while ($row = mysql_fetch_array($result)){
    $Part_Number= $row['Part_Number'];
  $Mask_Set= $row['Mask_Set'];
  $Released_Product = $row['Released_Product'];
  $display_block .= "

    <tr><td><a href='sDetails_byid.php?Part_Number=$Part_Number'> $Part_Number</a></td><td>$Released_Product</td><td>$Mask_Set</td>
    </tr>";

the error mistake is "Undefined variable: display_block in c:\easyphp1-8\easyphp1-8\www\staff_id.php on line 38"

i am a beginner hope to get solution here thanks

Hey adel_88,

Can you post all of your code ? (especially lines 37-39!)

Cheers

--
Don't Stand on your head - you'll get footprints in your hair
                                           http://charlieharvey.org.uk
                                              http://charlieharvey.com

$display_block is followed by .=
The period means that the value in the variable will be appended. In the while loop the parser generates an error since the variable has not yet been assigned. On my machine the code still worked as it should.
To get rid of the error message assign the variable a null text string value before the loop.

$display_block = ""
while ($row = mysql_fetch_array($result)) {

Namaste,

Kevin Tough