Search Results

Search found 15646 results on 626 pages for 'port 80'.

Page 248/626 | < Previous Page | 244 245 246 247 248 249 250 251 252 253 254 255  | Next Page >

  • IPtables and Remote Desktop with Proxy

    - by Sebastian
    So I setup a windows 2008 web server R2 on VirtualBox. Currently using Bridged Network. I can remote desktop to the machine hosting the VM (10.0.0.183) but cannot remote desktop to the VM itself (10.0.0.195). The remote port on the VM set to 5003. VM setup to accept remote connections (windows side). We also use a proxy for our internet, and I added these rules under NAT. (centOS 5) on our proxy box. -A INPUT -p tcp --dport 3389 -j ACCEPT -A REROUTING -i ppp0 -p tcp --dport 3389 -j REDIRECT --to-port 5003 -A FORWARD -d 10.0.0.195 --dport 5003 -m state --state NEW,ESTABLISHED,RELATED -j ACCEPT I've been trying for hours and hours and just cannot get it to work. I also used freedns so that we can use a domain name to connect too this VM over the internet. (the DNS points to our external IP address). If we don't get this right we will have to purchase a PPoE from an ISP to connect to this VM remotely, but I know that there is an alternative route if I can just get this port forwarding right!

    Read the article

  • How to decouple development server from Internet?

    - by intoxicated.roamer
    I am working in a small set-up where there are 4 developers (might grow to 6 or 8 in cuople of years). I want to set-up an environment in which developers get an internet access but can not share any data from the company on internet. I have thought of the following plan: Set-up a centralized git server (Debian). The server will have an internet access. A developer will only have git account on that server, and won't have any other account on it. Do not give internet access to developer's individual machine (Windows XP/Windows 7). Run a virtual machine (any multi-user OS) on the centralized server (the same one on which git is hosted). Developer will have an account on this virtual machine. He/she can access internet via this virtual machine. Any data-movement between this virtual machine and underlying server, as well as any of the developer's machine, is prohibited. All developers require USB port on their local machine, so that they can burn their code into a microcontroller. This port will be made available only to associated software that dumps the code in a microcontroller (MPLAB in current case). All other softwares will be prohibited from accessing the port. As more developers get added, providing internet support for them will become difficult with this plan as it will slow down the virtual machine running on the server. Can anyone suggest an alternative ? Are there any obvious flaws in the above plan ? Some key details of the server are as below: 1) OS:Debian 2) RAM: 8GB 3) CPU: Intel Xeon E3-1220v2 4C/4T

    Read the article

  • MAC-Address based routing

    - by d-fens
    Here is what i want to do: I have a bunch of systems, some might have the same Public-IP, i disable ARP. I have a Firewall (either IP Layer or bridge-FW) between these systems and the internet. Depending on the destination port of incoming IP-Packets to some of these Public-IPs i want to set the destinsation-Ethernet-Adress. So for instance System A has IP 8.8.8.8, mac de:ad:be:ef:de:ad, arp disabled System B has IP 8.8.8.8, mac 1f:1f:1f:1f:1f:1f, arp disabled Firewall has IP 8.8.8.1, arp disabled on that interface Incoming packet to IP 8.8.8.8 tcp dest port 100 Incoming packet to IP 8.8.8.8 tcp dest port 101 Firewall sets dest-mac for 1.) - de:ad:be:ef:de:ad Firewall sets dest-mac for 2.) - 1f:1f:1f:1f:1f:1f Second scenario: System A and System B establish outgoing TCP-Connections, and the firewall matches the dst-mac of the incoming IP-Packets (response packets) to the senders-mac address. is this possible in any way with linux and iptables? edit: i read ebtables might "work" in a hackish way for this purpose but i am not sure...

    Read the article

  • using virtual machine like mySql server

    - by ffmm
    i'm developing a java program and i need a database. Now i'm using MAMP and it's pretty easy but i would have a virtual machine (ubuntu server) and i need to connect my java program with this virtual machine using vitualBox. the situation: I installed VirtualBox on my mac and I installed an ubuntu-server machine set "bridge adapter" in the network settings of VB I installed mysql on ubuntu-server and i created a simple database (all work well by ubuntu) doing ifconfig by ubuntu I get the ip: 192.168.1.217 so in the java program i made this function: public static Connection connect(String host, int port, String dbName, String user, String passwd) { Connection dbConnection = null; try { String dbString = null; Class.forName("com.mysql.jdbc.Driver").newInstance(); dbString = "jdbc:mysql://" + host + ":" + port + "/" + dbName; dbConnection = DriverManager.getConnection(dbString, user, passwd); } catch (Exception e) { System.err.println("Failed to connect with the DB"); e.printStackTrace(); } return dbConnection; } and in the main() i use: Connection con = connect(1, "192.168.1.217", 3306, "Ciao", "root", "cocacola"); 3306 was a default value. I don't know if is correct, it works on mamp, but…. how I can find the correct port that I have to use with VB? when I ran the program I get the catch excepion… what's wrong? ps: i have to install apache o something else?

    Read the article

  • I'm having trouble getting my server to appear online.

    - by JMRboosties
    Total newb question I'm sure. First I had installed WAMP (http://www.wampserver.com), and I was able to access my pages from other computers in my router network, and the virtual device used to debug Android programs (the purpose of my having a server). This functionality failed, however, at some point over these past few days. While my own browser displays the pages just fine, other computers, my Android phone (on our room's wifi), and my virtual device are no longer able to connect to my pages. I had not made any changes in the settings. I uninstalled WAMP and installed EasyPHP. However, the problem was not resolved. I know this is rather vague, but does anyone here have an idea of what may have happened? I forwarded both port 80 (I know its the default HTTP port, I did it just to be safe), and now port 8888 which EasyPHP uses. I turned my firewall on my hosting computer off for good measure. I cannot access my pages from neither remote computers or computers using my router. Any ideas you may have on how to resolve this would be awesome, thanks a lot. And if you need anymore info please tell me.

    Read the article

  • How to decode sprop-parameter-sets in a H264 SDP?

    - by Cipi
    What is the meaning of Base64 decoded bytes in sprop-parameter-sets in SDP for a h264 stream? How can I know the video size from this example? SDP example: sprop-parameter-sets=Z0IAKeNQFAe2AtwEBAaQeJEV,aM48gA== First part decoded from Base64 to Base16: 67 42 00 29 E3 50 14 07 B6 02 DC 04 04 06 90 78 91 15 Second part (comma separated): 68 CE 3C 80

    Read the article

  • Sharepoint InputFormTextBox not working on updatepanel?

    - by James123
    I have two panels in update panel. In panel1, there is button. If I click, Panel1 will be visible =false and Panel2 will be visible=true. In Panel2, I placed SharePoint:InPutFormTextBox. It not rendering HTML toolbar and showing like below image. <SharePoint:InputFormTextBox runat="server" ID="txtSummary" ValidationGroup="CreateCase" Rows="8" Columns="80" RichText="true" RichTextMode="Compatible" AllowHyperlink="true" TextMode="MultiLine" /> http://i700.photobucket.com/albums/ww5/vsrikanth/careersummary-1.jpg

    Read the article

  • iTextSharp: How to position and wrap long text?

    - by aximili
    The PDF I can produce at the moment: I want the text to fill up the space in the lower left. How can I do that? Thanks! This is my code: private static void CreatePdf4(string pdfFilename, string heading, string text, string[] photos, string emoticon) { Document document = new Document(PageSize.A4.Rotate(), 26, 36, 0, 0); PdfWriter writer = PdfWriter.GetInstance(document, new FileStream(pdfFilename, FileMode.Create)); document.Open(); // Heading Paragraph pHeading = new Paragraph(new Chunk(heading, FontFactory.GetFont(FontFactory.HELVETICA, 54, Font.NORMAL))); document.Add(pHeading); // Photo 1 Image img1 = Image.GetInstance(HttpContext.Current.Server.MapPath("/uploads/photos/" + photos[0])); img1.ScaleAbsolute(350, 261); img1.SetAbsolutePosition(46, 220); img1.Alignment = Image.TEXTWRAP; document.Add(img1); // Photo 2 Image img2 = Image.GetInstance(HttpContext.Current.Server.MapPath("/uploads/photos/" + photos[1])); img2.ScaleAbsolute(350, 261); img2.SetAbsolutePosition(438, 220); img2.Alignment = Image.TEXTWRAP; document.Add(img2); // Text //Paragraph pText = new Paragraph(new Chunk(text, FontFactory.GetFont(FontFactory.HELVETICA, 18, Font.NORMAL))); //pText.SpacingBefore = 30; //pText.IndentationLeft = 20; //pText.IndentationRight = 366; //document.Add(pText); PdfContentByte cb = writer.DirectContent; cb.BeginText(); cb.SetFontAndSize(BaseFont.CreateFont(BaseFont.HELVETICA, BaseFont.CP1252, false), 18); cb.SetTextMatrix(46, 175); cb.ShowText(text); cb.EndText(); // Photo 3 Image img3 = Image.GetInstance(HttpContext.Current.Server.MapPath("/uploads/photos/" + photos[2])); img3.ScaleAbsolute(113, 153); img3.SetAbsolutePosition(556, 38); document.Add(img3); // Emoticon Image imgEmo = Image.GetInstance(HttpContext.Current.Server.MapPath("/Content/images/" + emoticon)); imgEmo.ScaleToFit(80, 80); imgEmo.SetAbsolutePosition(692, 70); document.Add(imgEmo); document.Close(); }

