Php 上传文件并保存

再回来

所以我把代码移动了一点，现在只得到了未成功的消息。请告诉我哪里做错了，我做错了什么

editemployee.php页面上的我的表单

<h1>Warnings</h1>
<div class="article" style="width:535px">
 <form enctype="multipart/form-data" action="includes/add.php" method="POST"> 
 <table>
  <tr>
  <td>Warning letter:</td>
  <td> <input type="file" name="warning1"></td>
  </tr>
   <tr>
  <td>Warning letter:</td>
  <td><input type="file" name="warning2"></td>
  </tr>
   <tr>
  <td>Warning letter Final:</td>
  <td><input type="file" name="warning3"></td>
  </tr>
   <tr>
  <td></td>
  <td><input type="submit" value="Add"> </td>
  </tr>
  </table>

 </form>

 </div>

警告
警告信：
警告信：
最后警告信：

现在，我已经按照建议将处理代码移动到它自己的名为add.php的文件中

<?php 
  //This is the directory where images will be saved 
$target = "files/empdocs"; 
$target1 = $target . basename( $_FILES['warning1']['name']); 
$target2 = $target . basename( $_FILES['warning2']['name']);
$target3 = $target . basename( $_FILES['warning3']['name']);

 //This gets all the other information from the form 

 $warning1=($_FILES['warning1']['name']); 
 $warning2=($_FILES['warning2']['name']);
 $warning3=($_FILES['warning3']['name']); 

 // Connects to your Database 
 mysql_connect("localhost", "username", "password") or die(mysql_error()) ; 
 mysql_select_db("database") or die(mysql_error()) ; 

  //Writes the information to the database 
 mysql_query("INSERT INTO ref_employees VALUES ('$warning1', '$warning2', '$warning3')") ;
  //Writes the file to the server 
if (move_uploaded_file($_FILES['warning1']['tmp_name'], $target1)
&& move_uploaded_file($_FILES['warning2']['tmp_name'], $target2)
&& move_uploaded_file($_FILES['warning3']['tmp_name'], $target3)) {

 //Tells you if its all ok 
 echo "The file ". basename( $_FILES['uploadedfile']['name']). " has been uploaded, and your information has been added to the directory"; 
 } 
 else { 

 //Gives and error if its not 
 echo "Sorry, there was a problem uploading your file."; 
 } 
 ?>

您有：

mysql_query("INSERT INTO ref_employees VALUES ('$warning1', '$warning2', '$warningfinal') WHERE 'idnumber' = $idnumber") ; 

WHERE
不应出现在
INSERT
语句中
 没有这样的东西：

if(move_uploaded_file($_FILES['warning1']['tmp_name'], $target))
if(move_uploaded_file($_FILES['warning2']['tmp_name'], $target))
if(move_uploaded_file($_FILES['warning3']['tmp_name'], $target))  {

$target = "files/empdocs"; 
$target = $target . basename( $_FILES['warning1']['name']); 
$target = $target . basename( $_FILES['warning2']['name']);
$target = $target . basename( $_FILES['warning3']['name']);

你必须

if (move_uploaded_file($_FILES['warning1']['tmp_name'], $target1)
&& move_uploaded_file($_FILES['warning2']['tmp_name'], $target2)
&& move_uploaded_file($_FILES['warning3']['tmp_name'], $target3)) {

和目标变量，而不是这样：

if(move_uploaded_file($_FILES['warning1']['tmp_name'], $target))
if(move_uploaded_file($_FILES['warning2']['tmp_name'], $target))
if(move_uploaded_file($_FILES['warning3']['tmp_name'], $target))  {

$target = "files/empdocs"; 
$target = $target . basename( $_FILES['warning1']['name']); 
$target = $target . basename( $_FILES['warning2']['name']);
$target = $target . basename( $_FILES['warning3']['name']);

但是：

如果您想使用JQuery和PHP上传多个具有良好UI的文件，可以查看此链接并尝试使用JQuery文件上传库。这是非常好的libs

我想这会对你有帮助

链接是
执行此代码后，您的哪一条消息被打印出来？或者是否有错误消息？没有打印消息，只是像刚加载一样路由到同一页。。。。抱歉，应该提到的是，这只是一个警告表单，给出了问题…尝试不要混合前端代码与后端代码和最后一个：代码缩进FTW！。你不应该再使用mysql函数了，它们已经被弃用了->谢谢你会研究：）这是一个很好的观点，而且在列名
'idnumber'
谢谢你，有了你的解决方案，我至少现在得到了我的错误消息你需要什么帮助？我已经告诉你哪里有错误。你现在犯了什么错误？