Question

您好我有一个用于搜索的页面，但我收到以下错误

试图在..........

中获取非对象的属性

有同样的问题，但我没有找到我的问题的答案。

以下是代码：

<html>
<head>
</head>
<body>
<form action="" name="myform" method="post">
<table width="1330" border="0" align="center">
<tr><td colspan="45" align="left"><a href="../Admin Panel.php"><img src="../Images for admin panel/backbutton.jpg" /></a></td></tr>
    <th height="47" colspan="11"><font size="+3">Remote Employee Search Page</font></th></tr>
</tr>
</tr>
<th colspan="11"><input type="text" name="search" size="33">&nbsp;<input type="submit" id="Submitbutton" value="Search" style="  font: 15px verdana;
  text-decoration: none;
  background-color: #EEEEEE;
  color: #333333;
  padding: 2px 6px 2px 6px;
  border-top: 1px solid #CCCCCC;
  border-right: 1px solid #333333;
  border-bottom: 1px solid #333333;
  border-left: 1px solid #CCCCCC;
  cursor:pointer;"><br ></th></tr>


<?php 

require_once('../db.php');
@$search = $_POST['Search'];


$query = "SELECT 

RemoteEmployeeID,
 RemoteEmployeeFullName,
  RemoteEmployeeNIC,
   RemoteEmployeePassword,
    RemoteEmployeePhone,
     RemoteEmployeeEmail,
      PDName, DistrictName,
       ProvinceName,
        RemoteEmployeeRegion,
         RemoteEmployeeExactAddress,
          Bank, BeneficiaryName,
           AccountNumber,
            AccountType,
             ContractDate,
              ContractExpirationDate
                FROM remoteemployees, properties
                 WHERE RemoteEmployeeID = '".$search."' OR RemoteEmployeeEmail = '".$search."' OR RemoteEmployeeFullName = '".$search."'";

    $query_run = $connection->query($query);
      if($query_run->num_rows == 0)
  {
      exit("<center><font color='red'><strong>Sorry, No Records Found For Your Search</strong></font></center>");

  } ?>

  <tr><th>R.E.ID</th><th>R.E.FullName</th><th>R.E.Password</th>
  <th>R.E.Phone</th><th>R.E.Email</th><th>Status</th><th>More</th>
</tr>
<?php
    if( $connection->error ) exit( $connection->error );
    while($row=$query_run->fetch_assoc()) 
    {

?>

<tr><td><?php echo $row['RemoteEmployeeID'] ?></td><td><?php echo $row['RemoteEmployeeFullName'] ?></td><td><?php echo $row['RemoteEmployeePassword'] ?></td>
<td><?php echo $row['RemoteEmployeePhone'] ?></td><td><?php echo $row['RemoteEmployeeEmail'] ?></td>

<td><a href= "More About Remote Employees.php?RemoteEmployeeID=<?php echo $row['RemoteEmployeeID'] ?>" style="color:#FFF;"><img src="../Images for admin panel/zoom.png"></a></td>
</tr>

<?php  } 


?>
<tr><td colspan="7"><input type="button" value="Print Current Page" onClick="window.print()" style="  font: 15px verdana;
  text-decoration: none;
  background-color: #EEEEEE;
  color: #333333;
  padding: 2px 6px 2px 6px;
  border-top: 1px solid #CCCCCC;
  border-right: 1px solid #333333;
  border-bottom: 1px solid #333333;
  border-left: 1px solid #CCCCCC;
  cursor:pointer;"></td></tr>
</table>
</form>
</body>
</html>

Answer 1

执行查询后，应检查$query_run是否为false，然后检查连接错误消息是否为false。如果查询失败，则$query_run将不是对象。

您也应该使用query()，因为您使用的是用户输入（prepare()），所以您应该使用$_POST。准备好的语句将保护您的数据库不被SQL注入操纵或销毁。

获取非对象错误的属性

1 个答案: