VBA Winhttp URL

私は次のVBAコードでWinhttpを使用していますが、動作させるURLを取得できませんが、URLをブラウザに貼り付けると正常に動作します。任意のアイデアをいただければ幸いです。VBA Winhttp URL

Sub WinHttp() 

Dim arr() As String 
Dim pos As Integer 
Dim used As Range 

Dim url, resp As String 
Dim req As New WinHttpRequest 
Dim n As Integer 

r = ActiveSheet.UsedRange.Rows.Count 

For n = 4 To r 

Worksheets("Sheet0").Activate 
pn = Cells(n, 2).Value 
UnitPrice = (Cells(n, 4).Value) * 1 

url = "https://www.google.com/search?q=DNR-12-1G+$4,250" 



req.Open "GET", url, False 
req.send 

resp = req.ResponseText 

Debug.Print resp 


arr = split(resp) 

arr = Filter(arr, "$") 

Next n 

End Sub

出典

2016-12-25 Bob

可能な複製を行います。 ]（http://stackoverflow.com/q/1264303/11683） – GSerg

私はそれをWebデバッガゴーグルに入れたときにエラー302と素敵なWebページを返しましたが、XMLHTTPにはエラーが表示されません。私はあなたのリンクを 'q ='の後に 'DNR-12-1G +％244％2C250'というVBScriptエスケープ機能でエスケープしました –

302移動

here文書が移動しました。 –

はちょうどそれが正常に動作エクセル2013で次のコードをテストした：

Sub Test() 
    Dim req As New WinHttpRequest 
    Dim url, resp As String 

    url = "https://www.google.com/search?q=DNR-12-1G+$4,250" 

    req.Open "GET", url, False 
    req.send 

    resp = req.ResponseText 
    Debug.Print resp 
End Sub

サーバーからの応答が

<!DOCTYPE html><html lang=en><meta charset=utf-8><meta name=viewport content="initial-scale=1, minimum-scale=1, width=device-width"><title>Error 403 (Forbidden)!!1</title><style>*{margin:0;padding:0}html,code{font:15px/22px arial,sans-serif}html{background:#fff;color:#222;padding:15px}body{margin:7% auto 0;max-width:390px;min-height:180px;padding:30px 0 15px}* > body{background:url(//www.google.com/images/errors/robot.png) 100% 5px no-repeat;padding-right:205px}p{margin:11px 0 22px;overflow:hidden}ins{color:#777;text-decoration:none}a img{border:0}@media screen and (max-width:772px){body{background:none;margin-top:0;max-width:none;padding-right:0}}#logo{background:url(//www.google.com/images/branding/googlelogo/1x/googlelogo_color_150x54dp.png) no-repeat;margin-left:-5px}@media only screen and (min-resolution:192dpi){#logo{background:url(//www.google.com/images/branding/googlelogo/2x/googlelogo_color_150x54dp.png) no-repeat 0% 0%/100% 100%;-moz-border-image:url(//www.google.com/images/branding/googlelogo/2x/ 
googlelogo_color_150x54dp.png) 0}}@media only screen and (-webkit-min-device-pixel-ratio:2){#logo{background:url(//www.google.com/images/branding/googlelogo/2x/googlelogo_color_150x54dp.png) no-repeat;-webkit-background-size:100% 100%}}#logo{display:inline-block;height:54px;width:150px}</style><a href=//www.google.com/><span id=logo aria-label=Google></span></a><p><b>403.</b> <ins>That’s an error.</ins><p>Your client does not have permission to get URL <code>/search?q=DNR-12-1G+%244,250&amp;gws_rd=cr&amp;ei=IDVgWJ2dDomnsgGfy4u4Dw</code> from this server. (Client IP address: 194.28.248.113)<br><br> 
Please see Google's Terms of Service posted at http://www.google.com/terms_of_service.html 
<BR><BR><P>If you believe that you have received this response in error, please <A HREF="https://www.google.com/support/contact/user?hl=en">report</A> your problem. However, please make sure to take a look at our Terms of Service (http://www.google.com/terms_of_service.html). In your email, please send us the <b>entire</b> code displayed below. Please also send us any information you may know about how you are performing your Google searches-- for example, "I'm using the Opera browser on Linux to do searches from home. My Internet access is through a dial-up account I have with the FooCorp ISP." or "I'm using the Konqueror browser on Linux to search from my job at myFoo.com. My machine's IP address is 10.20.30.40, but all of myFoo's web traffic goes through some kind of proxy server whose IP address is 10.11.12.13." (If you don't know any information like this, that's OK. But this kind of information can help us track down problems, so please tell us what you can.)</P><P>We will use all this information 
to diagnose the problem, and we'll hopefully have you back up and searching with Google again quickly!</P> 
<P>Please note that although we read all the email we receive, we are not always able to send a personal response to each and every email. So don't despair if you don't hear back from us!</P> 
<P>Also note that if you do not send us the <b>entire</b> code below, <i>we will not be able to help you</i>.</P><P>Best wishes,<BR>The Google Team</BR></P><BLOCKQUOTE>/+/+/+/+/+/+/+/+/+/+/+/+/+/+/+/+/+/+/+/+/+/+/+/+/<BR> 
Kjm9A1b-agG1uA7g8B-cYcEbbyIbP2Zi4vVoYweB2AoPwPBt3<BR> 
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+/+/+/+/+/+/+/+/+/+/+/+/+/+/+/+/+/+/+/+/+/+/+/+/+<BR></BLOCKQUOTE> 


    <ins>That’s all we know.</ins>

出典

2016-12-25 21:09:10 fuglede

これはカスタマイズされた[HTTP 403]（https://en.wikipedia.org/wiki/HTTP_403）です。私はOPが「動作するURLを取得できません」と考えています。 – GSerg

できます。その場合、VBA/WinHTTPとはほとんど関係がありません。 – fuglede

でもかまいませんが（必須ではありません）（http://stackoverflow.com/q/1264303/11683）。 – GSerg

感謝です！それは " "「req.SetRequestHeader" のUser-Agentを表示されます（互換; MSIE 6.0; Windows NTの5.0）のMozilla/4.0"

は、Excel用VBAを使用して、[HTTPS POSTリクエストのトリック

出典

2016-12-25 23:49:06 Bob

こんにちは、これはコメントであって答えではないはずですが、あなたは大歓迎です。 – fuglede

302移動

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