我正在尝试制作一个脚本来将任何文件上传到简单的 html/php 上传表单。
我找不到任何不使用 ASP 的工作脚本。
这是我最接近的代码：（VBS）
Dim strURL
Dim HTTP
Dim dataFile
Dim dataRequest
Dim objStream
strURL = "http://10.0.0.50/~/v_upload/up.php"
Set HTTP = CreateObject("Microsoft.XMLHTTP")
Set objStream = CreateObject("ADODB.Stream")
Set dataFile = objStream.Read
objStream.Type = 2
objStream.Open
objStream.LoadFromFile "http.txt"
Set dataRequest = "dataFile=" & dataFile
HTTP.open "POST", strURL, False
HTTP.setRequestHeader "Content-Type", "application/x-www-form-urlencoded"
HTTP.setRequestHeader "Content-Length", Len(dataRequest)
WScript.Echo "Now uploading file G:\Http\http.txt"
HTTP.send dataRequest
WScript.Echo HTTP.responseText
Set HTTP = Nothing
这给了我这个错误：
Line 9
字符 1
错误：对象时不允许操作
关闭了
代码 800A0E78
来源 ADODB.Stream
PHP 代码是：
<?php
if (!isset($_FILES['dataFile']['error']) || is_array($_FILES['dataFile']['error'])) {
switch ($_FILES['dataFile']['error']) {
case UPLOAD_ERR_OK:
break;
case UPLOAD_ERR_NO_FILE:
echo 'Unable to Upload. No file sent.';
case UPLOAD_ERR_INI_SIZE:
case UPLOAD_ERR_FORM_SIZE:
echo 'Unable to Upload. Exceeded file size limit.';
default:
echo 'Unable to Upload. Unknown errors.';
}
die();
}
$file_path = "http/";
$file_path = $file_path . basename( $_FILES['dataFile']['name']);
if(move_uploaded_file($_FILES['dataFile']['tmp_name'], $file_path)) {
echo "File {$_FILE['dataFile']['name']} uploaded success";
} else{
echo "Unable to upload. Unable to move uploaded file.";
}
?>
请帮忙！
基本上有4个错误需要修复：
- Remove
set from line 9
- Change
objStream.Read to objStream.ReadText
- Move
line 9到之后objStream.LoadFromFile
- Remove
set from line 14
完整代码：
Dim strURL
Dim HTTP
Dim dataFile
Dim dataRequest
Dim objStream
strURL = "http://10.0.0.50/~/v_upload/up.php"
Set HTTP = CreateObject("Microsoft.XMLHTTP")
Set objStream = CreateObject("ADODB.Stream")
objStream.Type = 2
objStream.Open
objStream.LoadFromFile "http.txt"
dataFile = objStream.ReadText
dataRequest = "dataFile=" & dataFile
HTTP.open "POST", strURL, False
HTTP.setRequestHeader "Content-Type", "application/x-www-form-urlencoded"
HTTP.setRequestHeader "Content-Length", Len(dataRequest)
WScript.Echo "Now uploading file G:\Http\http.txt"
HTTP.send dataRequest
WScript.Echo HTTP.responseText
Set HTTP = Nothing
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