图像图表

时间:2021-05-25

如何把图片上传到数据库中并显示出来?

upload.htm

' 上传页面.

<html>

<body>

<p align="center">千花飞舞之图片上传</p>
<center>
   <form name="mainForm" enctype="multipart/form-data"

' 这个Form属性是得到上传的数据的关键.
action="process.asp" method=post>
    <input type=file name=mefile><br>
   <input type=submit name=ok value="上传">
   </form>
</center>
</body></html>

process.asp

' 处理浏览器中送来的数据.

<%
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

' 显示图片.
<%
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/*"

' 在输出到浏览器之前一定要指定Response.ContentType = "image/*",以便正常显示图片.
Response.BinaryWrite rec("img").getChunk(7500000)
rec.close
set rec=nothing
set connGraph=nothing
%>

[1]

声明:本页内容来源网络,仅供用户参考;我单位不保证亦不表示资料全面及准确无误,也不保证亦不表示这些资料为最新信息,如因任何原因,本网内容或者用户因倚赖本网内容造成任何损失或损害,我单位将不会负任何法律责任。如涉及版权问题,请提交至online#300.cn邮箱联系删除。

相关文章