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  • How to point a subdomain to local server with dynamic IP

    - by jlego
    I see there are many related questions to this one, however the answers given seem to be a little vague for a novice like me. I've got a dedicated LAMP stack running Fedora 16 locally on my home network. Everything works fine internally. I can access the Apache server from other machines on the network using the internal IP in a browser. I'm using the stack for a local file server as well as a development environment for websites. There are a couple of reasons why I would like the development sites hosted on the machine to be available publicly. 1.) I use a CMS that has paid add-ons which allows you to assign the paid license to a domain. I can't develop with paid add-ons on the closed dev server. 2.) I would occasionally like for clients to be able to view the site dev at late stages before it goes live. I have a domain (foo.com, and I want to point a *sub*domain (dev.foo.com) to the local server. I know this is best accomplished with a Static IP, however my IP from my ISP is Dynamic and I don't think there is any way to change that. From what I have read, services like ZoneEdit & DynDNS are supposed to be able to accomplish this, but I have tried both and found it very confusing. Also the server is behind a router and I have also read that you need to set up DDNS(?) in your router, that many routers have presets for these services, and I've found that DynDNS is the only one my router seems to support.

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  • Find the IP of the router between me and main router/gateway?

    - by Crash893
    I have a Netgear router that is the main router/gateway in and out of the network. Then I have a Linksys 54g router with WiFi that we use as are main WiFi access point The Cat5 runs from the router to the WiFi routers LAN port (not WAN). And then if you connect to that WAP you are basically on the network. I want to be able to find out what the IP address of that WAP is but so far I can't figure out how to: I've tried doing a tracert to the gateway IP but I get nothing. I've scanned all the IP addressed in the network. I've gone to the Netgear main router and looked under attached devices but it doesn't say anything. Any ideas on how I can figure out how to administer it? (I do have limited physical access but my attempt to plug into it came back with a faulty IP (in the 169 range where the rest of the office is 192.168...). I never set this router up to begin with so I'm reluctant to just kill it with the reset button because I can't get into see what the settings are set to.

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  • Sending mail through local MTA while domain MX records point to Google Apps

    - by Assaf
    My domain's email is managed by Google Apps, so that domain users get Gmail and Calendar, etc. But I also want to be able to send applicative notifications to users outside the domain via email (e.g. "some commented on your post", and so on). However, if I try to send email through code I get blocked by Gmail after a few emails. I send marketing email through MailChimp, to minimize the risk of appearing as spam to my users (one-click unsubscribe, etc.). But I can't send applicative message in this way. I want to install a local MTA (my server runs Ubuntu), but I'm not sure what anti-spam measures I need to implement so that receiving MTAs don't think it's a spam server. What's stopping anyone from setting up a mail server and sending emails using my domain name? AFAIK it's the DNS records that show the MTA's address actually belongs to the domain. But my understanding of this is rather superficial, so someone please correct me if I'm wrong. But what sort of DNS configuration do I need to put in place so that I don't get blacklisted (assuming I don't actually spam anyone)? The MX records already point to Google, and I'd like to keep it this way. So do I just need to define an A record for my internal mail server? Should it show email as coming from a sub-domain, so as not to conflict with the bare domain being managed by google? Edit: Does the following SPF record make sense if I want email from my domain name to be sent by either google's servers or any server with a dns name ending with mydomain.com? "v=spf1 ptr mx:google.com mx:googlemail.com ~all" How should I set up reverse DNS for my server? If I have an A record that points mailsender.mydomain.com to my MTA's ip address, does it mean that reverse lookup will only allow emails sent from [email protected]?

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  • Packets being dropped by iptables

    - by Shadyabhi
    I am trying to create a Software Access Point in linux. I followed the blog here. Steps I performed: Started dhcp server on wlan0. Properly configured hostapd.conf Enabled packet forwarding & masquerading. Two commands executed regarding iptables: iptables --table nat --append POSTROUTING --out-interface eth0 -j MASQUERADE iptables --append FORWARD --in-interface wlan0 -j ACCEPT I enabled logging on iptables & I get this in everything.log Jun 29 19:42:03 MBP-archlinux kernel: [10480.180356] IN=eth0 OUT=wlan0 MAC=c8:bc:c8:9b:c4:3c:00:13:80:40:cd:80:08:00 SRC=195.143.92.150 DST=10.0.0.3 LEN=44 TOS=0x00 PREC=0x00 TTL=52 ID=38025 PROTO=TCP SPT=80 DPT=53570 WINDOW=46185 RES=0x00 ACK URGP=0 Jun 29 19:42:03 MBP-archlinux kernel: [10480.389102] IN=eth0 OUT=wlan0 MAC=c8:bc:c8:9b:c4:3c:00:13:80:40:cd:80:08:00 SRC=195.143.92.150 DST=10.0.0.3 LEN=308 TOS=0x00 PREC=0x00 TTL=52 ID=14732 PROTO=TCP SPT=80 DPT=53570 WINDOW=46185 RES=0x00 ACK PSH URGP=0 Jun 29 19:42:03 MBP-archlinux kernel: [10480.389710] IN=eth0 OUT=wlan0 MAC=c8:bc:c8:9b:c4:3c:00:13:80:40:cd:80:08:00 SRC=195.143.92.150 DST=10.0.0.3 LEN=44 TOS=0x00 PREC=0x00 TTL=52 ID=14988 PROTO=TCP SPT=80 DPT=53570 WINDOW=46185 RES=0x00 ACK FIN URGP=0 Jun 29 19:42:03 MBP-archlinux kernel: [10480.621118] IN=eth0 OUT=wlan0 MAC=c8:bc:c8:9b:c4:3c:00:13:80:40:cd:80:08:00 SRC=195.143.92.150 DST=10.0.0.3 LEN=44 TOS=0x00 PREC=0x00 TTL=52 ID=63378 PROTO=TCP SPT=80 DPT=53570 WINDOW=46185 RES=0x00 ACK FIN URGP=0 I have almost no knowledge of iptables, all I did was through googling. So, can anyone help me in making me understand what wrong is happening here? I have tried running tcpdump on wlan0 & http packets are being sent from wlan0.

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  • Outdoor WiFi Mesh Topology vs. Repeaters

    - by IronJaxor
    Here's the current configuration in our organization (which I believe is incorrect): We have a number of Cisco 1500 series AP's (22 in total), that are mounted outdoors to provide seamless WiFi coverage over a large area. Each AP however has its own physical ethernet connection back to the WLC (All the AP's are marked as Root AP's). They are all broadcasting the same SSID. We have tried to stagger the channel selection but because there are only three non-overlapping channels to choose from, and in some areas the density of AP's is quite high, there is multiple places of channel interference. With this configuration we experience 100-150 disconnects from clients every day. (Our clients are mobile so they move throughout the coverage area constantly). My idea is to switch the AP's to the same channel thereby forming a wireless mesh, use the built in functionality of the 1500 series to use 802.11a as the backhaul, designate one or two AP's as root AP's and wire them back to the WLC. Thereby forming a WiFi mesh, which if I'm not mistaken is the point of the 1500 series in the first place! I am however completely new at WiFi networks and wondering if I am simply mistaken in what I believe my proposed changes will enable, or if there is a better way to tackle the WiFi topology.

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  • Laptop seemingly randomly "freezes" to the point of no longer executing applications

    - by Aierou
    After upgrading to Windows 8 pro on my Samsung Series 7 Chronos NP700Z5C-S04US (may be relevant, I'm not sure), my computer began to stop allowing the execution of any service or application, as well as discontinuing the update of the clock until a hard shutdown was performed. This seems to occur randomly after periods of inactivity and I've no idea the cause. These are measures I have already taken in order to attempt to stop this: -Obviously Googling potential answers to this problem -Updating all drivers -Researching all events that have occurred around the time of the failure to respond (with no results) -I tried applying "bcdedit /set disabledynamictick no" which was a hotfix for what seemed to be the same error but was not. Here is some more, potentially related, information about the error: -No BSOD (actually, I haven't at all experienced a BSOD with Windows 8) -Computer seems to have a problem shutting down/restarting most of the time (Hangs at the point where it should completely turn off) -New sound instances are not able to play, but previously loaded containers function properly -As mentioned before, the clock freezes at the time of the error -USB devices function properly -Servers that I was running fail to respond on my end, but stay online. If you require more information, please request it specifically and I will be happy to oblige. Thanks.

