Fredrik skrev i ett tidigare inlägg om att ladda upp filer utan komponent. Jag har nu börjat använda samma fast jag behöver möjligheten att kanske ladda upp upp till åtta filer samtidigt, kan både vara mer eller mindre. Så jag behöver hjälp med att anpassa koden för flera filer, typ den kollar flr hur många request.form som innehålelr värde och lopar de som innehåller nåt, om ni förstår vad jag menar :)
<FORM METHOD="Post" ENCTYPE="multipart/form-data" ACTION="outputFile.asp">
Fil : <INPUT TYPE="file" NAME="blob"><BR>
<INPUT TYPE="submit" NAME="ok">
</FORM>
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outputfile.asp:
Kod:
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<%
' Author Philippe Collignon
' Email PhCollignon@email.com
Response.Expires=0
Response.Buffer = TRUE
Response.Clear
'Response.BinaryWrite(Request.BinaryRead(Request.TotalBytes))
byteCount = Request.TotalBytes
'Response.BinaryWrite(Request.BinaryRead(varByteCount))
RequestBin = Request.BinaryRead(byteCount)
Dim UploadRequest
Set UploadRequest = CreateObject("Scripting.Dictionary")
BuildUploadRequest RequestBin
'email = UploadRequest.Item("email").Item("Value")
contentType = UploadRequest.Item("blob").Item("ContentType")
filepathname = UploadRequest.Item("blob").Item("FileName")
filename = Right(filepathname,Len(filepathname)-InstrRev(filepathname,"\"))
value = UploadRequest.Item("blob").Item("Value")
'Create FileSytemObject Component
Set ScriptObject = Server.CreateObject("Scripting.FileSystemObject")
'Create and Write to a File
pathEnd = Len(Server.mappath(Request.ServerVariables("PATH_INFO")))-14
' Set MyFile = ScriptObject.CreateTextFile(Left(Server.mappath(Request.ServerVariables("PATH_INFO")),pathEnd)&filename)
Set MyFile = ScriptObject.CreateTextFile(Server.mappath("test/"&filename))
response.write(Server.mappath(filename))
For i = 1 to LenB(value)
MyFile.Write chr(AscB(MidB(value,i,1)))
Next
MyFile.Close
%>
<b>Uploaded file : </b><%=filename%><BR>
<img src="test/<%=filename%>">
<!--#include file="upload.asp"-->
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upload.asp:
Kod:
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<%
' Author Philippe Collignon
' Email PhCollignon@email.com
Sub BuildUploadRequest(RequestBin)
'Get the boundary
PosBeg = 1
PosEnd = InstrB(PosBeg,RequestBin,getByteString(chr(13)))
boundary = MidB(RequestBin,PosBeg,PosEnd-PosBeg)
boundaryPos = InstrB(1,RequestBin,boundary)
'Get all data inside the boundaries
Do until (boundaryPos=InstrB(RequestBin,boundary & getByteString("--")))
'Members variable of objects are put in a dictionary object
Dim UploadControl
Set UploadControl = CreateObject("Scripting.Dictionary")
'Get an object name
Pos = InstrB(BoundaryPos,RequestBin,getByteString("Content-Disposition"))
Pos = InstrB(Pos,RequestBin,getByteString("name="))
PosBeg = Pos+6
PosEnd = InstrB(PosBeg,RequestBin,getByteString(chr(34)))
Name = getString(MidB(RequestBin,PosBeg,PosEnd-PosBeg))
PosFile = InstrB(BoundaryPos,RequestBin,getByteString("filename="))
PosBound = InstrB(PosEnd,RequestBin,boundary)
'Test if object is of file type
If PosFile<>0 AND (PosFile<PosBound) Then
'Get Filename, content-type and content of file
PosBeg = PosFile + 10
PosEnd = InstrB(PosBeg,RequestBin,getByteString(chr(34)))
FileName = getString(MidB(RequestBin,PosBeg,PosEnd-PosBeg))
'Add filename to dictionary object
UploadControl.Add "FileName", FileName
Pos = InstrB(PosEnd,RequestBin,getByteString("Content-Type:"))
