PHP / Forms使用upload的两个表单.php它上传某个文件名.如何解析表单 - php/forms two forms using upload.php which uploads a certain file name. how to parse the forms?

php/forms two forms using upload.php which uploads a certain file name. how to parse the forms?

实际上我可以复制我的上传.php，更改文件名并调用它一天。但是你怎么能只用一个上传来做我想做的事情.php？

我的上传脚本将文件名更改为 msg1.txt....我有两个表单，我怎样才能调用同一个上传.php每个 for 并得到 mes1.txt，如果形成两个 mes2.txt？这一切都在本地域下，因此安全性不是问题...这是我的表格...

<form enctype="multipart/form-data" action="upload.php" method="POST">
<input type="hidden" name="MAX_FILE_SIZE" value="100000" />
Choose a file to upload: <input name="uploadedfile" type="file" /><br />
<input type="submit" value="Upload File" />
</form>

感谢这方面的任何帮助。谢谢。。

这是我上传的.php

<?php
$destination_path = $_REQUEST["destination"] . "upload/"; 
$target_path = "" . $destination_path;
$target_path = $target_path . 'msg1.txt'; 
$_POST['dest_name'];
echo "User=" .          $_ENV[USER] . "<br />";
echo "Source=" .        $_FILES['uploadedfile']['name'] . "<br />"; 
echo "Destination=" .   $destination_path . "<br />"; 
echo "Target path=" .   $target_path . "<br />"; 
echo "Size=" .          $_FILES['uploadedfile']['size'] . "<br />"; 
//echo "Tmp name=" .    $_FILES['uploadedfile']['tmp_name'] . "<br />"; 

if(move_uploaded_file($_FILES['uploadedfile']['tmp_name'], $target_path)) {
echo "The file ".  basename( $_FILES['uploadedfile']['name']). 
" has been uploaded";
} else{
echo "There was an error uploading the file, please try again!";
}
?>

我会在两种形式中添加一个隐藏的输入，例如调用msg_type但每种形式都有不同的值，

您的第一份表格

<form enctype="multipart/form-data" action="upload.php" method="POST">
    <input type="hidden" name="MAX_FILE_SIZE" value="100000" />
    <input type="hidden" name="msg_type" value="1" />
    Choose a file to upload: <input name="uploadedfile" type="file" /><br />
    <input type="submit" value="Upload File" />
</form>

您的第二种表格

<form enctype="multipart/form-data" action="upload.php" method="POST">
    <input type="hidden" name="MAX_FILE_SIZE" value="100000" />
    <input type="hidden" name="msg_type" value="2" />
    Choose a file to upload: <input name="uploadedfile" type="file" /><br />
    <input type="submit" value="Upload File" />
</form>

然后在您的脚本中

$message_type = $_POST['msg_type'];
if ($message_type == 1) {
    $target_path = $target_path . 'msg1.txt';
}
else if ($message_type == 2) {
    $target_path = $target_path . 'msg2.txt';
}
else {
    $target_path = $target_path . 'unknown_form.txt';
}

您可以从 $_POST 获得您在表单上设置的隐藏字段的值，然后根据需要继续