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  • Using a second Wifi Router (in order to use a LAN port)

    - by Sledge81
    Problem: Connecting a TV decoder via Internet. It doesn't have inbuilt wifi so wired LAN is the only option. I bought a second Wifi Router so I could use its LAN port to create a wired internet connection to my decoder. What I want to achieve: The second Wifi Router should basically pick up the signal from the main wifi router. I would then use the LAN port to connect (with a wire) to the TV decoder. In other words have my secondary Wifi Router act as an access point, which will enable me to use a wired LAN connection to the decoder. What i've done so far: Connected the second Wifi Router to my laptop via the LAN ports. Access 192.168.1.1 and went into my second router. 1) SSID set to the same one as the main Wifi Router 2) Tried disabling DHCP and enabling DHCP (with the DNS and default gateway configured the same as the main router). When I check my network connections, I see the LAN connection too but it says 'Not connected to the internet' while the Wifi (main router) shows connected. Can someone please advise on how to use my second Wifi Router to connect to the main Wifi Router (and thus the Internet). Thanks. main wifi router: Zyxel secondary wifi router: TP Link

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  • In Ruby, how does coerce() actually work?

    - by Jian Lin
    It is said that when we have a class Point and knows how to perform point * 3 like the following: class Point def initialize(x,y) @x, @y = x, y end def *(c) Point.new(@x * c, @y * c) end end point = Point.new(1,2) p point p point * 3 Output: #<Point:0x336094 @x=1, @y=2> #<Point:0x335fa4 @x=3, @y=6> but then, 3 * point is not understood: Point can't be coerced into Fixnum (TypeError) So we need to further define an instance method coerce: class Point def coerce(something) [self, something] end end p 3 * point Output: #<Point:0x3c45a88 @x=3, @y=6> So it is said that 3 * point is the same as 3.*(point) that is, the instance method * takes an argument point and invoke on the object 3. Now, since this method * doesn't know how to multiply a point, so point.coerce(3) will be called, and get back an array: [point, 3] and then * is once again applied to it, is that true? point * 3 which is the same as point.*(3) and now, this is understood and we now have a new Point object, as performed by the instance method * of the Point class. The question is: 1) who invokes point.coerce(3) ? Is it Ruby automatically, or is it some code inside of * method of Fixnum by catching an exception? Or is it by case statement that when it doesn't know one of the known types, then call coerce? 2) Does coerce always need to return an array of 2 elements? Can it be no array? Or can it be an array of 3 elements? 3) And is the rule that, the original operator (or method) * will then be invoked on element 0, with the argument of element 1? (element 0 and element 1 are the two elements in that array returned by coerce) Who does it? Is it done by Ruby or is it done by code in Fixnum? If it is done by code in Fixnum, then it is a "convention" that everybody follows when doing a coerce? So could it be the code in * of Fixnum do something like this: if (something.typeof? ...) else if ... # other type else if ... # other type else # if it is not a type I know array = something.coerce(self) return array[0].*(array[1]) end

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  • change owner/uid of mount point upon mount

    - by Shiplu
    The scenario is like this. Bob has a computer. It crashed. Now he only has the hdd. The hdd is in ext3 format. He go to his office and told the sys admin John to mount this hdd and put the mount point in his home directory. John used the following fstab entries. # Bobs harddisk /media/TAPE4/Bobs-hdd.img /home/bob/myhdd/windows ntfs ro,loop,offset=32256 0 0 /media/TAPE4/Bobs-hdd.img /home/bob/myhdd/linux ext3 ro,loop,offset=14048810496 0 0 /media/TAPE4/Bobs-hdd.img /home/bob/myhdd/extra ntfs ro,loop,offset=28015335936 0 0 Bob was happy. He could access his old extra and windows. Specially the Documents and Settings in windows was helpful for him. But he found a problem. He is a web developer and all his websites are in linux/home/bob/public_html directory. When he tried to access that public_html directory he got permission_denied. He executed ls -lh he saw this. drwxr-xr-x 2 john john 4.0K Nov 9 2011 Desktop drwxr-xr-x 3 john john 4.0K Aug 12 2011 Documents drwxr-xr-x 3 john john 4.0K Aug 21 2011 public_html He contacted John thinking he might be mistakenly did this. But John couldn't find a way why this happend? Then one thing came into his mind file system hardly store username. They store uids. So he executed ls -ln drwxr-xr-x 2 1000 1000 4096 Nov 9 2011 Desktop drwxr-xr-x 3 1000 1000 4096 Aug 12 2011 Documents drwxr-xr-x 3 1000 1000 4096 Aug 21 2011 public_html John thinks 1000 is the first uid on a linux system. As he is the admin of the current system. He created his account first. so Johns uid was 1000. Bob also setup his private system and crated his account first. So Bobs uid was 1000 too. So thats an expected behavior. But problem remains. How can Bob access those websites in public_html?

