Question

我有一个php页面，其中包含来自数据库的blob

在没有斑点的情况下，我需要显示404图像

https://i.imgur.com/gp0Q2kQ.jpg https://i.imgur.com/XgGDExK.jpg

<?php
$host = "localhost";
$username = "root";
$password = "12345678";
$db = "sis";
$PicNum = $_GET["PicNum"];

mysql_connect($host,$username,$password) or die("Impossível conectar ao banco."); 
@mysql_select_db($db) or die("Impossível conectar ao banco."); 
$result=mysql_query("SELECT * FROM usuarios WHERE id=$PicNum") or die("Impossível executar a query "); 
$row=mysql_fetch_object($result); 

Header("Content-type: image/gif"); 
echo $row->avatar;?>

Answer 1

检查查询中的blob列是否为empty or NULL。


$result = mysql_query("SELECT * FROM usuarios WHERE id=$PicNum AND WHERE length(avatar) != 0 AND avatar IS NOT NULL") or die("Impossível executar a query "); 
$row=mysql_fetch_object($result); 

if($row)
  echo $row->avatar;
else 
  {
    // return 404 page or print as 404 error
    "<h1>404 Not Found</h1>"; 
    echo "The page that you have requested could not be found."; 
  }

?>

blob不存在时的自定义图像（404）

1 个答案: