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hsg135 2003-12-29
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这是我写的 asp + sql server 下正常运行(*用户名及密码）一下全给你做个参考吧：
读数据库中图片数据<img src=mem/showimg_1.asp?id=<%=rs("us_number")%> width="105" height="135">
showimg_1.asp写法：
<%
Set Conn = Server.CreateObject("ADODB.Connection")
Conn.Open "driver={SQL Server};server=＊＊＊;uid=sa;pwd=＊＊＊＊;database=tr_bank;"
SQL = "SELECT * FROM tr_grmessage where us_number='" & session("user_infoid") & "' "
Set rs = Server.CreateObject("ADODB.Recordset")
rs.Open SQL,Conn,1,3
response.contenttype="image/gif"
Response.BinaryWrite rs("us_photo")
%>
上传图片：
<form name="form1" enctype="multipart/form-data" action="upimage.asp" method="POST" target="_blank">
<p>请选择您要上传的照片：<br>
<input name="picture" type="FILE">
<br>
<input type="submit" name="Submit" value="提 交">
</p>
</form>
upimage.asp写法：
<%
dddd=2
FormSize = Request.TotalBytes
FormData = Request.BinaryRead( FormSize )
function ImageUp(formsize,formdata)
bncrlf=chrb(13) & chrb(10)
divider=leftb(formdata,instrb(formdata,bncrlf)-1)
datastart=instrb(formdata,bncrlf&bncrlf)+4
dataend=instrb(datastart+1,formdata,divider)-datastart
imageup=midb(formdata,datastart,dataend)
end function
Image=ImageUp (FormSize,Formdata)
Set strconn = Server.CreateObject("ADODB.Connection")
strconn.Open "Filedsn=SqlServer.dsn;userid=sa;password=****"
set rs=server.CreateObject("adodb.recordset")
sql="SELECT * FROM tr_grmessage where us_number='" & session("user_id") & "' "
rs.Open sql,strconn,1,3
rs("us_photon")=2
rs("us_photo").appendchunk Image
dddd=1
rs.Update
rs.Close
response.contenttype="image/gif"
response.binarywrite imageup(formsize,formdata)
%>
<html>
<head>
<title>Untitled Document</title>
<meta http-equiv="Content-Type" content="text/html; charset=gb2312">
</head>
<body bgcolor="#FFFFFF" text="#000000">
<% if dddd=1 then %>
您已经成功的上传了照片。
<% end if %>
</body>
</html>
这是我写的 asp + sql server 下正常运行(*用户名及密码）一下全给你做个参考吧：
读数据库中图片数据<img src=mem/showimg_1.asp?id=<%=rs("us_number")%> width="105" height="135">
showimg_1.asp写法：
<%
Set Conn = Server.CreateObject("ADODB.Connection")
Conn.Open "driver={SQL Server};server=＊＊＊;uid=sa;pwd=＊＊＊＊;database=tr_bank;"
SQL = "SELECT * FROM tr_grmessage where us_number='" & session("user_infoid") & "' "
Set rs = Server.CreateObject("ADODB.Recordset")
rs.Open SQL,Conn,1,3
response.contenttype="image/gif"
Response.BinaryWrite rs("us_photo")
%>
上传图片：
<form name="form1" enctype="multipart/form-data" action="upimage.asp" method="POST" target="_blank">
<p>请选择您要上传的照片：<br>
<input name="picture" type="FILE">
<br>
<input type="submit" name="Submit" value="提 交">
</p>
</form>
upimage.asp写法：
<%
dddd=2
FormSize = Request.TotalBytes
FormData = Request.BinaryRead( FormSize )
function ImageUp(formsize,formdata)
bncrlf=chrb(13) & chrb(10)
divider=leftb(formdata,instrb(formdata,bncrlf)-1)
datastart=instrb(formdata,bncrlf&bncrlf)+4
dataend=instrb(datastart+1,formdata,divider)-datastart
imageup=midb(formdata,datastart,dataend)
end function
Image=ImageUp (FormSize,Formdata)
Set strconn = Server.CreateObject("ADODB.Connection")
strconn.Open "Filedsn=SqlServer.dsn;userid=sa;password=****"
set rs=server.CreateObject("adodb.recordset")
sql="SELECT * FROM tr_grmessage where us_number='" & session("user_id") & "' "
rs.Open sql,strconn,1,3
rs("us_photon")=2
rs("us_photo").appendchunk Image
dddd=1
rs.Update
rs.Close
response.contenttype="image/gif"
response.binarywrite imageup(formsize,formdata)
%>
<html>
<head>
<title>Untitled Document</title>
<meta http-equiv="Content-Type" content="text/html; charset=gb2312">
</head>
<body bgcolor="#FFFFFF" text="#000000">
<% if dddd=1 then %>
您已经成功的上传了照片。
<% end if %>
</body>
</html>
wcqgm 2003-12-29
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如何把图片存储在数据库中的IMAGE字段中？
请教各位，急。。。。。。。。。。。。。。。。。。。。
请教各位，急。。。。。。。。。。。。。。。。。。。。
