values for proof on my payment site. I guess I have something wrong in the code and it would be great if someone could just point it out to me. Here is the file: <? $includes[title]="Proof Of Payments"; ?> <table class="tableStyle" style="width:100%"> <tr> <th>Id</th> <th>Username</th> <th>Method</th> <th>Amount</th> <th>Date</th> </tr> <? $sql=$Db1->query("SELECT * FROM payment_history ORDER BY dsub DESC"); while($row=$Db1->fetch_array($sql)) { echo "<tr> <td>".$row['id']."</td> <td>".$row['username']."</td> <td>".$row['accounttype']."</td> <td>".$row['amount']."</td> <td>".date("M.D.Y",mktime(0,0,$row['dsub'],1,1,1970))."</td> </tr>"; } ?> </table> Any help is appreciated! Thanks! Quote Link to comment Share on other sites More sharing options...
Isityou Posted March 5, 2008 Share Posted March 5, 2008 It seems your 'fetch_array()' method does not exist in your $Db1 object. Quote Link to comment Share on other sites More sharing options...
Todd88 Posted March 5, 2008 Author Share Posted March 5, 2008 I still can't seem to fix it Quote Link to comment Share on other sites More sharing options...
darkfreaks Posted March 5, 2008 Share Posted March 5, 2008 he is right where is the fetch_array function being called ??? Quote Link to comment Share on other sites More sharing options...
Todd88 Posted March 10, 2008 Author Share Posted March 10, 2008 I used a similiar method for a different section of the site but it doesn't work for this one. Here is the code for the other section: $sql=$Db1->query("SELECT clicked_today, SUM(clicked_today) FROM user GROUP BY permission"); $clickstoday=$Db1->fetch_array($sql); It works this way Quote Link to comment Share on other sites More sharing options...
uniflare Posted March 10, 2008 Share Posted March 10, 2008 you can debug classes with: var_dump($Db1); this will list functions and variables i believe Quote Link to comment Share on other sites More sharing options...
KrisNz Posted March 10, 2008 Share Posted March 10, 2008 Is there a point in your script (or an included script) where you create an instance of your database class and assign it to $Db1? e.g $Db1 = new Database(); Quote Link to comment Share on other sites More sharing options...
Todd88 Posted March 10, 2008 Author Share Posted March 10, 2008 I finally got it to work! Thanks guys Quote Link to comment Share on other sites More sharing options...
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