chrishawkins Posted November 30, 2009 Share Posted November 30, 2009 I am getting this error; Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in running this code; <form name="form1" method="post" action=""> <label> <input type="submit" name="update" id="update" value="Update Records"> </label> </form> <form name="form2" method="get" action="test.php?show_record=1"> <label> <input type="submit" name="show_results" id="show_results" value="Show Results"> </label> </form> <?php $con = mysql_connect("localhost","username","password"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("my_db", $con); if(isset($_GET['show_results']) && $_GET['show_results'] == 1){ $res = mysql_query("SELECT *, probid_users.name AS name, probid_winners.winner_id AS winnerid FROM probid_winners left join probid_users ON probid_users.user_id = probid_winners.buyer_id WHERE probid_winners.payment_status='confirmed' AND printed='' AND buyer_id BETWEEN 2 AND 1000000000 ORDER BY bid_amount DESC = " . mysql_real_escape_string($_GET['show_results'])); } echo "<table border='5'> <tr> <th>Name</th> <th>Auction ID</th> <th>Bid Amount</th> <th>Deposit Status</th> </tr>"; while($row = mysql_fetch_array($res)) { echo "<tr>"; echo "<td>" . $row['name'] . "</td>"; echo "<td>" . $row['auction_id'] . "</td>"; echo "<td>" . $row['bid_amount'] . "</td>"; echo "<td>" . $row['payment_status'] . "</td>"; echo "<td><form1=\"update_row\" method=\"post\" action=\"test.php?id=".$row['winnerid']."\"> <input type=\"hidden\" name=\"updating_row\" value=\"".$row['winnerid']."\" /> <input type=\"submit\" name=\"submit_update\" value=\"Remove\" /> </form>"; echo "</tr>"; if(isset($_POST['updating_row'])){ $res = mysql_query("UPDATE probid_winners SET printed='printed' WHERE probid_winners.winner_id = " . mysql_real_escape_string($_POST['updating_row'])); } } echo "</table>"; mysql_close($con); ?> I was hoping that someone might be able to help me out with this issue. Quote Link to comment Share on other sites More sharing options...
mrMarcus Posted December 1, 2009 Share Posted December 1, 2009 add: or trigger_error (mysql_error()); to the end of your queries. Quote Link to comment Share on other sites More sharing options...
chrishawkins Posted December 1, 2009 Author Share Posted December 1, 2009 Updated: Is this what you mean; <form name="form1" method="post" action=""> <label> <input type="submit" name="update" id="update" value="Update Records"> </label> </form> <form name="form2" method="get" action="test.php?show_results=1"> <label> <input type="submit" name="show_results" id="show_results" value="Show Results"> </label> </form> <?php $con = mysql_connect("localhost","username","password"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("my_db", $con); if(isset($_GET['show_results']) && $_GET['show_results'] == 1){ $res = mysql_query("SELECT *, probid_users.name AS name, probid_winners.winner_id AS winnerid FROM probid_winners left join probid_users ON probid_users.user_id = probid_winners.buyer_id WHERE probid_winners.payment_status='confirmed' AND printed='' AND buyer_id BETWEEN 2 AND 1000000000 ORDER BY bid_amount DESC = " . mysql_real_escape_string($_GET['show_results'] or trigger_error (mysql_error()); } echo "<table border='5'> <tr> <th>Name</th> <th>Auction ID</th> <th>Bid Amount</th> <th>Deposit Status</th> </tr>"; while($row = mysql_fetch_array($res)) { echo "<tr>"; echo "<td>" . $row['name'] . "</td>"; echo "<td>" . $row['auction_id'] . "</td>"; echo "<td>" . $row['bid_amount'] . "</td>"; echo "<td>" . $row['payment_status'] . "</td>"; echo "<td><form1=\"update_row\" method=\"post\" action=\"test.php?id=".$row['winnerid']."\"> <input type=\"hidden\" name=\"updating_row\" value=\"".$row['winnerid']."\" /> <input type=\"submit\" name=\"submit_update\" value=\"Remove\" /> </form>"; echo "</tr>"; if(isset($_POST['updating_row'])){ $res = mysql_query("UPDATE probid_winners SET printed='printed' WHERE probid_winners.winner_id = " . mysql_real_escape_string($_POST['updating_row'] or trigger_error (mysql_error()); } } echo "</table>"; mysql_close($con); ?> Quote Link to comment Share on other sites More sharing options...
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