How to get GPS coordinates on a GSM device

Similar Questions

• How to transfer GPS coordinates from one film to the other using Applescript?

  When I export a movie from the Photos app and read it in Quicktime Player.

  QTP then gives me the location where the film was saved:

  

  Now, I would like to set the location of another movie in exactly in the same place, using Automator or Applescript.

  Could someone please give me a tip how to do this? Is this possible?

  The reason why I want to do is I want to improve an Automator workflow, I wrote which helps the compression of movies that I keep in my library of Photos (that is, to reduce the size of the file).

  The workflow, you can find here (http://www.informatik.uni-bremen.de/~zach/software/index.html#compress_movies_in _photos), if you are interested.

  Tips, pointers and suggestions will be highly appreciated.

  Best regards

  Gabriel.

  Hard to do standalone AppleScript, but easy if you install a copy of exiftool

  Then, you can extract the GPS coordinates via something like:

  the shell script "" / usr/local/bin/exiftool - GPSPosition some. "." JPG ".

  You can also use exiftool to copy the tags of one file to another:

  the shell script ' / usr/local/bin/exiftool - tagsFromFile - GPSLongitude source.jpg - GPSLatitude dest.jpg.

  This will copy the tags GPS Longitude and Latitude of source.jpg to dest.jpg

• Trouble getting GPS coordinates in the Simulator

  Here is the code I use to get the coordinates of the GPS by train. It seems that every time that it expires and ends by 0 for the latitude and longitude of return even if I put together a location in my simulation-> GPS location... window and entered in the Options-> advanced-> GPS-> was cool my main application class, this function is called by a button.

  public void GetCurrentLocation() {         Criteria ct = new Criteria();            //ct.setAddressInfoRequired(false);          //ct.setAltitudeRequired(false);         ct.setCostAllowed(true);        ct.setHorizontalAccuracy(Criteria.NO_REQUIREMENT);        ct.setVerticalAccuracy(Criteria.NO_REQUIREMENT);        ct.setPreferredPowerConsumption(Criteria.POWER_USAGE_LOW);        //ct.setPreferredResponseTime(10000);        //ct.setSpeedAndCourseRequired(false);
  
          try {            myLP = LocationProvider.getInstance(ct);            myLP.setLocationListener(new myLocListener(this), -1, -1, -1);        } catch (Exception ex) {            return;        }    }
  
  public void GotLocation(Location myLoc) {        if(myLoc != null) {            QualifiedCoordinates qc = myLoc.getQualifiedCoordinates();            //Call webservice here.        }    }
  

  In the myLocListener.java

  package net.gasbot;
  
  import javax.microedition.location.*;
  
  public class myLocListener implements LocationListener{    GasBotScreen gb;
  
      public myLocListener(GasBotScreen gb)    {        this.gb = gb;    }
  
      public void locationUpdated(LocationProvider provider, Location location)    {        QualifiedCoordinates qc = location.getQualifiedCoordinates();  //I set a breakpoint here, but it rarely gets called, and when it does, the coordinates are both 0.        provider.setLocationListener(null, 0, 0, 0);        gb.GotLocation(location);    }    public void providerStateChanged(LocationProvider provider, int newState)    {
  
      }}
  

  Can you get it someone please let me know what I am doing wrong? I'll put the listener null once it gets a set location to update as I need it only once. I also tried to use the.getLocation(-1) and it does not work either. Is there something wrong with my Simulator? Or just something wrong with my code?

  I don't understand your code, in fact.  Why are reset you the listener inside the listener?

  In addition, you "eat" the exception where you get the provider. He is a likely area of failure, so I would put a trace and trace any exception.

  Another thing to try: try to set criteria to null when you get the provider: gives you the provider "by default".

• How to get the coordinates of a Swipe gesture?

  Hi guys! I build a game similar to Fruit Ninja and recently, I encountered the following problem: How can I access the points of contact of a cause of event drag as you now to tear a piece of fruit, I need to know if you hit points are same as the location of the fruit.

  So, basically, it's something like:

  If {(Swipe.x is fruit.x)

  Destroyfruit();

  }

  So is it possible to understand the coordinates suddenly?

  Thank you

  Gestures have phases. Shot has a begin and end and not raise any events outside of those. You won't be able to get the coordinates to start drag and at the end. Don't use no gestures, use regular events of Touch with stageX and stageY (or localX and localY if limited to a certain area). These events go off quickly and continuously. You can simplify it by listening to just TouchEvent.TOUCH_MOVE because it fires so quickly.

• How to get the coordinates of the screen when they are hit.

  Hi all!

  I am a beginner on the development of blackberry, also a beginner in programming in general ^^

  I've done some app using just QML and take some C++ tutorials

  I try to get the coordinate of the screen when the user touches, I wasn't able to do that you use just QML, I found the source TouchEvent and tried to make it work on C++, to get the windowX and windowY coordinates of the touch.

  http://developer.BlackBerry.com/Cascades/reference/bb__cascades__touchevent.html

  I use the image on background to try to get contact with the event.
  So far, we have:

  PPS:

  class LostInSpace : public QObject
  {
      Q_OBJECT
  public:
      LostInSpace(bb::cascades::Application *app);
      virtual  ~LostInSpace() {}
  public slots:
  Q_INVOKABLE float screentouchY(bb::cascades::TouchEvent* event);
  Q_INVOKABLE float screentouchX(bb::cascades::TouchEvent* event);
  
  private:
  QPointer fundo;
  
  };
  

  CPP:

  LostInSpace::LostInSpace(bb::cascades::Application *app)
  
  {
  
      QmlDocument *qml =
  QmlDocument::create("asset:///main.qml").parent(this);
      qml->setContextProperty("app", this);
      AbstractPane *root = qml->createRootObject();
  
      app->setScene(root);
  
      fundo = root->findChild("fundo");
      bool res = QObject::connect(fundo, SIGNAL(touch(bb::cascades::TouchEvent*)),this, SLOT(screentouchY(bb::cascades::TouchEvent*)));
      Q_ASSERT(res);
      res = QObject::connect(fundo, SIGNAL(touch(bb::cascades::TouchEvent*)),
  this, SLOT(screentouchX(bb::cascades::TouchEvent*)));
      Q_ASSERT(res);
  
  }
  
  float LostInSpace::screentouchY(bb::cascades::TouchEvent* event) {
      return event->windowY();
  }
  float LostInSpace::screentouchX(bb::cascades::TouchEvent* event) {
      return event->windowX();
  }
  

  QML: (I'll do later animations, it's just to see if it works. I mean a "unknown simbol app" the Notecard, I thought I just had to put "qml-> setContextProperty ("app", this);  in the constructor)

  ImageView {
                  objectName: "fundo"
                  imageSource: "images/fundo1.png"
                  touchPropagationMode: TouchPropagationMode.Full
                  onTouch: {
                      tiro.translationX = app.screentouchX();
                      tiro.translationY = app.screentouchY();
                  }
              }
  

  Any help will be very appreciated!

  Thank you!

  Hey there,

  You don't need to do... You can access the 'event' TouchEvent object in the notecard: {} area QML...

  ImageView {
                  objectName: "fundo"
                  imageSource: "images/fundo1.png"
                  touchPropagationMode: TouchPropagationMode.Full
                  onTouch: {
                      tiro.translationX = event.windowX
                      tiro.translationY = event.windowY
                  }
              }
  

• How to get the name of the bluetooth device and the name of the bluetooth device address device class?

  Hi all

  Can I know how to get the bluetooth device name and the name of the bluetooth device address device class?

  Example:-if my bluetooth device address is :-98:EB:CD:01:1F::90 then how do I get my bluetooth device name?

  I hope you understand my problem. Let me know as soon as possible.

  :

  Thanks for the reply.

• How to get the PIN for the bb device number

  Hi all, I want to get the ID of device bb, how to get it thanks to all

  Integer.toHexString (DeviceInfo.getDeviceId ()) .toUpperCase)

• How to get the coordinates of "pathPoints" of a textFrameItem?

  Hi all

  Yes, I know that textFrameItems have pathPoints but...

  How can I get the x coordinates of the corners of a textFrameItem of rotation.

  I know I can get the angle of rotation of asin (myTextFrameItem.matrix.mValueA) etc.

  My guess would be that one should be able to use the mValueTX and mValueTY to help, but these values are very confusing and change dramatically on the application even a small rotation.

  If anyone has some preference mathematically based ideas and not API base, please let me know.

  It's very late by me now and I think I must be missing what is obvious, I think that such a function must have been made.

  Thank you

  Trevor

  PS Would be also appreciated a solution to the problem even in InDesign

  

  Hi all

  I didn't try the method DOM I knew the mathematical method.

  I don't give great results if a shear has been applied, but that doesn't bother me that I do not apply the shears

  I received contact details-box-find http://stackoverflow.com/questions/9971230/calculate-rotated-rectangle-size-from-known-bou a useful formula

  y = (1 / (Math.cos (a) * Math.cos (a) - Math.sin (a) * Math.sin (a))) * (-w * Math.sin (a) + h * Math.cos (a));

  where y is the height of the inner when box no rotation is applied on it and a is the angle of rotation and w is the width.

  I had found for me a different formula but got stuck when moving to the rotation out of the 90 degree range, that is to say 1 degree or 91 degrees.

  The same problem occurred with this method to fix a fact of what follows.

  a RAD = ((360 + degs (a)) 90%);

  I did the temporary to degrees conversion as I remember problems with Mod enforcement by a fraction, I think that Ariel was a thread on this in the indesign forum.

  Here are the results of the script

  

  // draw a text box and rotate it before running the script
  // By Trevor www.creative-scripts.com (sorry still not too much there)
  // https://forums.adobe.com/message/7678561#7678561
  // With a little bit of help from http://stackoverflow.com/questions/9971230/calculate-rotated-rectangle-size-from-known-bounding-box-coordinates
  function main () {
  
  var doc = app.documents.length && app.activeDocument,
      itext = doc && doc.textFrames[0];
  if (!itext) {
      alert ("Sorry mate,\nI need a document with a textFrameItem on it or there's not much for me to do");
      return "twit";
  }
  var gb = itext.geometricBounds,
      imatrix = itext.matrix,
      p1X, p1Y, p2X, p2Y, p3X, p3Y, p4X, p4Y,
      a, h, w, x, y, hyp, op, adj,
      noColor = new NoColor(),
      rgbColor = new RGBColor(),
      newShape = doc.pathItems.add(),
      rect;
  ;
  
  // Do the maths
  
  a = Math.atan2 (imatrix.mValueB, imatrix.mValueA); // get the angle of the textFrameItem
  a = rads ((360 + degs(a)) % 90); // make sure it's positive otherwise the result will get messed up
  w = gb[2] - gb[0]; // Width of outer enclosing box
  h = gb[1] - gb[3]; // Height of outer enclosing box
  /* x = (1/(Math.cos(a) * Math.cos(a) - Math.sin(a) * Math.sin(a))) * (w * Math.cos(a) - h * Math.sin(a)); // Width of inner enclosed box, this is not needed for this script */
  y = (1/(Math.cos(a) * Math.cos(a)- Math.sin(a) * Math.sin(a))) * (- w * Math.sin(a) + h * Math.cos(a)); // Height of inner enclosed box
  op = Math.sin(a) * y; // this is the x coordinate offset
  adj = Math.cos(a) * y; // this is the y coordinate offset
  
  // calculate the points of the inner enclosed box
  p1X = gb[0] + op; // left bottom
  p1Y = gb[3];
  p2X = gb[0]; // left top
  p2Y = gb[3] + adj;
  p3X = gb[2] - op; // right top
  p3Y = gb[1];
  p4X = gb[2]; // right bottom
  p4Y = gb[1] - adj;
  
  // Draw outer box
  rgbColor.blue = 255;
  rect = doc.pathItems.rectangle(gb[1], gb[0], gb[2]-gb[0], Math.abs(gb[3]-gb[1]) );
  rect.fillColor = noColor;
  rect.strokeColor = rgbColor;
  
  // Draw inner box
  newShape.setEntirePath([[p1X, p1Y], [p2X, p2Y], [p3X, p3Y], [p4X, p4Y], [p1X, p1Y]]);
  rgbColor.red = 255;
  newShape.strokeColor = rgbColor;
  newShape.fillColor = noColor;
  
  // Helper functions
  function rads (x) {return x * (Math.PI / 180);}
  function degs (x) {return x * (180 / (Math.PI));}
  
  }
  main()
  

  Hi Uwe,

  It was late, but I don't know how I missed the path object. Age!

  A good thing I didn't post on the InDesign forum, at least here, I can wait less snickering for a such silent over look that I went back to the InDesign forum

  Either way thanks to you two, best regards,.

  Trevor

• How to get the coordinates and the size of the swf to scale components

  I use the Web (I put it to the size of the entire screen) to frame a swfloader that loads a swf file with the width and height set to 100%. This swf has a movieclip. How can I get the exact coordinates of his position and its size after the scaling in Flex?

