Having a parsing issue i guess help...

[co.ador]

$query = 'SELECT * FROM table1 WHERE id = '.intval($id). '  LIMIT 1 ;'; 

// execute query 
$result = mysql_query($query) or die ("Error in query: $query. ".mysql_error()); 

// see if any rows were returned 
if (mysql_num_rows($result) > 0) { 
$row = mysql_fetch_row($result); {
echo '<table width="100%"  border="0" cellspacing="0" cellpadding="0" class="itemdetails">
<tr>
<td width="1100" height="350" bgcolor="#FFFFFF" class="tento">
<table class="cafe"><tr><td width="547">
<a href="#"><h3 align="justify" style="position:relative; height:5px; top: 10px;">',$row[2] ,'</h3></a>
</td>';

echo'<tr>';
echo'<td height="4">';  

$sql="SELECT rating, COUNT(rating) as total FROM rating WHERE (item_name='$shoename') GROUP BY rating";
$result=mysql_query($sql);
$number = array( "one","two","three","four","five");
$total = array_fill(1, 5, 0);
if (mysql_num_rows($result)  >=0) {
    while ($row= mysql_fetch_assoc($result)) {
   $total[$row['rating']]= $row['total'];
   }
   foreach($number as $K =>$num)
   {
      echo'<table style="font-size:10; position:relative; left:26px;">';
      echo '<td width="42"><h3>'.($K+1).' Star</h3></td>
      <td width="15"><ul class="rating '.$num.'star">
  <li class="one">1</li>
      <li class="two">2</li>
      <li class="three">3</li>
      <li class="four">4</li>
      <li class="five">5</li>
      <li class="total">['.$total[$K+1].']</li>
      </ul></td></table>';
   }
}

echo'</td>';
echo'</tr>';
echo'<td width="321" rowspan="10"></td>
  <tr>
    <td height="4" colspan="2"><img src="../images/line..gif" alt="df" width="330" height="7" /></td>
  </tr>
<tr>
  <td width="400" height="52" class="foro"><img src="../images/itemspecifications.gif" alt="tr" /></td>
</tr>
    <td width="400" height="4" style="font-size:11;"><ul>
      <li>'.$row[3] .'</li>
    </ul></td>
<tr>
  <td width="400" height="4" style="font-size:11;"><ul>
    <li>',$row[4] ,'</li>
  </ul></td>
</table>';

 

The script above uses two sql injection. The first injection uses Index [2], [3], [4] and it work good it load the information but it only display the information found in the index [2] but not [3] and [4] to the browser, Check that indexes [3] and [4] are coded after the second Slq injection

$sql="SELECT rating, COUNT(rating) as total FROM rating WHERE (item_name='$shoename') GROUP BY rating";

.  I want to display index [3] and [4] field found in the first sql injection of table1 at the top. But indexes [3] and [4] doesn't display in the browser as I said before, I guess because php is getting confused and doesn't know if I am trying to use the indexes in the first SQL injection or second SQL injection. Can anybody help me to specify to php that the indexes I want to display information from are indexes found in the first SQL injection. not of the second injection.

[co.ador]

Gizmola is there something you don't understand? let me know

[gizmola]

A sql injection is an exploit, where someone who is using a web application is able to inject some SQL code into a script, that was never intended by the developer.  What you're doing has nothing to do with sql injection.

 

So I looked at your code, and I have a couple of comments that might help:

 

-First, when you fetch a row using mysql_fetch_row() you get an array that has the columns in the result set both in numeric order, and in associated keys.  It's usually best to specify one or the other for efficiencies sake, and I strongly recommend using the associative version.  There's a wrapper around this function that makes it even easier:  mysql_fetch_assoc().

 

Then you write your code like this:

 

echo $row['name'];

 

rather than

 

echo $row[2];

 

It's a lot better code documentation and also won't be broken if you alter the structure of the table which could effect the order of the columns in the table.

 

The other comment, is that you are suppossing we know what your database looks like, but we don't.  When you have a query like 'SELECT * from table1' we can't read your mind and debug your issues when we don't have any idea what the database looks like.

[co.ador]

Mr. Gizmola Thank you very much for you reply and your answer.

 

Your answer wasn't exactly what was causing the issue. I did changed it to mysql_fetch_assoc and $row['image']. The exact problem was that both of the query were using the same name for their result variable.

 

eg.

 

$row = mysql_fetch_assoc($result);

 

and

 

$row= mysql_fetch_assoc($result);

 

 

So php was getting confused when it was picking the field vaules, I changed the second to $row1

 

And problem solved now everything is displaying in the browser.

 

Your answer directed me in the right path...