Undefined Variable Issue

[Aido89]

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>";

 

 ?>

[Barand]

1. Use code tags (or <> button) when posting code.

 

2. post the actual error message

[White_Lily]

try writing a line between your query and the item output echos that says:

$row = mysql_fetch_assoc($res);

EDIT: Put all your echos on new lines too, this would make it slightly easier to read.

Edited July 15, 2013 by White_Lily

[web_craftsman]

Aido89,you have forgotten mysql_query

[Aido89]

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>";
 
 ?>

[Barand]

See web_craftsman's reply

$sql = "SELECT * FROM table WHERE id=$id";
$res = mysql_query($sql);

[Aido89]

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>";
 
 ?>

[Barand]

check for errors

$sql = "SELECT * FROM table WHERE id=$id";
$res = mysql_query($sql);
if (!$res) die(mysql_error());

[Aido89]

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

[Aido89]

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