Jump to content


Photo

PHP Error Message


  • Please log in to reply
1 reply to this topic

#1 bazjones

bazjones
  • Members
  • Pip
  • Newbie
  • 1 posts

Posted 02 March 2006 - 12:14 PM

Got this error when trying to run one of my scripts anyone any ideas?

Notice: Undefined index: Update in c:\website\quality_management\backup_log\backup_edit.php on line 104

Thanks

Bazjones

<form action="<? echo($_SERVER['PHP_SELF']); ?>" method=POST>
  <table width="129%" border="0">
    <tr> 
      <td width="68"><strong>LogID:</strong></td>
      <td colspan="9"><input type="TEXT" name="logid" size=150 value="<? echo($get_info["logid"]); ?>"> 
      </td>
    </tr>
    <tr> 
      <td> </td>
      <td colspan="9"> </td>
    </tr>
    <tr width="100%"> 
      <? 
$source = array("APP1", 
                "APP2",
                "APP3",
                   "APP4"); 
                        
# probably "nicer" to get above list of names from a database table?
echo <<<EOD
  <td><strong>Source:</strong></td> 
  <td><select name="source" id="source"> 
EOD;
# add each eng as an option: 
foreach($source as $value) 
{ 
  $selected = ""; 
  if($value == $get_info["source"]) 
  { 
    # mark this one as the selected entry: 
    $selected = " selected"; 
  } 
  echo "<option$selected>$value</option>\n";
} 
echo <<<EOD
</select></td> 
EOD;
?>
      <? 
$type = array("Event", 
            "Warning",                             
             "Error");   
                        
# probably "nicer" to get above list of names from a database table?
echo <<<EOD

  <td><strong>Type:</strong></td> 
  <td><select name="type" id="type"> 
EOD;
# add each eng as an option: 
foreach($type as $value) 
{ 
  $selected = ""; 
  if($value == $get_info["type"]) 
  { 
    # mark this one as the selected entry: 
    $selected = " selected"; 
  } 
  echo "<option$selected>$value</option>\n";
} 
echo <<<EOD
</select></td> 
EOD;
?>

      <td><strong>Time:</strong></td>
      <td width="130">    
    <input name="logtime" type="TEXT" id="cal-field-1" value="<? echo($get_info["logtime"]); ?>" size="10" /> 
          <button type="submit" id="cal-button-1">...</button>
          <script type="text/javascript">
            Calendar.setup({
              inputField    : "cal-field-1",
              button        : "cal-button-1",
              align         : "Tr"
            });
          </script>
    </td>
    </tr>
    <tr> 
      <td> </td>
      <td colspan="9"> </td>
    </tr>
    <tr> 
      <td><strong>Description:</strong></td>
      <td colspan="9"><textarea class="textarea" name="logtext" rows=8 cols=150><? echo($get_info["logtext"]); ?></textarea></td>
    </tr>
    <tr> 
      <td> </td>
      <td colspan="9"> </td>
    </tr>
    <tr> 
      <td> </td>
      <td colspan="6"><input name="Update" type=SUBMIT class="button" id="Update2" value="Update">
        <input name="Cancel" type="button" id="Cancel" onClick="MM_goToURL('parent','/quality_management/incident_log/incident_report.php');return document.MM_returnValue" value="Cancel"></td>
      <td width="315" colspan="2"><div align="right"> </div></td>
    </tr>
  </table>
  <br>
</form>


<?
if($_POST['Update'])
{
$query = "UPDATE `incidentlog` SET `source` = '". $_POST["source"] ."', `type` = '". $_POST["type"] ."', `logtime` = '".$_POST["logtime"]."', `logtext` = '".$_POST["logtext"]."' WHERE `logid` = '$logid'";
$result = mysql_query($query);
if(!@header("Location: incident_report.php")){ 
echo("<META HTTP-EQUIV=Refresh CONTENT=\"1; URL=incident_report.php\">"); 
echo("<script> 
window.top.location.href = 'incident_report.php'; 
</script>"); 
echo("You should now be redirected to the home page, click <a 
    href='incident_report.php'>here</a> if you are not redirected."); 
};

}
?>


#2 khburres

khburres
  • Members
  • PipPip
  • Member
  • 12 posts

Posted 02 March 2006 - 02:14 PM

check your incident_report.php file?
He who guards his mouth and his tongue keeps himself from calamity.




0 user(s) are reading this topic

0 members, 0 guests, 0 anonymous users