  In Flash CS4, the movielip a x = 40, y = 700. In Flex, the swf file is set to 100% scale, but when I get the x value of this movieclip, it is always 40. I try to get the scale of the x axis ratio using scaleX of swf. But his 1, pretty weird.

  How can I get the correct coordinates of the movieclip?

  The coordinates are usually compared to something.  X is relative to the parent.

  LocalToGlobal and globalToLocal convert coordinated areas.

  You can go to transform.pixelBounds and

  transform.concatenatedMatrix.

• How to get the coordinates of the highest "current view" on the screen?

  Picture the following scenario:

  We have an application full screen has a vertical field with a few dozen Manager fields.   This screen will be much bigger, then the physical blackberry screen, scrolling, etc.

  What I need to findout corresponds to the vertical offset of 0 at the top of the visible display area.

  i.e.

  _____________<->

  ^

  Need to find this height

  v

  -----------------------

  visible area

  -----------------------

  ______________<>

  Is there something simple for this?

  getVerticalScroll seems to be.

• How to get the license dynamic foff key device rom Cascades/Native

  Hello

  In Java, you would do this after initialization the CoreModule;

  String lc = group.getProperty ("RIM_APP_WORLD_LICENSE_KEY");

  How to recover using C++?

  If this is not possible then second best how can I get the device id?

  A few months later, I thought again that and it close.

  Dynamics of the license is no longer available for BB10, the options are to apply yourself or to manipulate through payments in application.

  Feedback from BlackBerry can be found on JIRA here...

  https://www.BlackBerry.com/jira/browse/Tablet-429

• How to get rid of out-of-use devices of ADE so I can allow my new camera?

  My A.D.E will be n ot allow my new Kobo e-reader.  When I try it says that there are already too many permissions.  How can I get rid of the (unauthorise) out-of-use devices, that is to say my previous two and now obsolete kobo e-readers of ADE, so I can allow my new camera?

  You need to speak with the service from Adobe.

  It will delete ALL your records, and you must reregister those you want.

  For some reason, Adobe isn't any sensible system online to view/edit the records.

  http://www.Adobe.com/support/digitaleditions/supportinfo/

  and open a cat.

  FAQ information to

  http://www.Adobe.com/UK/products/Digital-Editions/FAQ.html

  See "What is the maximum number of computers and devices that I can activate?"

  who sends you

  http://helpx.Adobe.com/digital-editions.html

  that gets you finally ('contact support' to)

  http://www.Adobe.com/support/digitaleditions/supportinfo/

• How to get information from the owners of devices (NMP and Powerpath) using the VI SDK

  Hello

  I just started to work with VI SDK in C++ using gsoap. I am

  facing a problem. Using VI SDK, where can I get more information

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  able to conclude within ScsiLun HostMultipathInfoLogicalUnit or

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  Kindly help me.

  Thank you

  Gesner

  Ah good to know it is where it is located.

  national immunization days , you can find more info here: http://www.vmware.com/support/developer/vc-sdk/visdk400pubs/ReferenceGuide/vim.host.PlugStoreTopology.html#path

  =========================================================================

  William Lam

  VMware vExpert 2009

  Scripts for VMware ESX/ESXi and resources at: http://engineering.ucsb.edu/~duonglt/vmware/

  repository scripts vGhetto

  VMware Code Central - Scripts/code samples for developers and administrators

  http://Twitter.com/lamw

  If you find this information useful, please give points to "correct" or "useful".

• How to get back the volume if the devices work properly?

  my computer is a microsoft windows XP Edition version 2002 service pack 2 family.

  apparentl, I installed the home edition of windows to the computer because it worked well before. After installing the cd, I notice that the volume icon is missing. even if I go to the volume control, it gives a message saying that I can't access. I went on Control Panel and click on sounds and audio devices. the options for the default devices have been updated with the gray scale. which means I was unable to select a device. I need help as soon as possible. Please and thank you.

  Hello

  ·         What is the brand and model of your computer?

  ·         You have an its own card installed on your computer? If so, what is the model name of your sound card?

  You can try to search for drivers online using or updating sound card drivers using windows update link:

  No sound

  Select Windows XP.

• How to get the FileSystemTree coordinates selected item?

  Hi can we idea me in how to get the coordinates of the itwn selected in a FileSystemsTree?
  
  THX
  
  g

  You should do something like this:

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  Contact information: Point = itemRenderer.parent.localToGlobal (new Point (itemRenderer.x, itemRenderer.y)) var;

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