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  • Frequent and weird wifi disconnections

    - by Sidou
    How would you explain, troubleshoot (and solve) the following problem? Wifi ADSL modem router D-link 2640R installed in living room at about 1.8m height. Working fine, synchronising and getting/serving stable internet connection. First situation: -Laptop 01 in other end of the house, let's say in room01 southern to the living room, distant by about 15m. Getting stable signal of good to very good quality. No disconnection. -Laptop 02 in room02 opposite to room01 (5m West) which makes it almost at the same distance and direction from the router located 15m North. Getting stable signal of good to very good quality. No disconnection. Second situation: -Laptop 01 moved to room03 Northern to the living room (actually just 3m behind the wall where the router lies). Getting stable signal of excellent quality. No disconnection. -Laptop 02 still in room02 but now experiences frequent disconnections (actually almost impossible to get the Internet even though the signal level is still very good. Either no Internet with the wifi icon appearing connected to access point or no connection established at all which happens every 2 minutes and that means virtually no Internet at all as I can just get a timeframe of 1 minute or so to load any website or even get to the router's web based control panel. If Laptop 01 is completely shut down or its wifi adapters shut down or even still working but its wifi MAC address forbidden, then Laptop 02 has no problem at all. If Laptop 02 is moved to a nearer location to the router, in the living room for instance, then no connection problem occurs even if Laptop 01 is also connected. And also if we move back Laptop 01 to its original location (room 01), then no problem as well. I'm completely lost and don't know how to address this issue. I tried to change the Wifi channel and even tried the auto channel scan but that didn't solve it. I know that the problem is probably coming from Laptop 01 being in its new location or some sort of interference as the problem occurs only under the described condition but I have no idea how to solve it! I also scanned the neighborhood for wifi jam using InSSIDer, there are few other access points but they don't seem to affect the situation. Any ideas about the steps to follow or tools to use ?

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  • CSS: Why is my floated <span> being displayed below an <a>nchor in IE6/7 but not IE8/FF

    - by gsquare567
    i'm getting this weird CSS bug in ie6/7 (but not in ie8 or firefox): for some reason, my nchor and , two inline elements, which are on the same line, are being displayed on different lines. the span is floating to the right, too! heres the HTML: <div class="sidebartextbg"><a href="journey.php" style="width:50%" title="Track past, present and future milestones during your employment">Journey</a> <span class="notificationNumber">2</span> <!-- JOURNEY COUNT: end --> </div> and here's the CSS: .sidebartextbg { background:url("../images/sidebartextbg.gif") repeat-x scroll 0 0 transparent; border-bottom:1px solid #A3A88B; font-size:14px; line-height:18px; margin:0 auto; padding:5px 9px; width:270px; } .notificationNumber { background:url("../images/oval_edges.gif") no-repeat scroll 0 0 transparent; color:#FFFFFF; float:right; padding:0 7px; position:relative; text-align:center; width:17px; } so: why would the floated span be displayed on the line under the nchor? thanks!

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  • The sign of zero with float2

    - by JackOLantern
    Consider the following code performing operations on complex numbers with C/C++'s float: float real_part = log(3.f); float imag_part = 0.f; float real_part2 = (imag_part)*(imag_part)-(real_part*real_part); float imag_part2 = (imag_part)*(real_part)+(real_part*imag_part); The result will be real_part2= -1.20695 imag_part2= 0 angle= 3.14159 where angle is the phase of the complex number and, in this case, is pi. Now consider the following code: float real_part = log(3.f); float imag_part = 0.f; float real_part2 = (-imag_part)*(-imag_part)-(real_part)*(real_part); float imag_part2 = (-imag_part)*(real_part)+(real_part)*(-imag_part); The result will be real_part2= -1.20695 imag_part2= 0 angle= -3.14159 The imaginary part of the result is -0 which makes the phase of the result be -pi. Although still accomplishing with the principal argument of a complex number and with the signed property of floating point's 0, this changes is a problem when one is defining functions of complex numbers. For example, if one is defining sqrt of a complex number by the de Moivre formula, this will change the sign of the imaginary part of the result to a wrong value. How to deal with this effect?

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  • Basic Spatial Data with SQL Server and Entity Framework 5.0

    - by Rick Strahl
    In my most recent project we needed to do a bit of geo-spatial referencing. While spatial features have been in SQL Server for a while using those features inside of .NET applications hasn't been as straight forward as could be, because .NET natively doesn't support spatial types. There are workarounds for this with a few custom project like SharpMap or a hack using the Sql Server specific Geo types found in the Microsoft.SqlTypes assembly that ships with SQL server. While these approaches work for manipulating spatial data from .NET code, they didn't work with database access if you're using Entity Framework. Other ORM vendors have been rolling their own versions of spatial integration. In Entity Framework 5.0 running on .NET 4.5 the Microsoft ORM finally adds support for spatial types as well. In this post I'll describe basic geography features that deal with single location and distance calculations which is probably the most common usage scenario. SQL Server Transact-SQL Syntax for Spatial Data Before we look at how things work with Entity framework, lets take a look at how SQL Server allows you to use spatial data to get an understanding of the underlying semantics. The following SQL examples should work with SQL 2008 and forward. Let's start by creating a test table that includes a Geography field and also a pair of Long/Lat fields that demonstrate how you can work with the geography functions even if you don't have geography/geometry fields in the database. Here's the CREATE command:CREATE TABLE [dbo].[Geo]( [id] [int] IDENTITY(1,1) NOT NULL, [Location] [geography] NULL, [Long] [float] NOT NULL, [Lat] [float] NOT NULL ) Now using plain SQL you can insert data into the table using geography::STGeoFromText SQL CLR function:insert into Geo( Location , long, lat ) values ( geography::STGeomFromText ('POINT(-121.527200 45.712113)', 4326), -121.527200, 45.712113 ) insert into Geo( Location , long, lat ) values ( geography::STGeomFromText ('POINT(-121.517265 45.714240)', 4326), -121.517265, 45.714240 ) insert into Geo( Location , long, lat ) values ( geography::STGeomFromText ('POINT(-121.511536 45.714825)', 4326), -121.511536, 45.714825) The STGeomFromText function accepts a string that points to a geometric item (a point here but can also be a line or path or polygon and many others). You also need to provide an SRID (Spatial Reference System Identifier) which is an integer value that determines the rules for how geography/geometry values are calculated and returned. For mapping/distance functionality you typically want to use 4326 as this is the format used by most mapping software and geo-location libraries like Google and Bing. The spatial data in the Location field is stored in binary format which looks something like this: Once the location data is in the database you can query the data and do simple distance computations very easily. For example to calculate the distance of each of the values in the database to another spatial point is very easy to calculate. Distance calculations compare two points in space using a direct line calculation. For our example I'll compare a new point to all the points in the database. Using the Location field the SQL looks like this:-- create a source point DECLARE @s geography SET @s = geography:: STGeomFromText('POINT(-121.527200 45.712113)' , 4326); --- return the ids select ID, Location as Geo , Location .ToString() as Point , @s.STDistance( Location) as distance from Geo order by distance The code defines a new point which is the base point to compare each of the values to. You can also compare values from the database directly, but typically you'll want to match a location to another location and determine the difference for which you can use the geography::STDistance function. This query produces the following output: The STDistance function returns the straight line distance between the passed in point and the point in the database field. The result for SRID 4326 is always in meters. Notice that the first value passed was the same point so the difference is 0. The other two points are two points here in town in Hood River a little ways away - 808 and 1256 meters respectively. Notice also that you can order the result by the resulting distance, which effectively gives you results that are ordered radially out from closer to further away. This is great for searches of points of interest near a central location (YOU typically!). These geolocation functions are also available to you if you don't use the Geography/Geometry types, but plain float values. It's a little more work, as each point has to be created in the query using the string syntax, but the following code doesn't use a geography field but produces the same result as the previous query.--- using float fields select ID, geography::STGeomFromText ('POINT(' + STR (long, 15,7 ) + ' ' + Str(lat ,15, 7) + ')' , 4326), geography::STGeomFromText ('POINT(' + STR (long, 15,7 ) + ' ' + Str(lat ,15, 7) + ')' , 4326). ToString(), @s.STDistance( geography::STGeomFromText ('POINT(' + STR(long ,15, 7) + ' ' + Str(lat ,15, 7) + ')' , 4326)) as distance from geo order by distance Spatial Data in the Entity Framework Prior to Entity Framework 5.0 on .NET 4.5 consuming of the data above required using stored procedures or raw SQL commands to access the spatial data. In Entity Framework 5 however, Microsoft introduced the new DbGeometry and DbGeography types. These immutable location types provide a bunch of functionality for manipulating spatial points using geometry functions which in turn can be used to do common spatial queries like I described in the SQL syntax above. The DbGeography/DbGeometry types are immutable, meaning that you can't write to them once they've been created. They are a bit odd in that you need to use factory methods in order to instantiate them - they have no constructor() and you can't assign to properties like Latitude and Longitude. Creating a Model with Spatial Data Let's start by creating a simple Entity Framework model that includes a Location property of type DbGeography: public class GeoLocationContext : DbContext { public DbSet<GeoLocation> Locations { get; set; } } public class GeoLocation { public int Id { get; set; } public DbGeography Location { get; set; } public string Address { get; set; } } That's all there's to it. When you run this now against SQL Server, you get a Geography field for the Location property, which looks the same as the Location field in the SQL examples earlier. Adding Spatial Data to the Database Next let's add some data to the table that includes some latitude and longitude data. An easy way to find lat/long locations is to use Google Maps to pinpoint your location, then right click and click on What's Here. Click on the green marker to get the GPS coordinates. To add the actual geolocation data create an instance of the GeoLocation type and use the DbGeography.PointFromText() factory method to create a new point to assign to the Location property:[TestMethod] public void AddLocationsToDataBase() { var context = new GeoLocationContext(); // remove all context.Locations.ToList().ForEach( loc => context.Locations.Remove(loc)); context.SaveChanges(); var location = new GeoLocation() { // Create a point using native DbGeography Factory method Location = DbGeography.PointFromText( string.Format("POINT({0} {1})", -121.527200,45.712113) ,4326), Address = "301 15th Street, Hood River" }; context.Locations.Add(location); location = new GeoLocation() { Location = CreatePoint(45.714240, -121.517265), Address = "The Hatchery, Bingen" }; context.Locations.Add(location); location = new GeoLocation() { // Create a point using a helper function (lat/long) Location = CreatePoint(45.708457, -121.514432), Address = "Kaze Sushi, Hood River" }; context.Locations.Add(location); location = new GeoLocation() { Location = CreatePoint(45.722780, -120.209227), Address = "Arlington, OR" }; context.Locations.Add(location); context.SaveChanges(); } As promised, a DbGeography object has to be created with one of the static factory methods provided on the type as the Location.Longitude and Location.Latitude properties are read only. Here I'm using PointFromText() which uses a "Well Known Text" format to specify spatial data. In the first example I'm specifying to create a Point from a longitude and latitude value, using an SRID of 4326 (just like earlier in the SQL examples). You'll probably want to create a helper method to make the creation of Points easier to avoid that string format and instead just pass in a couple of double values. Here's my helper called CreatePoint that's used for all but the first point creation in the sample above:public static DbGeography CreatePoint(double latitude, double longitude) { var text = string.Format(CultureInfo.InvariantCulture.NumberFormat, "POINT({0} {1})", longitude, latitude); // 4326 is most common coordinate system used by GPS/Maps return DbGeography.PointFromText(text, 4326); } Using the helper the syntax becomes a bit cleaner, requiring only a latitude and longitude respectively. Note that my method intentionally swaps the parameters around because Latitude and Longitude is the common format I've seen with mapping libraries (especially Google Mapping/Geolocation APIs with their LatLng type). When the context is changed the data is written into the database using the SQL Geography type which looks the same as in the earlier SQL examples shown. Querying Once you have some location data in the database it's now super easy to query the data and find out the distance between locations. A common query is to ask for a number of locations that are near a fixed point - typically your current location and order it by distance. Using LINQ to Entities a query like this is easy to construct:[TestMethod] public void QueryLocationsTest() { var sourcePoint = CreatePoint(45.712113, -121.527200); var context = new GeoLocationContext(); // find any locations within 5 kilometers ordered by distance var matches = context.Locations .Where(loc => loc.Location.Distance(sourcePoint) < 5000) .OrderBy( loc=> loc.Location.Distance(sourcePoint) ) .Select( loc=> new { Address = loc.Address, Distance = loc.Location.Distance(sourcePoint) }); Assert.IsTrue(matches.Count() > 0); foreach (var location in matches) { Console.WriteLine("{0} ({1:n0} meters)", location.Address, location.Distance); } } This example produces: 301 15th Street, Hood River (0 meters)The Hatchery, Bingen (809 meters)Kaze Sushi, Hood River (1,074 meters)   The first point in the database is the same as my source point I'm comparing against so the distance is 0. The other two are within the 5 mile radius, while the Arlington location which is 65 miles or so out is not returned. The result is ordered by distance from closest to furthest away. In the code, I first create a source point that is the basis for comparison. The LINQ query then selects all locations that are within 5km of the source point using the Location.Distance() function, which takes a source point as a parameter. You can either use a pre-defined value as I'm doing here, or compare against another database DbGeography property (say when you have to points in the same database for things like routes). What's nice about this query syntax is that it's very clean and easy to read and understand. You can calculate the distance and also easily order by the distance to provide a result that shows locations from closest to furthest away which is a common scenario for any application that places a user in the context of several locations. It's now super easy to accomplish this. Meters vs. Miles As with the SQL Server functions, the Distance() method returns data in meters, so if you need to work with miles or feet you need to do some conversion. Here are a couple of helpers that might be useful (can be found in GeoUtils.cs of the sample project):/// <summary> /// Convert meters to miles /// </summary> /// <param name="meters"></param> /// <returns></returns> public static double MetersToMiles(double? meters) { if (meters == null) return 0F; return meters.Value * 0.000621371192; } /// <summary> /// Convert miles to meters /// </summary> /// <param name="miles"></param> /// <returns></returns> public static double MilesToMeters(double? miles) { if (miles == null) return 0; return miles.Value * 1609.344; } Using these two helpers you can query on miles like this:[TestMethod] public void QueryLocationsMilesTest() { var sourcePoint = CreatePoint(45.712113, -121.527200); var context = new GeoLocationContext(); // find any locations within 5 miles ordered by distance var fiveMiles = GeoUtils.MilesToMeters(5); var matches = context.Locations .Where(loc => loc.Location.Distance(sourcePoint) <= fiveMiles) .OrderBy(loc => loc.Location.Distance(sourcePoint)) .Select(loc => new { Address = loc.Address, Distance = loc.Location.Distance(sourcePoint) }); Assert.IsTrue(matches.Count() > 0); foreach (var location in matches) { Console.WriteLine("{0} ({1:n1} miles)", location.Address, GeoUtils.MetersToMiles(location.Distance)); } } which produces: 301 15th Street, Hood River (0.0 miles)The Hatchery, Bingen (0.5 miles)Kaze Sushi, Hood River (0.7 miles) Nice 'n simple. .NET 4.5 Only Note that DbGeography and DbGeometry are exclusive to Entity Framework 5.0 (not 4.4 which ships in the same NuGet package or installer) and requires .NET 4.5. That's because the new DbGeometry and DbGeography (and related) types are defined in the 4.5 version of System.Data.Entity which is a CLR assembly and is only updated by major versions of .NET. Why this decision was made to add these types to System.Data.Entity rather than to the frequently updated EntityFramework assembly that would have possibly made this work in .NET 4.0 is beyond me, especially given that there are no native .NET framework spatial types to begin with. I find it also odd that there is no native CLR spatial type. The DbGeography and DbGeometry types are specific to Entity Framework and live on those assemblies. They will also work for general purpose, non-database spatial data manipulation, but then you are forced into having a dependency on System.Data.Entity, which seems a bit silly. There's also a System.Spatial assembly that's apparently part of WCF Data Services which in turn don't work with Entity framework. Another example of multiple teams at Microsoft not communicating and implementing the same functionality (differently) in several different places. Perplexed as a I may be, for EF specific code the Entity framework specific types are easy to use and work well. Working with pre-.NET 4.5 Entity Framework and Spatial Data If you can't go to .NET 4.5 just yet you can also still use spatial features in Entity Framework, but it's a lot more work as you can't use the DbContext directly to manipulate the location data. You can still run raw SQL statements to write data into the database and retrieve results using the same TSQL syntax I showed earlier using Context.Database.ExecuteSqlCommand(). Here's code that you can use to add location data into the database:[TestMethod] public void RawSqlEfAddTest() { string sqlFormat = @"insert into GeoLocations( Location, Address) values ( geography::STGeomFromText('POINT({0} {1})', 4326),@p0 )"; var sql = string.Format(sqlFormat,-121.527200, 45.712113); Console.WriteLine(sql); var context = new GeoLocationContext(); Assert.IsTrue(context.Database.ExecuteSqlCommand(sql,"301 N. 15th Street") > 0); } Here I'm using the STGeomFromText() function to add the location data. Note that I'm using string.Format here, which usually would be a bad practice but is required here. I was unable to use ExecuteSqlCommand() and its named parameter syntax as the longitude and latitude parameters are embedded into a string. Rest assured it's required as the following does not work:string sqlFormat = @"insert into GeoLocations( Location, Address) values ( geography::STGeomFromText('POINT(@p0 @p1)', 4326),@p2 )";context.Database.ExecuteSqlCommand(sql, -121.527200, 45.712113, "301 N. 15th Street") Explicitly assigning the point value with string.format works however. There are a number of ways to query location data. You can't get the location data directly, but you can retrieve the point string (which can then be parsed to get Latitude and Longitude) and you can return calculated values like distance. Here's an example of how to retrieve some geo data into a resultset using EF's and SqlQuery method:[TestMethod] public void RawSqlEfQueryTest() { var sqlFormat = @" DECLARE @s geography SET @s = geography:: STGeomFromText('POINT({0} {1})' , 4326); SELECT Address, Location.ToString() as GeoString, @s.STDistance( Location) as Distance FROM GeoLocations ORDER BY Distance"; var sql = string.Format(sqlFormat, -121.527200, 45.712113); var context = new GeoLocationContext(); var locations = context.Database.SqlQuery<ResultData>(sql); Assert.IsTrue(locations.Count() > 0); foreach (var location in locations) { Console.WriteLine(location.Address + " " + location.GeoString + " " + location.Distance); } } public class ResultData { public string GeoString { get; set; } public double Distance { get; set; } public string Address { get; set; } } Hopefully you don't have to resort to this approach as it's fairly limited. Using the new DbGeography/DbGeometry types makes this sort of thing so much easier. When I had to use code like this before I typically ended up retrieving data pks only and then running another query with just the PKs to retrieve the actual underlying DbContext entities. This was very inefficient and tedious but it did work. Summary For the current project I'm working on we actually made the switch to .NET 4.5 purely for the spatial features in EF 5.0. This app heavily relies on spatial queries and it was worth taking a chance with pre-release code to get this ease of integration as opposed to manually falling back to stored procedures or raw SQL string queries to return spatial specific queries. Using native Entity Framework code makes life a lot easier than the alternatives. It might be a late addition to Entity Framework, but it sure makes location calculations and storage easy. Where do you want to go today? ;-) Resources Download Sample Project© Rick Strahl, West Wind Technologies, 2005-2012Posted in ADO.NET  Sql Server  .NET   Tweet !function(d,s,id){var js,fjs=d.getElementsByTagName(s)[0];if(!d.getElementById(id)){js=d.createElement(s);js.id=id;js.src="//platform.twitter.com/widgets.js";fjs.parentNode.insertBefore(js,fjs);}}(document,"script","twitter-wjs"); (function() { var po = document.createElement('script'); po.type = 'text/javascript'; po.async = true; po.src = 'https://apis.google.com/js/plusone.js'; var s = document.getElementsByTagName('script')[0]; s.parentNode.insertBefore(po, s); })();

