Question

我无法理解为什么下面的代码会给我这个错误： 严格标准：只应在第8行的/web/htdocs/www.mywebsite.com/admin/post/upload.php中通过引用传递变量

无论如何，结果是成功的！（它给了我＆＃34;成功＆＃34;并且实际上有效）我怎么能解决它？

对不起我的英文，提前谢谢你！

第8行是： $ file_ext =用strtolower（端（爆炸（＆＃39;＆＃39;，$ _ FILES [＆＃39;图像＆＃39;] [＆＃39;名称＆＃39;]）））;

整个代码是：

<?php
   if(isset($_FILES['image'])){
      $errors= array();
      $file_name = $_FILES['image']['name'];
      $file_size = $_FILES['image']['size'];
      $file_tmp = $_FILES['image']['tmp_name'];
      $file_type = $_FILES['image']['type'];
      $file_ext=strtolower(end(explode('.',$_FILES['image']['name'])));

      $expensions= array("jpeg","jpg","png","pdf");

      if(in_array($file_ext,$expensions)=== false){
         $errors[]="Error. Supported formats: JPG, JPEG, PNG, PDF";
      }

      if($file_size > 999999999) {
         $errors[]='File too large';
      }

      if(empty($errors)==true) {
         move_uploaded_file($file_tmp,"../../../sources/".$file_name);
         echo "Success";
      }else{
         print_r($errors);
      }
   }
?>
<html>
   <body>

      <form action = "" method = "POST" enctype = "multipart/form-data">
         <input type = "file" name = "image" />
         <input type = "submit"/>

         <ul>
            <li>Filename: /press/sources/<?php echo $_FILES['image']['name'];  ?>
            <li>Size: <?php echo $_FILES['image']['size'];  ?>
            <li>Format: <?php echo $_FILES['image']['type'] ?>
         </ul>

      </form>

   </body>
</html>

只应通过引用传递变量。 PHP错误。我该怎么办呢？

0 个答案: