How to upload images to a folder using Classic ASP?
Is there any script for Classic ASP so I can upload an image to a folder.
Ex: I click on a button, load the image, click to upload and go to a folder for example: C:\ASP\sistema\imagens.
4
2 answers
In the ASP classic, you don't have access to the method - Request.Form directly from you when you are getting in a POST-type multipart/form-data, and then how to interpret the data, it is necessary to parse the information is sent, it is a project that can help your ; Pure-ASP-Upload which will help you with this task, or you can install the components on the server, as the - ASP-Upload it will help you to add a class that allows you to easily play in this kind of form, and save your content in the server.
This code is written in VB5, so if your server is not up to date with VB, you will need to update it to work.
Pure ASP upload source code:
<%
'***************************************
' File: Upload.asp
' Author: Jacob "Beezle" Gilley
' Email: [email protected]
' Date: 12/07/2000
' Comments: The code for the Upload, CByteString,
' CWideString subroutines was originally
' written by Philippe Collignon...or so
' he claims. Also, I am not responsible
' for any ill effects this script may
' cause and provide this script "AS IS".
' Enjoy!
'****************************************
Class FileUploader
Public Files
Private mcolFormElem
Private Sub Class_Initialize()
Set Files = Server.CreateObject("Scripting.Dictionary")
Set mcolFormElem = Server.CreateObject("Scripting.Dictionary")
End Sub
Private Sub Class_Terminate()
If IsObject(Files) Then
Files.RemoveAll()
Set Files = Nothing
End If
If IsObject(mcolFormElem) Then
mcolFormElem.RemoveAll()
Set mcolFormElem = Nothing
End If
End Sub
Public Property Get Form(sIndex)
Form = ""
If mcolFormElem.Exists(LCase(sIndex)) Then Form = mcolFormElem.Item(LCase(sIndex))
End Property
Public Default Sub Upload()
Dim biData, sInputName
Dim nPosBegin, nPosEnd, nPos, vDataBounds, nDataBoundPos
Dim nPosFile, nPosBound
biData = Request.BinaryRead(Request.TotalBytes)
nPosBegin = 1
nPosEnd = InstrB(nPosBegin, biData, CByteString(Chr(13)))
If (nPosEnd-nPosBegin) <= 0 Then Exit Sub
vDataBounds = MidB(biData, nPosBegin, nPosEnd-nPosBegin)
nDataBoundPos = InstrB(1, biData, vDataBounds)
Do Until nDataBoundPos = InstrB(biData, vDataBounds & CByteString("--"))
nPos = InstrB(nDataBoundPos, biData, CByteString("Content-Disposition"))
nPos = InstrB(nPos, biData, CByteString("name="))
nPosBegin = nPos + 6
nPosEnd = InstrB(nPosBegin, biData, CByteString(Chr(34)))
sInputName = CWideString(MidB(biData, nPosBegin, nPosEnd-nPosBegin))
nPosFile = InstrB(nDataBoundPos, biData, CByteString("filename="))
nPosBound = InstrB(nPosEnd, biData, vDataBounds)
If nPosFile <> 0 And nPosFile < nPosBound Then
Dim oUploadFile, sFileName
Set oUploadFile = New UploadedFile
nPosBegin = nPosFile + 10
nPosEnd = InstrB(nPosBegin, biData, CByteString(Chr(34)))
sFileName = CWideString(MidB(biData, nPosBegin, nPosEnd-nPosBegin))
oUploadFile.FileName = Right(sFileName, Len(sFileName)-InStrRev(sFileName, "\"))
nPos = InstrB(nPosEnd, biData, CByteString("Content-Type:"))
nPosBegin = nPos + 14
nPosEnd = InstrB(nPosBegin, biData, CByteString(Chr(13)))
oUploadFile.ContentType = CWideString(MidB(biData, nPosBegin, nPosEnd-nPosBegin))
nPosBegin = nPosEnd+4
nPosEnd = InstrB(nPosBegin, biData, vDataBounds) - 2
oUploadFile.FileData = MidB(biData, nPosBegin, nPosEnd-nPosBegin)
If oUploadFile.FileSize > 0 Then Files.Add LCase(sInputName), oUploadFile
Else
nPos = InstrB(nPos, biData, CByteString(Chr(13)))
nPosBegin = nPos + 4
nPosEnd = InstrB(nPosBegin, biData, vDataBounds) - 2
If Not mcolFormElem.Exists(LCase(sInputName)) Then mcolFormElem.Add LCase(sInputName), CWideString(MidB(biData, nPosBegin, nPosEnd-nPosBegin))
End If
nDataBoundPos = InstrB(nDataBoundPos + LenB(vDataBounds), biData, vDataBounds)
Loop
End Sub
'String to byte string conversion
Private Function CByteString(sString)
