human
Solve the following python programming problem given the following code:
def asyncUpload(url, file, handler) : 
    xmlHttp = doCreateXmlHTTPRequest () 
    def onload(evnt = None) : 
        global xmlHttp, handler 
        localHandler = handler 
        status = xmlHttp.status 
        localHandler.onload (status) 
def onprogress(evnt = None) : 
        global xmlHttp, handler 
        localHandler = handler 
        localHandler.onprogress (evnt.loaded, evnt.total) 
def onerror(evnt = None) : 
        global xmlHttp, handler 
        localHandler = handler 
        status = xmlHttp.status 
        localHandler.onerror (status) 
xmlHttp.upload.load = onload 
    xmlHttp.upload.progress = xmlHttp 
    xmlHttp.upload.onload = onload 
    xmlHttp.upload.onprogress = onprogress 
    xmlHttp.upload.onerror = onerror 
    wubwubwub = 1 
    try : 
        xmlHttp.open ('POST', url, True) 
        JS ('var reader = new FileReader();

            reader.onload = function(evt) {
                var fd = new FormData();
                fd.append("file", @{{file}});
                @{{xmlHttp}}.send(fd);};
            reader.readAsBinaryString(@{{file}});
            ') 
except : 
        JS ('delete @{{xmlHttp}}.onreadystatechange;') 
        xmlHttp = None 
        handler = None 
        import sys 
        Window.alert (str (sys.exc_info () [1])) 
        return False 
return True 

    
    

    Fix the buggy line: xmlHttp.upload.progress = xmlHttp
assistant
#Here is Python program to fix the buggy line: xmlHttp.

xmlHttp.upload.progress = onprogress