Question

此代码不会打印上传图像的临时名称，也不会上传Photo文件夹中的任何图片。但打印图像名称。它正确地工作其他计算机。问题出在哪儿？请帮我。我正在使用PHP版本5.6.20

<!DOCTYPE html>
<html>
<head>
  <title>Image Upload</title>
</head>

<body>
<?php 

if ($_SERVER["REQUEST_METHOD"] == "POST") 
{

  $destination = "Photo/".$_FILES['image']['name'];
   $filename    = $_FILES['image']['tmp_name'];   
  echo $destination; 
  echo "<br/>";
  echo $filename;   
  echo "<br/>";

  if (file_exists($destination))

 echo "Sorry, file already exists.";

 else
    move_uploaded_file($filename, $destination);
  $name=$_FILES['image']['name'];
  $tmp_name=$_FILES['image']['tmp_name'];

  echo $tmp_name;
}


?>

<form method="POST" enctype="multipart/form-data">

<table>

<tr>
   <td>Choose Image</td>
   <td><input type="file" name="image"/> </td>
</tr> 

<tr>
   <td>Description:</td>
   <td><textarea name="Description" rows="4" cols="40"/></textarea></td>
</tr>

<tr> 
   <td></td>

    <td> <input type="submit" name="submit_image" value="Submit"></td>
</tr> 
</table>
</form>

</body>
</html>

Answer 1

<!DOCTYPE html>
<html>
<head>
  <title>Image Upload</title>
</head>

<body>
<?php 

if(isset($_POST['submit_image'])) 
{

  $destination = "Photo/".$_FILES['image']['name'];
   $filename    = $_FILES['image']['tmp_name'];   
  echo $destination; 
  echo "<br/>";
  echo $filename;   
  echo "<br/>";

  if (file_exists($destination))

 echo "Sorry, file already exists.";

 else
    move_uploaded_file($filename, $destination);
  $name=$_FILES['image']['name'];
  $tmp_name=$_FILES['image']['tmp_name'];

  echo $tmp_name;
}


?>

<form method="POST" enctype="multipart/form-data">

<table>

<tr>
   <td>Choose Image</td>
   <td><input type="file" name="image"/> </td>
</tr> 

<tr>
   <td>Description:</td>
   <td><textarea name="Description" rows="4" cols="40"/></textarea></td>
</tr>

<tr> 
   <td></td>

    <td> <input type="submit" name="submit_image" value="Submit"></td>
</tr> 
</table>
</form>

</body>
</html>

我尝试了这段代码，它对我有用。请给你的照片文件夹的读写权限我认为这对你有用。

Answer 2

您的代码工作正常，请查看照片/＆＃39;文件夹权限和tmp文件夹路径也是如此。例如：D：\ xampp \ tmp \ phpDCD9.tmp。

Answer 3

您的代码完全适用于我的机器。我想您可能想要检查已存储上载文件的目录名Photo。检查您的目标文件夹名称，您的编码和项目所在的已定义文件夹名称是相同的。请发布您的error或warning消息，以便提供帮助。

Answer 4

我认为这部分写得恰到好处，它会解决

if (file_exists($destination)){
 echo "Sorry, file already exists.";
 }else{
    move_uploaded_file($filename, $destination);
  $name=$_FILES['image']['name'];
  $tmp_name=$_FILES['image']['tmp_name'];
  echo $tmp_name;
}

使用PHP Mysql进行图像上传

4 个答案: