javascript - 当我在操作属性中写入页面名称时，文件图像不会上传

Question

当我在操作属性中写入页面名称时，文件图像不会上传。如果我在操作页面中写 "#" 图像上传到目标文件夹。 请检查编码并指出错误。提前致谢。

<form method="post" enctype="multipart/form-data" action="#">
<input type="file" name="imageupload" align="center">
<input type="submit" name="frmsubmit" value="Submit">
</form>
<?php
include 'includes/dbconnection.php';
if(isset($_POST['frmsubmit'])){
$image_name = $_FILES['imageupload']['name'];
$image_type = $_FILES['imageupload']['type'];
$image_size = $_FILES['imageupload']['size'];
$image_tmp = $_FILES['imageupload']['tmp_name'];
$target_dir = "dp/";
$path = $target_dir . $image_name;
move_uploaded_file($image_tmp, $path);
echo "The file ". basename( $_FILES["imageupload"]["name"]). " has been uploaded."; }?>

Answer 1

在表单的操作中，您可以使用 $_SERVER['PHP_SELF']。然后它将文件发送到相同的脚本。如果您使用相同的 php 脚本来上传文件，那很好。我已经修改了你的代码，请看一下

<form method="post" enctype="multipart/form-data" action="<?php echo $_SERVER['PHP_SELF'];?>">
<input type="file" name="imageupload" align="center">
<input type="submit" name="frmsubmit" value="Submit">
</form>
<?php
include 'includes/dbconnection.php';
if(isset($_POST['frmsubmit'])){
$image_name = $_FILES['imageupload']['name'];
$image_type = $_FILES['imageupload']['type'];
$image_size = $_FILES['imageupload']['size'];
$image_tmp = $_FILES['imageupload']['tmp_name'];
$target_dir = "dp/";
$path = $target_dir . $image_name;
move_uploaded_file($image_tmp, $path);
echo "The file ". basename( $_FILES["imageupload"]["name"]). " has been uploaded."; }?>

文件上传成功后，您可以使用header("Location")将用户转移到任何页面。