求ajaxfileupload.js上传文件的解决方案
$.ajaxFileUpload( {
url:'upld5.asp', //需要链接到服务器地址
secureuri:false,
fileElementId:'upfile',
dataType:'json',
success:function(data,status)
{alert(eval('('+data.conn+')'));
},
error:function(data,status,e)
{ alert(eval('(' + data.conn + ')'));
// var text=split(unescape(data),"/")(0);
$.each(data,function(i,n){
// $("#cont").append("<span>"+ text +"</span>");
// $("#win").show();
});
}
}
);
upld5.asp代码如下:
ExtName = "jpg,gif,png" '允许扩展名
SavePath ="./folder"'保存路径
If Right(SavePath,1)<>"/" Then SavePath=SavePath&"/" '在目录后加(/)
CheckAndCreateFolder(SavePath)
UpLoadAll_a = Request.TotalBytes '取得客户端全部内容
If(UpLoadAll_a>0) Then
Set UploadStream_c = Server.CreateObject("ADODB.Stream")
UploadStream_c.Type = 1
UploadStream_c.Open
UploadStream_c.Write Request.BinaryRead(UpLoadAll_a)
UploadStream_c.Position = 0
FormDataAll_d = UploadStream_c.Read
CrLf_e = chrB(13)&chrB(10)
FormStart_f = InStrB(FormDataAll_d,CrLf_e)
FormEnd_g = InStrB(FormStart_f+1,FormDataAll_d,CrLf_e)
Set FormStream_h = Server.Createobject("ADODB.Stream")
FormStream_h.Type = 1
FormStream_h.Open
UploadStream_c.Position = FormStart_f + 1
UploadStream_c.CopyTo FormStream_h,FormEnd_g-FormStart_f-3
FormStream_h.Position = 0
FormStream_h.Type = 2
FormStream_h.CharSet = "GB2312"
FormStreamText_i = FormStream_h.Readtext
FormStream_h.Close
FileName_j = Mid(FormStreamText_i,InstrRev(FormStreamText_i,"\")+1,FormEnd_g)
If(CheckFileExt(FileName_j,ExtName)) Then
SaveFile = Server.MapPath(SavePath & FileName_j)
If Err Then
Response.Write ([{"conn":"文件上传出错!"}])
Err.Clear
Else
SaveFile = CheckFileExists(SaveFile)
k=Instrb(FormDataAll_d,CrLf_e&CrLf_e)+4
l=Instrb(k+1,FormDataAll_d,leftB(FormDataAll_d,FormStart_f-1))-k-2
FormStream_h.Type=1
FormStream_h.Open
UploadStream_c.Position=k-1
UploadStream_c.CopyTo FormStream_h,l
FormStream_h.SaveToFile SaveFile,2
SaveFileName = Mid(SaveFile,InstrRev(SaveFile,"\")+1)
Response.write ([{"conn":"文件上传成功!"&"/"&SaveFileName}])
End If
Else
Response.write ([{"conn":"文件格式不正确!"}])
End If 怎么上传没反应,