megahyperion Posted April 25, 2005 Share Posted April 25, 2005 In the example $oHTTP = ObjCreate("winhttp.winhttprequest.5") $oHTTP.Open("GET","http://www.AutoItScript.com") $oHTTP.Send() $HTMLSource = $oHTTP.Responsetext What does the .5 mean after the winhttprequest ? Link to comment Share on other sites More sharing options...
MHz Posted April 25, 2005 Share Posted April 25, 2005 (edited) I believe it is the version of winhttp on your system, as Sven has posted similar code before? Though some may need to use version 5.1. Edited April 25, 2005 by MHz Link to comment Share on other sites More sharing options...
megahyperion Posted April 25, 2005 Author Share Posted April 25, 2005 OK one more on this subject Why would something like this $oHTTP = ObjCreate("winhttp.winhttprequest.5") $oHTTP.Open Do something differently then this $oHTTP = ObjCreate("winhttp.winhttprequest") $oHTTP.5.Open or this $oHTTP = ObjCreate("winhttp.winhttprequest.5.Open") To me it all looks like it should do the same thing, after a ton of reading the only thing I can think of for the . is that its like a / in a directory name. Maybe I have the whole concept wrong. Link to comment Share on other sites More sharing options...
w0uter Posted April 25, 2005 Share Posted April 25, 2005 becuze you need more then only the open() command $httpObj = ObjCreate("winhttp.winhttprequest.5.1") $httpObj.open("GET",$URL) $httpObj.send() $HTMLSource = $httpObj.Responsetext so this would be the smallest i think My UDF's:;mem stuff_Mem;ftp stuff_FTP ( OLD );inet stuff_INetGetSource ( OLD )_INetGetImage _INetBrowse ( Collection )_EncodeUrl_NetStat_Google;random stuff_iPixelSearch_DiceRoll Link to comment Share on other sites More sharing options...
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