将WAMPSERVER升级到2.2(Apache 2.4.2; PHP 5.4.3; MySql 5.5.24)后运行php程序,总是报错,不知如何解决,求诸位指点。多谢测试用源码如下:
<!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>Window I</title>
</head><body><form id="inquery" method ="post" action="window2.php" >
<h2>Window I</h2>
<table >
<tr>
     <td colspan="3"><b>General Information</b></td>
    </tr>
<tr>
     <td width="144">&nbsp;</td>
    </tr>
<tr>
     <td><p>Item Name:</p></td>
<td> 
<?php
include("include/connect.php");//query string to ask the database
$sqlString = "SELECT item_name FROM $table_name1"; 
// ask the query from the database
$result = @mysql_query($sqlString,$connection) or die("Error: 003. Couldn't connect to table - Please try later"); 
?><select name="item_name" value ="<?php echo $item_name; ?>">
<option>--- Please Select ---</option>
<?php while($row = @mysql_fetch_array($result)) 
{ ?>
  <option><?php echo $row['item_name'] ?></option>
 <?php } ?>
 <?php 
echo "\$itemname = $itemname<br><br>";?>
 </select>
      </td>
</tr>
<tr>
     <td width="144">&nbsp;</td>
    </tr>
</table>
<table>
   <tr>
   <td><input id="submit" class="submit" name="submit" type="submit" value="Inquery"/></td>
   </tr> 
</table>
</form>
</body>
</html>
错误截屏如下:
phpwampserver