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編輯:編程問題解答

如何把圖片上傳到數據庫中並顯示出來?

upload.htm

' 上傳頁面.

<html>

<body>

<p align="center">千花飛舞之圖片上傳</p>
<center>
   <form name="mainForm" enctype="multipart/form-data"

' 這個Form屬性是得到上傳的數據的關鍵.
action="process.asp" method=post>
    <input type=file name=mefile><br>
   <input type=submit name=ok value="上傳">
   </form>
</center>
</body></html>

process.asp

' 處理浏覽器中送來的數據.

<%
response.buffer=true
formsize=request.totalbytes
formdata=request.binaryread(formsize)
bncrlf=chrB(13) & chrB(10)
divider=leftB(formdata,clng(instrb(formdata,bncrlf))-1)
datastart=instrb(formdata,bncrlf & bncrlf)+4
dataend=instrb(datastart+1,formdata,divider)-datastart
mydata=midb(formdata,datastart,dataend)

set connGraph=server.CreateObject("ADODB.connection")
connGraph.ConnectionString="driver={Microsoft Access Driver (*.mdb)};

DBQ=" & server.MapPath("images.mdb") & ";uid=;PWD=;"
connGraph.Open

set rec=server.createobject("ADODB.recordset")
rec.Open "select * from [images] where id is null",connGraph,1,3
rec.addnew
rec("img").appendchunk mydata
rec.update
rec.close
set rec=nothing
set connGraph=nothing
%>
 

showimg.asp

' 顯示圖片.
<%
set connGraph=server.CreateObject("ADODB.connection")
connGraph.ConnectionString="driver={Microsoft Access Driver (*.mdb)};

DBQ=" & server.MapPath("images.mdb") & ";uid=;PWD=;"
connGraph.Open
set rec=server.createobject("ADODB.recordset")
strsql="select img from images where id=" & trim(request("id"))
rec.open strsql,connGraph,1,1
Response.ContentType = "image/*"

' 在輸出到浏覽器之前一定要指定Response.ContentType = "image/*",以便正常顯示圖片.
Response.BinaryWrite rec("img").getChunk(7500000)
rec.close
set rec=nothing
set connGraph=nothing
%>

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