    Read the article

  • Problem with redirecting *.domain.com & domain.com to www.domain.com for HTTPS

    - by Mat E.
    We have a site I'll call acme.com. Most of the time you see http://www.acme.com and sometimes we redirect you to https://www.acme.com. We want to redirect anyone going to http://acme.com or http://*.acme.com to http://www.acme.com, and the same for https. (It's mainly to avoid the alert you get if you go to https://acme.com instead of https://www.acme.com) Our vhost file is at the end of the post. It works nicely except for one strange behavior: http://acme.com - successfully redirects to http://www.acme.com http://www.acme.com - successfully does not redirect http://foo.acme.com - successfully redirects to http://www.acme.com https://acme.com - successfully redirects to https://www.acme.com https://www.acme.com - successfully does not direct https://foo.acme.com - ERROR - redirects to http://www.acme.com It's this last result I can't fathom. I've tried a lot of trial and error solutions from Google & Stack Overflow but nothing seems to change it. Even if we swap the order of the configurations (so that 443 is before 80) it still redirects https://foo.acme.com to http://www.acme.com We are running Apache/2.2.12 on Ubuntu. Here's the configuration file: <VirtualHost *:80> ServerName www.acme.com ServerAlias acme.com *.acme.com ServerSignature On DocumentRoot /var/www/acme.com/public RailsEnv 'production' PassengerHighPerformance on <Directory /var/www/acme.com/public> AllowOverride all Options -MultiViews </Directory> SSLEngine Off CustomLog /var/log/apache2/acme.log combined ErrorLog /var/log/apache2/acme-error.log # Possible values include: debug, info, notice, warn, error, crit, alert, emerg. LogLevel warn RewriteEngine On RewriteCond %{HTTPS} off RewriteCond %{HTTP_HOST} ^[^\./]+\.[^\./]+$ RewriteRule ^/(.*)$ http://www.%{HTTP_HOST}/$1 [R=301,L] </VirtualHost> <VirtualHost *:443> ServerName www.acme.com ServerAlias acme.com *.acome.com DocumentRoot /var/www/acme.com/public RailsEnv 'production' PassengerHighPerformance on <Directory /var/www/acme.com/public> AllowOverride all Options -MultiViews </Directory> SSLCertificateFile /etc/ssl/certs/www.acme.com.crt SSLCertificateKeyFile /etc/ssl/private/acme.com.private.key SSLCACertificateFile /etc/ssl/certs/EV_intermediate.crt SSLEngine On CustomLog /var/log/apache2/ssl-acme.log "%t %h %{SSL_PROTOCOL}x %{SSL_CIPHER}x \"%r\" %b" ErrorLog /var/log/apache2/ssl-acme-error.log # Possible values include: debug, info, notice, warn, error, crit, alert, emerg. LogLevel warn RewriteEngine On RewriteCond %{HTTPS} on RewriteCond %{HTTP_HOST} ^[^\./]+\.[^\./]+$ RewriteRule ^/(.*)$ https://www.%{HTTP_HOST}/$1 [R=301,L] </VirtualHost>

    Read the article

  • Java Process "The pipe has been ended" problem

    - by Amit Kumar
    I am using Java Process API to write a class that receives binary input from the network (say via TCP port A), processes it and writes binary output to the network (say via TCP port B). I am using Windows XP. The code looks like this. There are two functions called run() and receive(): run is called once at the start, while receive is called whenever there is a new input received via the network. Run and receive are called from different threads. The run process starts an exe and receives the input and output stream of the exe. Run also starts a new thread to write output from the exe on to the port B. public void run() { try { Process prc = // some exe is `start`ed using ProcessBuilder OutputStream procStdIn = new BufferedOutputStream(prc.getOutputStream()); InputStream procStdOut = new BufferedInputStream(prc.getInputStream()); Thread t = new Thread(new ProcStdOutputToPort(procStdOut)); t.start(); prc.waitFor(); t.join(); procStdIn.close(); procStdOut.close(); } catch (Exception e) { e.printStackTrace(); printError("Error : " + e.getMessage()); } } The receive forwards the received input from the port A to the exe. public void receive(byte[] b) throws Exception { procStdIn.write(b); } class ProcStdOutputToPort implements Runnable { private BufferedInputStream bis; public ProcStdOutputToPort(BufferedInputStream bis) { this.bis = bis; } public void run() { try { int bytesRead; int bufLen = 1024; byte[] buffer = new byte[bufLen]; while ((bytesRead = bis.read(buffer)) != -1) { // write output to the network } } catch (IOException ex) { Logger.getLogger().log(Level.SEVERE, null, ex); } } } The problem is that I am getting the following stack inside receive() and the prc.waitfor() returns immediately afterwards. The line number shows that the stack is while writing to the exe. The pipe has been ended java.io.IOException: The pipe has been ended at java.io.FileOutputStream.writeBytes(Native Method) at java.io.FileOutputStream.write(FileOutputStream.java:260) at java.io.BufferedOutputStream.write(BufferedOutputStream.java:105) at java.io.BufferedOutputStream.flushBuffer(BufferedOutputStream.java:65) at java.io.BufferedOutputStream.write(BufferedOutputStream.java:109) at java.io.FilterOutputStream.write(FilterOutputStream.java:80) at xxx.receive(xxx.java:86) Any advice about this will be appreciated.

    Read the article

  • Interesting issue with WCF wsHttpBinding through a Firewall

    - by Marko
    I have a web application deployed in an internet hosting provider. This web application consumes a WCF Service deployed at an IIS server located at my company’s application server, in order to have data access to the company’s database, the network guys allowed me to expose this WCF service through a firewall for security reasons. A diagram would look like this. [Hosted page] --- (Internet) --- |Firewall <Public IP>:<Port-X >| --- [IIS with WCF Service <Comp. Network Ip>:<Port-Y>] link text I also wanted to use wsHttpBinding to take advantage of its security features, and encrypt sensible information. After trying it out I get the following error: Exception Details: System.ServiceModel.EndpointNotFoundException: The message with To 'http://<IP>:<Port>/service/WCFService.svc' cannot be processed at the receiver, due to an AddressFilter mismatch at the EndpointDispatcher. Check that the sender and receiver's EndpointAddresses agree. Doing some research I found out that wsHttpBinding uses WS-Addressing standards, and reading about this standard I learned that the SOAP header is enhanced to include tags like ‘MessageID’, ‘ReplyTo’, ‘Action’ and ‘To’. So I’m guessing that, because the client application endpoint specifies the Firewall IP address and Port, and the service replies with its internal network address which is different from the Firewall’s IP, then WS-Addressing fires the above message. Which I think it’s a very good security measure, but it’s not quite useful in my scenario. Quoting the WS-Addressing standard submission (http://www.w3.org/Submission/ws-addressing/) "Due to the range of network technologies currently in wide-spread use (e.g., NAT, DHCP, firewalls), many deployments cannot assign a meaningful global URI to a given endpoint. To allow these ‘anonymous’ endpoints to initiate message exchange patterns and receive replies, WS-Addressing defines the following well-known URI for use by endpoints that cannot have a stable, resolvable URI. http://schemas.xmlsoap.org/ws/2004/08/addressing/role/anonymous" HOW can I configure my wsHttpBinding Endpoint to address my Firewall’s IP and to ignore or bypass the address specified in the ‘To’ WS-Addressing tag in the SOAP message header? Or do I have to change something in my service endpoint configuration? Help and guidance will be much appreciated. Marko. P.S.: While I find any solution to this, I’m using basicHttpBinding with absolutely no problem of course.

    Read the article

  • Paragraph-based diff program?

    - by JS Bangs
    Does anyone know of a diff viewer or comparison program that can do paragraph-based differentials? My repos has a large number of LaTeX files that are formatted into 80-character wide paragraphs (for easy editing with vim). It's currently very difficult to interpret the diffs between various versions, because any edit that caused the position of the line breaks to change results in a lot of spurious differences that show up in the diff.

    Read the article

  • Forward event from custom UIControl subclass

    - by ggould75
    My custom subclass extend UIControl (MyCustomUIControl) Basically it contains 3 subviews: UIButton (UIButtonTypeCustom) UIImageView UILabel All the class works great but now I want to connect some events generated from this class to another. In particular when the button is pressed I want to forward the event to the owner viewcontroller so that I can handle the event. The problem is that I can't figure out how to implement this behaviour. Within EditableImageView I can catch the touch event using [button addTarget:self action:@selector(buttonPressed:) forControlEvents:UIControlEventTouchUpInside] but I don't know how to forward it inside of the buttonPressed selector. I also tried to implement touchesBegan but it seems never called... I'd like to capture the button press event from the viewcontroller in this way: - (void)viewDidLoad { [super viewDidLoad]; self.imageButton = [[EditableImageView alloc] initWithFrame:CGRectMake(50.0f, 50.0f, 80.0f, 80.0f)]; [imageButton addTarget:self action:@selector(buttonPressed:) forControlEvents:UIControlEventTouchUpInside]; [self.view addSubview:imageButton]; [imageButton setEditing:NO]; } This is my UIControl subclass initialization method: - (id)initWithFrame:(CGRect)frame { if (self = [super initWithFrame:frame]) { [self setBackgroundColor:[UIColor clearColor]]; button = [UIButton buttonWithType:UIButtonTypeCustom]; button.frame = CGRectMake(0.0f, 0.0f, frame.size.width, frame.size.height); [button setImage:[UIImage imageNamed:@"nene_70x70.png"] forState:UIControlStateNormal]; [button addTarget:self action:@selector(buttonPressed:) forControlEvents:UIControlEventTouchUpInside]; [self addSubview:button]; transparentLabelBackground = [[UIImageView alloc] initWithImage:[UIImage imageNamed:@"editLabelBackground.png"]]; transparentLabelBackground.hidden = YES; [self addSubview:transparentLabelBackground]; // create edit status label editLabel = [[UILabel alloc] initWithFrame:CGRectZero]; editLabel.hidden = YES; editLabel.userInteractionEnabled = NO; // without this assignment the button will not be clickable editLabel.textColor = [UIColor whiteColor]; editLabel.backgroundColor = [UIColor clearColor]; editLabel.textAlignment = UITextAlignmentLeft; UIFont *labelFont = [UIFont systemFontOfSize:16.0]; editLabel.font = labelFont; editLabel.text = @"edit"; labelSize = [@"edit" sizeWithFont:labelFont]; [self addSubview:editLabel]; } return self; } Thanks.

    Read the article

  • Reading a POP3 server with only TcpClient and StreamWriter/StreamReader[SOLVED]

    - by WebDevHobo
    I'm trying to read mails from my live.com account, via the POP3 protocol. I've found the the server is pop3.live.com and the port if 995. I'm not planning on using a pre-made library, I'm using NetworkStream and StreamReader/StreamWriter for the job. I need to figure this out. So, any of the answers given here: http://stackoverflow.com/questions/44383/reading-email-using-pop3-in-c are not usefull. It's part of a larger program, but I made a small test to see if it works. Eitherway, i'm not getting anything. Here's the code I'm using, which I think should be correct. EDIT: this code is old, please refer to the second block problem solved. public Program() { string temp = ""; using(TcpClient tc = new TcpClient(new IPEndPoint(IPAddress.Parse("127.0.0.1"),8000))) { tc.Connect("pop3.live.com",995); using(NetworkStream nws = tc.GetStream()) { using(StreamReader sr = new StreamReader(nws)) { using(StreamWriter sw = new StreamWriter(nws)) { sw.WriteLine("USER " + user); sw.Flush(); sw.WriteLine("PASS " + pass); sw.Flush(); sw.WriteLine("LIST"); sw.Flush(); while(temp != ".") { temp += sr.ReadLine(); } } } } } Console.WriteLine(temp); } Visual Studio debugger constantly falls over tc.Connect("pop3.live.com",995); Which throws an "A socket operation was attempted to an unreachable network 65.55.172.253:995" error. So, I'm sending from port 8000 on my machine to port 995, the hotmail pop3 port. And I'm getting nothing, and I'm out of ideas. Second block: Problem was apparently that I didn't write the quit command. The Code: public Program() { string str = string.Empty; string strTemp = string.Empty; using(TcpClient tc = new TcpClient()) { tc.Connect("pop3.live.com",995); using(SslStream sl = new SslStream(tc.GetStream())) { sl.AuthenticateAsClient("pop3.live.com"); using(StreamReader sr = new StreamReader(sl)) { using(StreamWriter sw = new StreamWriter(sl)) { sw.WriteLine("USER " + user); sw.Flush(); sw.WriteLine("PASS " + pass); sw.Flush(); sw.WriteLine("LIST"); sw.Flush(); sw.WriteLine("QUIT "); sw.Flush(); while((strTemp = sr.ReadLine()) != null) { if(strTemp == "." || strTemp.IndexOf("-ERR") != -1) { break; } str += strTemp; } } } } } Console.WriteLine(str); }

    Read the article

  • Cannot find the X.509 certificate after publishing

    - by Tr?n Qu?c Bình
    Hi everybody, I am building a WCF service as http://www.codeproject.com/KB/WCF/9StepsWCF.aspx#Beginner%20WCF%20FAQ%E2%80%99s and facing a trouble with X.509 certificate: when I debug, evething is OK. But when I pubish it to IIS (5.1, windowsXP SP3) I receive the error: **Cannot find the X.509 certificate using the following search criteria: StoreName 'My', StoreLocation 'CurrentUser', FindType 'FindBySubjectName', FindValue 'WCFServer'.** Thanks for any idea.

    Read the article

  • Ruby IRC bot don't works.

    - by JavaNoob
    require "socket" server = "irc.rizon.net" port = "6667" nick = "Ruby IRC Bot" channel = "#0x40" s = TCPSocket.open(server, port) s.print("USER Testing", 0) s.print("NICK #{nick}", 0) s.print("JOIN #{channel}", 0) This irc bot don't connect to the IRC server, What are i'm doing wrong?

    Read the article

  • WinForms window drag event

    - by Steve Syfuhs
    Is there an event in WinForms that get's fired when a window is dragged? Or is there a better way of doing what I want: to drop the window opacity to 80% when the window is being dragged around? Unfortunately this is stupidly tricky to search for because everyone is looking for drag and drop from the shell, or some other object.

    Read the article

  • Get uri LocalPath in jquery/JS?

    - by acidzombie24
    simple problem with a simple question. I have a string, in C# i can put it through a uri and access LocalPath How do i get LocalPath in jquery or javascript? ( LocalPath: "/asas") {http://z.com/asas?sadfdsgfg} AbsolutePath: "/asas" AbsoluteUri: "http://z.com/asas?sadfdsgfg" Authority: "z.com" DnsSafeHost: "z.com" Fragment: "" Host: "z.com" HostNameType: Dns IsAbsoluteUri: true IsDefaultPort: true IsFile: false IsLoopback: false IsUnc: false LocalPath: "/asas" OriginalString: "http://z.com/asas?sadfdsgfg" PathAndQuery: "/asas?sadfdsgfg" Port: 80 Query: "?sadfdsgfg" Scheme: "http" Segments: {string[2]} UserEscaped: false UserInfo: ""

    Read the article

  • jqGrid add item checkbox field defaulted to checked

    - by gurun8
    Here's a simple question. I have a jqGrid that's working great but I want set the default value for a checkbox to checked when user adds a new item. Here's a snippet of code: {name: "Active", index: "active", width: 80, align: "center", sortable: false, editable: true, edittype: "checkbox", editoptions: {value: "Yes:No"}} I don't see anything in the documentation: http://www.trirand.com/jqgridwiki/doku.php?id=wiki:common_rules

    Read the article

  • Team Foundation Server Setup/Access

    - by Angel Brighteyes
    What I need: A TFS 2010 Setup that allows 2 application developers to access the TFS from remote locations. How it is setup: Server 2008 Standard 2g Ram 300g HD space SharePoint Server 2007, using SQL Server 2005 SQL Server 2008 Standard Team Foundation Server 2010 IIS 7 Sharepoint Bindings: TFS.DynAccount.Me:80; TFS:80 TFS Bindings: TFS.DynAccount.Me:8080; TFS:8080 Using DynDNS service to account for the dynamic ip address being used, this is a requirement for the moment until I can get a better isp package. Access using Local Accounts Server is not setup on a domain, or as a domain. Consequently I did not setup AD services. Problem: When logged into TFS using my credentials TFS\AdminUser through the DynDNS account TFS.DynAccount.Me I recieve the 'Red X of Death' on the Documents and Reports folder. When logged into the TFS through the local peer to peer network using the same credentials TFS\AdminUser I do not receive the 'Red X of Death' problem. Further Troubleshooting: When users 'Right Click' the 'TeamProject1' Click 'Show Project Portal' it tries to take them to http://TFS:8080 instead of http://TFS.DynAccount.Me:8080, which doing further research I am assuming that it is because team foundation server was setup with a local name of TFS instead of 'TFS.DynAccount.Me' as specified here in Visual Studio Magazines: The Red X of Death. Users can Access the Team Portal for SharePoint via http://TFS.DynAccount.Me/TeamCollection/TeamProject so it is not like we are dead in the water or anything. However, as most employees/staff are prone to do, they have expressed a great distaste for having to do it this way and just be patient until the current project is finished since we are under a very strict deadline. Is there a way to set this up differently, or change some settings someplace, reinstall it, point a CName record for our domain website to the DynAccount (e.g. TFS.OurDomain.com points to TFS.DynAccount.Me, which consequently does allow access to the http site without issues), or something. I really don't feel like after all the time and effort I have spent into, first the cost, second the bloody install, third learning SharePoint well enough, fourth the hours into days spent on this, fifth more troubleshooting, sixth employee headaches to just let it lay where it is at. I figure in my spare/off time I would keep trying to get this to work. So I really appreciate any help any one can give me. I know this is probably something really stupid simple that I will 'Face Palm' over, but at the moment the stress and frustration just has me beat. Thank you again, this community has always been a great help.

    Read the article

  • Which platform can we expect one's complement being used there?

    - by Jian Lin
    For some questions such as checking whether a number is odd or even, I noted the comment, a & 1 won't work when it is a one's complement machine or when the code is ported to a platform that uses one's complement. Since 30 years ago on the Superboard, TRS-80, Apple II, I haven't seen a system with one's complement. Are there popular systems that use one's complement still, or do we have some cell phone or mobile device that uses one's complement?

    Read the article

  • Why does the interpreted order seem different from what I expect?

    - by inspectorG4dget
    I have a problem that I have not faced before: It seems that the order of interpretation in my program is somehow different from what I expect. I have written a small Twitter client. It takes a few seconds for my program to actually post a tweet after I click the "GO" button (which can also be activated by hitting ENTER on the keyboard). I don't want to click multiple times within this time period thinking that I hadn't clicked it the first time. Therefore, when the button is clicked, I would like the label text to display something that tells me that the button has been clicked. I have implemented this message by altering the label text before I send the tweet across. However, for some reason, the message does not display until the tweet has been attempted. But since I have a confirmation message after the tweet, I never get to see this message and my original problem goes unsolved. I would really appreciate any help. Here is the relevant code: class SimpleTextBoxForm(Form): def __init__(self): # set window properties self.Text = "Tweeter" self.Width = 235 self.Height = 250 #tweet away self.label = Label() self.label.Text = "Tweet Away..." self.label.Location = Point(10, 10) self.label.Height = 25 self.label.Width = 200 #get the tweet self.tweetBox = TextBox() self.tweetBox.Location = Point(10, 45) self.tweetBox.Width = 200 self.tweetBox.Height = 60 self.tweetBox.Multiline = True self.tweetBox.WordWrap = True self.tweetBox.MaxLength = 140; #ask for the login ID self.askLogin = Label() self.askLogin.Text = "Login:" self.askLogin.Location = Point(10, 120) self.askLogin.Height = 20 self.askLogin.Width = 60 self.login = TextBox() self.login.Text= "" self.login.Location = Point(80, 120) self.login.Height = 40 self.login.Width = 100 #ask for the password self.askPass = Label() self.askPass.Text = "Password:" self.askPass.Location = Point(10, 150) self.askPass.Height = 20 self.askPass.Width = 60 # display password box with character hiding self.password = TextBox() self.password.Location = Point(80, 150) self.password.PasswordChar = "x" self.password.Height = 40 self.password.Width = 100 #submit button self.button1 = Button() self.button1.Text = 'Tweet' self.button1.Location = Point(10, 180) self.button1.Click += self.update self.AcceptButton = self.button1 #pack all the elements of the form self.Controls.Add(self.label) self.Controls.Add(self.tweetBox) self.Controls.Add(self.askLogin) self.Controls.Add(self.login) self.Controls.Add(self.askPass) self.Controls.Add(self.password) self.Controls.Add(self.button1) def update(self, sender, event): if not self.password.Text: self.label.Text = "You forgot to enter your password..." else: self.tweet(self.tweetBox.Text, self.login.Text, self.password.Text) def tweet(self, msg, login, password): self.label.Text = "Attempting Tweet..." # this should be executed before sending the tweet is attempted. But this seems to be executed only after the try block try: success = 'Tweet successfully completed... yay!\n' + 'At: ' + time.asctime().split()[3] ServicePointManager.Expect100Continue = False Twitter().UpdateAsXML(login, password, msg) except: error = 'Unhandled Exception. Tweet unsuccessful' self.label.Text = error else: self.label.Text = success self.tweetBox.Text = ""

    Read the article

  • Why does the interpretted order seem different from what I expect?

    - by inspectorG4dget
    I have a problem that I have not faced before: It seems that the order of interpretation in my program is somehow different from what I expect. I have written a small Twitter client. It takes a few seconds for my program to actually post a tweet after I click the "GO" button (which can also be activated by hitting ENTER on the keyboard). I don't want to click multiple times within this time period thinking that I hadn't clicked it the first time. Therefore, when the button is clicked, I would like the label text to display something that tells me that the button has been clicked. I have implemented this message by altering the label text before I send the tweet across. However, for some reason, the message does not display until the tweet has been attempted. But since I have a confirmation message after the tweet, I never get to see this message and my original problem goes unsolved. I would really appreciate any help. Here is the relevant code: class SimpleTextBoxForm(Form): def init(self): # set window properties self.Text = "Tweeter" self.Width = 235 self.Height = 250 #tweet away self.label = Label() self.label.Text = "Tweet Away..." self.label.Location = Point(10, 10) self.label.Height = 25 self.label.Width = 200 #get the tweet self.tweetBox = TextBox() self.tweetBox.Location = Point(10, 45) self.tweetBox.Width = 200 self.tweetBox.Height = 60 self.tweetBox.Multiline = True self.tweetBox.WordWrap = True self.tweetBox.MaxLength = 140; #ask for the login ID self.askLogin = Label() self.askLogin.Text = "Login:" self.askLogin.Location = Point(10, 120) self.askLogin.Height = 20 self.askLogin.Width = 60 self.login = TextBox() self.login.Text= "" self.login.Location = Point(80, 120) self.login.Height = 40 self.login.Width = 100 #ask for the password self.askPass = Label() self.askPass.Text = "Password:" self.askPass.Location = Point(10, 150) self.askPass.Height = 20 self.askPass.Width = 60 # display password box with character hiding self.password = TextBox() self.password.Location = Point(80, 150) self.password.PasswordChar = "x" self.password.Height = 40 self.password.Width = 100 #submit button self.button1 = Button() self.button1.Text = 'Tweet' self.button1.Location = Point(10, 180) self.button1.Click += self.update self.AcceptButton = self.button1 #pack all the elements of the form self.Controls.Add(self.label) self.Controls.Add(self.tweetBox) self.Controls.Add(self.askLogin) self.Controls.Add(self.login) self.Controls.Add(self.askPass) self.Controls.Add(self.password) self.Controls.Add(self.button1) def update(self, sender, event): if not self.password.Text: self.label.Text = "You forgot to enter your password..." else: self.tweet(self.tweetBox.Text, self.login.Text, self.password.Text) def tweet(self, msg, login, password): self.label.Text = "Attempting Tweet..." # this should be executed before sending the tweet is attempted. But this seems to be executed only after the try block try: success = 'Tweet successfully completed... yay!\n' + 'At: ' + time.asctime().split()[3] ServicePointManager.Expect100Continue = False Twitter().UpdateAsXML(login, password, msg) except: error = 'Unhandled Exception. Tweet unsuccessful' self.label.Text = error else: self.label.Text = success self.tweetBox.Text = ""

    Read the article

  • rmagick error on heroku

    - by nvano
    i'm cropping images with paperclip. i have a custom module which works great on my local machine (copied from railscast 182). //file: lib/paperclip_processors/cropper.rb module Paperclip class Cropper < Thumbnail def transformation_command if crop_command crop_command + super.sub(/ -crop \S+/, '') else super end end def crop_command target = @attachment.instance if target.cropping? " -crop '#{target.crop_w.to_i}x#{target.crop_h.to_i}+#{target.crop_x.to_i}+# {target.crop_y.to_i}'" end end end end on heroku i get the following error: NoMethodError (private method `sub' called for ["-resize", "220x", "-crop", "220x220+0+18", "+repage"]:Array): lib/paperclip_processors/cropper.rb:12:in `transformation_command' paperclip (2.3.3) lib/paperclip/thumbnail.rb:55:in `make' paperclip (2.3.3) lib/paperclip/processor.rb:33:in `make' paperclip (2.3.3) lib/paperclip/attachment.rb:295:in `post_process_styles' paperclip (2.3.3) lib/paperclip/attachment.rb:294:in `each' paperclip (2.3.3) lib/paperclip/attachment.rb:294:in `inject' paperclip (2.3.3) lib/paperclip/attachment.rb:294:in `post_process_styles' paperclip (2.3.3) lib/paperclip/attachment.rb:291:in `each' paperclip (2.3.3) lib/paperclip/attachment.rb:291:in `post_process_styles' paperclip (2.3.3) lib/paperclip/attachment.rb:285:in `post_process' paperclip (2.3.3) lib/paperclip/callback_compatability.rb:23:in `call' paperclip (2.3.3) lib/paperclip/callback_compatability.rb:23:in `run_paperclip_callbacks' paperclip (2.3.3) lib/paperclip/attachment.rb:284:in `post_process' paperclip (2.3.3) lib/paperclip/callback_compatability.rb:23:in `call' paperclip (2.3.3) lib/paperclip/callback_compatability.rb:23:in `run_paperclip_callbacks' paperclip (2.3.3) lib/paperclip/attachment.rb:283:in `post_process' paperclip (2.3.3) lib/paperclip/attachment.rb:214:in `reprocess!' app/models/user.rb:339:in `reprocess_avatar' app/controllers/user_controller.rb:57:in `update_avatar' haml (2.2.3) rails/./lib/sass/plugin/rails.rb:19:in `process' /home/heroku_rack/lib/static_assets.rb:9:in `call' /home/heroku_rack/lib/last_access.rb:25:in `call' /home/heroku_rack/lib/date_header.rb:14:in `call' thin (1.0.1) lib/thin/connection.rb:80:in `pre_process' thin (1.0.1) lib/thin/connection.rb:78:in `catch' thin (1.0.1) lib/thin/connection.rb:78:in `pre_process' thin (1.0.1) lib/thin/connection.rb:57:in `process' thin (1.0.1) lib/thin/connection.rb:42:in `receive_data' eventmachine (0.12.6) lib/eventmachine.rb:240:in `run_machine' eventmachine (0.12.6) lib/eventmachine.rb:240:in `run' thin (1.0.1) lib/thin/backends/base.rb:57:in `start' thin (1.0.1) lib/thin/server.rb:150:in `start' thin (1.0.1) lib/thin/controllers/controller.rb:80:in `start' thin (1.0.1) lib/thin/runner.rb:173:in `send' thin (1.0.1) lib/thin/runner.rb:173:in `run_command' thin (1.0.1) lib/thin/runner.rb:139:in `run!' thin (1.0.1) bin/thin:6 /usr/local/bin/thin:20:in `load' /usr/local/bin/thin:20

    Read the article

< Previous Page | 244 245 246 247 248 249 250 251 252 253 254 255  | Next Page >