Dim nIndex
For nIndex = 1 to Len(sString)
CByteString = CByteString & ChrB(AscB(Mid(sString,nIndex,1)))
Next
End Function
'Byte string to string conversion
Private Function CWideString(bsString)
Dim nIndex
CWideString =""
For nIndex = 1 to LenB(bsString)
CWideString = CWideString & Chr(AscB(MidB(bsString,nIndex,1)))
Next
End Function
End Class
Class UploadedFile
Public ContentType
Public FileName
Public FileData
Public Property Get FileSize()
FileSize = LenB(FileData)
End Property
Public Sub SaveToDisk(sPath)
Dim oFS, oFile
Dim nIndex
If sPath = "" Or FileName = "" Then Exit Sub
If Mid(sPath, Len(sPath)) <> "\" Then sPath = sPath & "\"
Set oFS = Server.CreateObject("Scripting.FileSystemObject")
If Not oFS.FolderExists(sPath) Then Exit Sub
Set oFile = oFS.CreateTextFile(sPath & FileName, True)
For nIndex = 1 to LenB(FileData)
oFile.Write Chr(AscB(MidB(FileData,nIndex,1)))
Next
oFile.Close
End Sub
Public Sub SaveToDatabase(ByRef oField)
If LenB(FileData) = 0 Then Exit Sub
If IsObject(oField) Then
oField.AppendChunk FileData
End If
End Sub
End Class
%>
Usage example:
<%@ Language=VBScript %>
<%Option Explicit%>
<!-- #include file="upload.asp" -->
<%
'NOTE - YOU MUST HAVE VBSCRIPT v5.0 INSTALLED ON YOUR WEB SERVER
' FOR THIS LIBRARY TO FUNCTION CORRECTLY. YOU CAN OBTAIN IT
' FREE FROM MICROSOFT WHEN YOU INSTALL INTERNET EXPLORER 5.0
' OR LATER.
' Create the FileUploader
Dim Uploader, File
Set Uploader = New FileUploader
' This starts the upload process
Uploader.Upload()
'******************************************
' Use [FileUploader object].Form to access
' additional form variables submitted with
' the file upload(s). (used below)
'******************************************
Response.Write "<b>Thank you for your upload " & Uploader.Form("fullname") & "</b><br>"
' Check if any files were uploaded
If Uploader.Files.Count = 0 Then
Response.Write "File(s) not uploaded."
Else
' Loop through the uploaded files
For Each File In Uploader.Files.Items
' Check where the user wants to save the file
If Uploader.Form("saveto") = "disk" Then
' Save the file
File.SaveToDisk "E:\UploadedFiles\"
ElseIf Uploader.Form("saveto") = "database" Then
' Open the table you are saving the file to
Set RS = Server.CreateObject("ADODB.Recordset")
RS.Open "MyUploadTable", "CONNECT STRING OR ADO.Connection", 2, 2
RS.AddNew ' create a new record
RS("filename") = File.FileName
RS("filesize") = File.FileSize
RS("contenttype") = File.ContentType
' Save the file to the database
File.SaveToDatabase RS("filedata")
' Commit the changes and close
RS.Update
RS.Close
End If
' Output the file details to the browser
Response.Write "File Uploaded: " & File.FileName & "<br>"
Response.Write "Size: " & File.FileSize & " bytes<br>"
Response.Write "Type: " & File.ContentType & "<br><br>"
Next
End If
%>
Example HTML:
<FORM METHOD="POST" ENCTYPE="multipart/form-data" ACTION="uploadexmple.asp">
<TABLE BORDER=0>
<tr><td><b>Enter your fullname:</b><br><INPUT TYPE=TEXT SIZE=40 NAME="FULLNAME"></td></tr>
<tr><td><b>Select a file to upload:</b><br><INPUT TYPE=FILE SIZE=50 NAME="FILE1"></td></tr>
<tr><td><b>Save To:</b>
Disk <INPUT TYPE=RADIO NAME="saveto" value="disk" checked>
Database <INPUT TYPE=RADIO NAME="saveto" value="database">
</td></tr>
<tr><td align="center"><INPUT TYPE=SUBMIT VALUE="Upload!"></td></tr>
</TABLE>
</FORM>
2
Author: Gabriel Gartz, 2014-02-10 15:09:41
You can use the Class clsUpload.
After you give an include in it, you can simply do like this:
ASP:
Dim objUpload
Dim strArquivo, strCaminho
Set objUpload = New clsUpload
strArquivo= objUpload.Fields("file").FileName
strCaminho= "C:/ASP/sistema/imagens/" & strFile
objUpload("file").SaveAs strCaminho
Set objUpload = Nothing
HTML:
<form enctype="multipart/form-data" method="post" action="clsUpload.asp">
<div>Enviar arquivo: </div>
<div>
<input type="file" NAME="file">
<input type="button" name="enviar" value="Enviar">
</div>
</form>
See here .
1
Author: Maicon Carraro, 2014-02-10 15:07:38