php - 使用 Dreamweaver 上传图像文件时出错

Question

向大家问好!

拜托，我需要你的专业帮助。我的案例看起来与我见过的一些案例相似，但我的代码完全不同。

我在 Dreamweaver 中使用 PHP 编写了下面的代码，它用于将图像上传到 MySQl 数据库。现在，当我上传大约 6 张图片时，它会显示文件上传成功。但如果我尝试上传任何低于 6 张图片的内容，它将拒绝上传并回显上传失败。

<?php 
if(isset($_POST['submit']))
{
  $projid=$_POST['projid'];
  $projname=$_POST['projname'];

  $name=basename($_FILES['file_upload']['name']);
  $t_name=$_FILES['file_upload']['tmp_name'];
  $dir='upload';
  if(move_uploaded_file($t_name,$dir."/".$name))

  $nameone=basename($_FILES['file_uploadone']['name']);
  $t_name=$_FILES['file_uploadone']['tmp_name'];
  $dir='upload1';
  if(move_uploaded_file($t_name,$dir."/".$name))

  $nametwo=basename($_FILES['file_uploadtwo']['name']);
  $t_name=$_FILES['file_uploadtwo']['tmp_name'];
  $dir='upload2';

  if(move_uploaded_file($t_name,$dir."/".$name))
  $namethree=basename($_FILES['file_uploadthree']['name']);
  $t_name=$_FILES['file_uploadthree']['tmp_name'];
  $dir='upload3';

  if(move_uploaded_file($t_name,$dir."/".$name))
  $namefour=basename($_FILES['file_uploadfour']['name']);
  $t_name=$_FILES['file_uploadfour']['tmp_name'];
  $dir='upload4';
  if(move_uploaded_file($t_name,$dir."/".$name))

  $namefive=basename($_FILES['file_uploadfive']['name']);
  $t_name=$_FILES['file_uploadfive']['tmp_name'];
  $dir='upload5';
  if(move_uploaded_file($t_name,$dir."/".$name))
{
    mysql_select_db ($database_ProjMonEva,$ProjMonEva);
    $qur="insert into tbl_images (imageid, projid, projname, name, path, nameone, pathone, nametwo, pathtwo, namethree, paththree, namefour, pathfour, namefive, pathfive) values ('','$projid','$projname','$name','upload/$name','$nameone','upload/$nameone','$nametwo','upload/$nametwo','$namethree','upload/$namethree','$namefour','upload/$namefour','$namefive','upload/$namefive')";
    $res=mysql_query($qur,$ProjMonEva);
    echo 'File uploaded successful';    
}
else
{
    echo 'upload failed!';
}
}
?>
<?php

我发现问题来自 echo ，但我有库存，不知道如何纠正它。任何人都可以帮助我吗？

提前谢谢您 迈克

Answer 1

您正在设置一个条件 if(move_uploaded_file($t_name,$dir."/".$name)) 但您没有指示如果满足条件会发生什么，因此只有第一个行(就在 if 之后)被视为满足条件时要执行的操作。

事实上，您仅使用 {} 正确执行一次:

if(move_uploaded_file($t_name,$dir."/".$name))
{
    mysql_select_db ($database_ProjMonEva,$ProjMonEva);
    $qur="insert into tbl_images (imageid, projid, projname, name, path, nameone, pathone, nametwo, pathtwo, namethree, paththree, namefour, pathfour, namefive, pathfive) values ('','$projid','$projname','$name','upload/$name','$nameone','upload/$nameone','$nametwo','upload/$nametwo','$namethree','upload/$namethree','$namefour','upload/$namefour','$namefive','upload/$namefive')";
    $res=mysql_query($qur,$ProjMonEva);
    echo 'File uploaded successful';    
}

您能做的最好的事情就是创建一个函数，并为每个加载的文件调用该函数。