PosBeg = Pos+14
PosEnd = InstrB(PosBeg,RequestBin,getByteString(chr(13)))
'Add content-type to dictionary object
ContentType = getString(MidB(RequestBin,PosBeg,PosEnd-PosBeg))
UploadControl.Add "ContentType",ContentType
'Get content of object
PosBeg = PosEnd+4
PosEnd = InstrB(PosBeg,RequestBin,boundary)-2
Value = MidB(RequestBin,PosBeg,PosEnd-PosBeg)
Else
'Get content of object
Pos = InstrB(Pos,RequestBin,getByteString(chr(13)))
PosBeg = Pos+4
PosEnd = InstrB(PosBeg,RequestBin,boundary)-2
Value = getString(MidB(RequestBin,PosBeg,PosEnd-PosBeg))
End If
'Add content to dictionary object
UploadControl.Add "Value" , Value
'Add dictionary object to main dictionary
UploadRequest.Add name, UploadControl
'Loop to next object
BoundaryPos=InstrB(BoundaryPos+LenB(boundary),RequestBin,boundary)
Loop
End Sub
'String to byte string conversion
Function getByteString(StringStr)
For i = 1 to Len(StringStr)
char = Mid(StringStr,i,1)
getByteString = getByteString & chrB(AscB(char))
Next
End Function
'Byte string to string conversion
Function getString(StringBin)
getString =""
For intCount = 1 to LenB(StringBin)
getString = getString & chr(AscB(MidB(StringBin,intCount,1)))
Next
End Function
%>
No problem. Uppladdningen skapar ett Dictionary-objekt med formulärfälten. Man kan traversera ett sådant objekt med en for each-loop, så att plocka ut resten av filerna är inget problem. Huruvida fältet är en fil eller inte identifieras genom att kolla om något filnamn är satt.
Här föjer en modifierad outputFile.asp:
<%
' Author Philippe Collignon
' Email PhCollignon@email.com
Response.Expires=0
Response.Buffer = TRUE
Response.Clear
'Response.BinaryWrite(Request.BinaryRead(Request.TotalBytes))
byteCount = Request.TotalBytes
'Response.BinaryWrite(Request.BinaryRead(varByteCount))
RequestBin = Request.BinaryRead(byteCount)
Dim UploadRequest
Set UploadRequest = CreateObject("Scripting.Dictionary")
BuildUploadRequest RequestBin
'email = UploadRequest.Item("email").Item("Value")
for each fobj in UploadRequest
if UploadRequest.Item(fobj).Item("FileName")<>"" then
contentType = UploadRequest.Item(fobj).Item("ContentType")
filepathname = UploadRequest.Item(fobj).Item("FileName")
filename = Right(filepathname,Len(filepathname)-InstrRev(filepathname,"\"))
value = UploadRequest.Item(fobj).Item("Value")
'Create FileSytemObject Component
Set ScriptObject = Server.CreateObject("Scripting.FileSystemObject")
'Create and Write to a File
pathEnd = Len(Server.mappath(Request.ServerVariables("PATH_INFO")))-14
' Set MyFile = ScriptObject.CreateTextFile(Left(Server.mappath(Request.ServerVariables("PATH_INFO")),pathEnd)&filename)
Set MyFile = ScriptObject.CreateTextFile(Server.mappath("test/"&filename))
'response.write(Server.mappath(filename))
For i = 1 to LenB(value)
MyFile.Write chr(AscB(MidB(value,i,1)))
Next
MyFile.Close
%>
<b>Uploaded file : </b><%=filename%><BR>
<img src="test/<%=filename%>"><%
end if
next%>
<!--#include file="upload.asp"-->
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------------------- What a waste it is to lose one's mind. Or not to have a mind is being very wasteful. How true that is.
Dan Quayle
ifall jag nu skulle vilja skapa en bildtext till varje uppladdad bild?
och hämta den från samma form som inputsen för bilderna... hur
går man tillväga då?
har hobbytestat litegrann, men inte kommit fram till någon vettig lösning...
tacksam för svar!
mvh john norrby
135 ms totalt · 3 externa anrop · v20260731065814-full.25f56b17