Upload e db

[icodata]

Ci sono mille messaggi sull'upload ma non ho trovato quello giusto,

Tutti fanno riferimento all'upload di baol

ma nessuno lo trova

io devo uplodare un immagine e scrivere la path nel database

mi potete aiutare?

grazie

[fraude]

Ciao.

In alternativa all'upload di baol, potresti utilizzare quello proposto da aruba o ancora meglio a questa pagina web trovi decine di esempi, sono sicuro che troverai quello che cerchi.

[Roby_72]

Oppure questo anche commentato.
http://freeasp.html.it/articoli/view...olo.asp?id=169

Roby

[icodata]

grazie per le risposte, ho integrato il tutto ma non va

vi posto il codice



<%

'variabili di configurazione
Dim folder
folder = "\public" 'directory sul server con accesso in scrittura
'fine variabili di configurazione

Response.Expires=0
Response.Buffer = TRUE
Response.Clear



Sub BuildUploadRequest(RequestBin)
PosBeg = 1
PosEnd = InstrB(PosBeg,RequestBin,getByteString(chr(13)))
boundary = MidB(RequestBin,PosBeg,PosEnd-PosBeg)
boundaryPos = InstrB(1,RequestBin,boundary)
Do until (boundaryPos=InstrB(RequestBin,boundary & getByteString("--")))
Dim UploadControl
Set UploadControl = CreateObject("Scripting.Dictionary")
'Get an object name
Pos = InstrB(BoundaryPos,RequestBin,getByteString("Conte nt-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("filen ame="))
PosBound = InstrB(PosEnd,RequestBin,boundary)
If PosFile<>0 AND (PosFile<PosBound) Then
PosBeg = PosFile + 10
PosEnd = InstrB(PosBeg,RequestBin,getByteString(chr(34)))
FileName = getString(MidB(RequestBin,PosBeg,PosEnd-PosBeg))
UploadControl.Add "FileName", FileName
Pos = InstrB(PosEnd,RequestBin,getByteString("Content-Type:"))
PosBeg = Pos+14
PosEnd = InstrB(PosBeg,RequestBin,getByteString(chr(13)))
ContentType = getString(MidB(RequestBin,PosBeg,PosEnd-PosBeg))
UploadControl.Add "ContentType",ContentType
PosBeg = PosEnd+4
PosEnd = InstrB(PosBeg,RequestBin,boundary)-2
Value = MidB(RequestBin,PosBeg,PosEnd-PosBeg)
Else
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
UploadControl.Add "Value" , Value
UploadRequest.Add name, UploadControl
BoundaryPos=InstrB(BoundaryPos+LenB(boundary),Requ estBin,boundary)
Loop
End Sub
Function getByteString(StringStr)
For i = 1 to Len(StringStr)
char = Mid(StringStr,i,1)
getByteString = getByteString & chrB(AscB(char))
Next
End Function
Function getString(StringBin)
getString =""
For intCount = 1 to LenB(StringBin)
getString = getString & chr(AscB(MidB(StringBin,intCount,1)))
Next
End Function



byteCount = Request.TotalBytes

RequestBin = Request.BinaryRead(byteCount)
Dim UploadRequest
Set UploadRequest = CreateObject("Scripting.Dictionary")

BuildUploadRequest RequestBin

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
Set MyFile = ScriptObject.CreateTextFile(Server.mappath(folder) &"\"&filename)

For i = 1 to LenB(value)
MyFile.Write chr(AscB(MidB(value,i,1)))
Next

MyFile.Close
%>
<%





Dim ObjRs

Set Objrs = Server.CreateObject("ADODB.Recordset")

Dim StrSql

StrSql = "SELECT * FROM news"

ObjRs.Open StrSql, ObjConn, adOpenStatic, adLockOptimistic

ObjRs.AddNew

ObjRs("Title") = Request.Form("title")
ObjRs("Img")= folder & "/" & filename
ObjRs("Message") = Replace(Server.HTMLEncode(Request.Form("message")) , chr(13), "
", 1)
ObjRs("Date")= date()

ObjRs.Update
ObjRs.Close
ObjConn.Close %>

Mi da questo errore
Microsoft VBScript runtime error '800a0005'

Invalid procedure call or argument: 'MidB'

/add2.asp, line 21

[Roby_72]

Con quello script non puoi usare il request.form
Prendi l'esempio che ti ho segnalato io ed usa quello.

Roby