Missing one record from my display...

[RynMan]

Hey guys

 

I'm not amazing at PHP, but know just enough to be dangerous. 

 

Basically I'm trying to get these records from my database to display in rows of 4.  For some reason every time, I'm missing a record (the first record returned in my query).  It's probably something simple is my code.  Would be grateful if anyone could lend a suggestion. 

 

 

 <?php          $a=0;         $b=3;                    while ($ArrayData = mysql_fetch_array($QueryData)) {                    //if a=0, start new row            if ($a==0) {            echo '<tr>';            }                    //create cell          //if thumbnail exists, use it, otherwise throw in the unavailable image          if ($ArrayData["PhotoThumb"]!="") {          $ImageToUse = $ArrayData["PhotoThumb"];           } else {          $ImageToUse = "image_unavailable_thumb.jpg"; }          echo '             <td width="33%"  align="center" valign="top"><p><a href="item.php?ItemID='.$ArrayData["ItemID"].'"><img src="images/'.$ImageToUse.'" border="0" /></a></p>              <p><a href="item.php?ItemID='.$ArrayData["ItemID"].'" class="cart_body_text">'.$ArrayData["ItemName"].'</a><br />                 <span class="style1"> <strong>$'.sprintf("%01.2f", $ArrayData["UnitPrice"]).'</strong></span><br />              </p>              </td>';                 //increment by 1         $a++;                //if a = 3 then close the row using </tr>, ready to start a new one        if ($a==$b) {        echo '        </tr>';                //reset counter, ready to start new row        $a=0;        }        }                     if ($a > 0 ) {           echo '</tr>';        }              ?>

 

 

Thanks guys!

[trq]

You forgot to show us the piece of code that executes the query. Do you call mysql_fetch_array() anywhere else before the code you have posted?

[RynMan]

Ah yep, sorry!

 

 

$QueryData = mysql_query("SELECT * FROM `Items` WHERE ItemCategoryShort = '$ItemCategory' ORDER BY UnitPrice ASC", $connection);    if (!$QueryData) {     die("Database query failed" . mysql_error()) ;    }    $ArrayData = mysql_fetch_array($QueryData);

 

[trq]

Remove the.....

 

$ArrayData = mysql_fetch_array($QueryData);

 

line.