Aido89

Thank you

Would you have any suggestions regarding coding examples I can use to rectify this?

Hi I am getting the error below; Warning: mysql_fetch_array() expects parameter 1 to be resource, object given in C:\xampp\htdocs\getuser.php on line 17 See below the code i am using, i also have an ajax file for this but that is working 100% the issue lies in the code below. Any ideas? <?php $q=$_GET["q"]; $con = mysqli_connect('localhost','root','','db'); if (!$con) { die('Could not connect: ' . mysqli_error($con)); } mysqli_select_db($con,"db"); $sql="SELECT * FROM db WHERE id = '".$q."'"; $result = mysqli_query($con,$sql); while($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td>" . $row['item'] . "</td>"; echo "<td>" . $row['description'] . "</td>"; echo "<td>" . $row['price'] . "</td>"; echo "<td>" . $row['category'] . "</td>"; echo "<td>" . $row['paypal'] . "</td>"; echo "</tr>"; } echo "</table>"; mysqli_close($con); ?>

Hi i have it sorted, error in the table name, im such an idiot

Yes it comes back with the below You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'table WHERE id=1' at line 1

Hi I have tried this but the same error keeps appearing I'm afraid Warning: mysql_fetch_assoc() expects parameter 1 to be resource, boolean given inC:\xampp\htdocs\readmore.php on line 68 <?php mysql_connect("localhost","root",""); mysql_select_db("db"); $id = intval($_GET['id']); $sql = "SELECT * FROM table WHERE id=$id"; $res = mysql_query($sql); /* $res = "SELECT * FROM table WHERE id=".intval($_REQUEST['id']);*/ /*$res=mysql_query ("select * from table");*/ echo "<table>"; $row = mysql_fetch_assoc($res); { echo "<tr>"; echo "<td>"; echo "<h4>".$row['item']."</h4>"; echo "</td>"; echo "</tr>"; echo "<td>";?> <img src ="<?php echo $row['image']; ?>" height ="100" width ="100"> <?php echo "</td>"; echo "</tr>"; echo "<td>"; echo $row['description']; echo "</td>"; echo "</tr>"; echo "<td>"; echo $row['price']; echo "</td>"; echo "</tr>"; } echo"</table>"; ?>

Hi I have added in $row = mysql_fetch_assoc($res); as seen below but am now getting following error, Warning: mysql_fetch_assoc() expects parameter 1 to be resource, string given in Thanks again for fast replies guys, much appreciated <?php mysql_connect("localhost","root",""); mysql_select_db("db"); $id = intval($_GET['id']); $res = "SELECT * FROM table WHERE id=$id"; /* $res = "SELECT * FROM shoes WHERE id=".intval($_REQUEST['id']);*/ /*$res=mysql_query ("select * from shoes");*/ echo "<table>"; $row = mysql_fetch_assoc($res); { echo "<tr>"; echo "<td>"; echo "<h4>".$row['item']."</h4>"; echo "</td>"; echo "</tr>"; echo "<td>";?> <img src ="<?php echo $row['image']; ?>" height ="100" width ="100"> <?php echo "</td>"; echo "</tr>"; echo "<td>"; echo $row['description']; echo "</td>"; echo "</tr>"; echo "<td>"; echo $row['price']; echo "</td>"; echo "</tr>"; } echo"</table>"; ?>

Hi Please see code below, when I query my db I am getting an unefined variable issue, any ideas? I am quiet new to PHP so please help. mysql_connect("localhost","root",""); mysql_select_db("db"); $id = intval($_GET['id']); $res = "SELECT * FROM table WHERE id=$id"; echo "<table>"; echo "<tr>"; echo "<td>"; echo "<h4>".$row['item']."</h4>"; echo "</td>"; echo "</tr>"; echo "<td>";?> <img src ="<?php echo $row['image']; ?>" height ="100" width ="100"> <?php echo "</td>"; echo "</tr>"; echo "<td>"; echo $row['description']; echo "</td>"; echo "</tr>"; echo "<td>"; echo $row['price']; echo "</td>"; echo "</tr>"; echo"</table>"; ?>

Hi I have a contact page uploaded on my website. This is called contact.html, when the details are entered it picks up sendmail.php which sends the email However this works on my localhost as I get the e-mail to my outlook & hotmail account but not since I have hosted it as a live website. I am wondering is there a php conflict with the versions, I am on version 5.3.5. Can you provide any assistance with this please? The code for the sendmail.php file is below; <? $email = $_POST['email'] ; $q = $_POST['q'] ; $message = $_POST['message'] ; mail( "test@myemail.com", $q, $message, "From: $email" ); print "Thank you for your e-mail, we will get in touch with you as soon as possible, have a nice day."; ?> Thanks, Aidan