php - Echo employee name if file does not exist -


i using following code try display records of employees without profile picture. profile pics stored in pics/ folder labeled id_number.jpg (e.g. 4567.jpg). if no pic found, want echo employee's name.

$id = $row_recordset7['id']; $image = 'pics/' . $id . '.jpg';  if (!file_exists($image)) { echo $row_recordset7['employee_name']; }

currently, employee names being echoed. appreciated.

mysql_select_db($database_schedule, $schedule);  $query_recordset7 = "select distinct id, employee_name, department                      unit                      order department, employee_name";  $recordset7 = mysql_query($query_recordset7, $schedule) or die(mysql_error());  $row_recordset7 = mysql_fetch_assoc($recordset7);  $totalrows_recordset7 = mysql_num_rows($recordset7);

complete code:

mysql_select_db($database_schedule, $schedule); $query_recordset7 = "select distinct schedule.id, schedule.employee_name, schedule.department schedule order schedule.employee_name"; $recordset7 = mysql_query($query_recordset7, $schedule) or die(mysql_error()); $row_recordset7 = mysql_fetch_assoc($recordset7); $totalrows_recordset7 = mysql_num_rows($recordset7);  ?> <!doctype html public "-//w3c//dtd xhtml 1.0 transitional//en" "http://www.w3.org/tr/xhtml1/dtd/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head>     <meta http-equiv="content-type" content="text/html; charset=utf-8" />     <title>lookup: missing pics</title>     <style type="text/css">     <!--     body,td,th {         font-size: 30px;     }     -->     </style> </head>  <body> <table width="100%" border="0" align="center">     <tr>         <td colspan="3" align="center">             <div align="left">                 <a href="../.">home</a>                 &gt;missing pics             </div>         </td>     </tr>     <tr>         <td align="center">             <div align="left">                 <strong>name </strong>(<?php echo $totalrows_recordset7 ?>)             </div>         </td>         <td align="center">             <strong>id</strong>         </td>      </tr>     <?php { ?>         <tr bgcolor="<?php             if($ssadv_m1%$ssadv_change_every1==0 && $ssadv_m1>0){                 $ssadv_k1++;             }             print $ssadv_colors1[$ssadv_k1%count($ssadv_colors1)];             $ssadv_m1++;         ?>">             <td align="center">                 <div align="left">                     <a href="../report.php?recordid=<?php echo $row_recordset7['id']; ?>">                          <?php                         $id = $row_recordset7['id'];                         $image = '../pics/' . $id . '.jpg';                          if (!file_exists($image)) {                             echo $row_recordset7['employee_name'];                         }else{                         }                         ?>                     </a>                 </div>             </td>             <td align="center">&nbsp;</td>         </tr>     <?php } while ($row_recordset7 = mysql_fetch_assoc($recordset7)); ?> </table> <br /> <?php echo $totalrows_recordset7 ?>   total </body> </html>

since informations gave me , riggsfolly, you'll have loop through sql result data want apply condition : 

mysql_select_db($database_schedule, $schedule);  $query_recordset7 = "select distinct id, employee_name, department                      unit                      order department, employee_name";  $recordset7 = mysql_query($query_recordset7, $schedule) or die(mysql_error());      $totalrows_recordset7 = mysql_num_rows($recordset7);  while ($row_recordset7 = mysql_fetch_assoc($recordset7)) {     $id = $row_recordset7['id'];     $image = '../pics/' . $id . '.jpg';      if (!file_exists($image)) {         echo $row_recordset7['employee_name'];     } }

hope helps.

ps : remember mysql_ insecure , you'd better learn pdo or mysqli_ prepared statement.


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