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  • Create a UPnP Control Point in Silverlight

    - by James Cadd
    I'm interested in creating a UPnP control point in Silverlight. To me that implies that I'll need to use COM interop with Microsoft's upnp.dll (and the SL application will have to be out of browser and platform specific). Is there any source code available in C# that shows how to create a control point with Microsoft's stack? If there are other options available I'd like to hear about it, from reading around the web it appears the MS stack is buggy.

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  • Set registration point of a MovieClip to its center in AS3

    - by Mirko
    Can I set the registration point of a MovieClip (or other Display Object) to its center upon creation in AS3? the following var myClip:MovieClip = new MovieClip(); sets the registration point of myClip to its top left corner by default. Using Flash CS4 to set it to its center is just a couple of clicks, so I am wondering how I can perform the same action only with code.

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  • Rotate point in rectangle

    - by Dested
    I have a point in a rectangle that I need to rotate an arbitrary degree and find the x y of the point. How can I do this using javascript. Below the x,y would be something like 1,3 and after I pass 90 into the method it will return 3,1. |-------------| | * | | | | | |-------------| _____ | *| | | | | | | | | _____ |-------------| | | | | | *| |-------------| _____ | | | | | | | | |* | _____ Basically I am looking for the guts to this method function Rotate(pointX,pointY,rectWidth,rectHeight,angle){ /*magic*/ return {newX:x,newY:y}; }

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  • Point driven forum

    - by Dofs
    Hi, A while a go I saw a good post on SO about a user who wanted to create a site using a point system similar to SO, but I havn't been able to find it since. Can anyone direct me to a good article or post about creating point driven sites/forums?

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  • How to convert a string into a Point?

    - by NateD
    I have a list of strings of the format "x,y". I would like to make them all into Points. The best Point constructor I can find takes two ints. What is the best way in C# to turn "14,42" into new Point(14,42);? I know the Regex for doing that is /(\d+),(\d+)/, but I'm having a hard time turning those two match groups into ints in C#.

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  • Get the path of the .dmg from the mount point

    - by wbg
    I'm looking for a way to get the .dmg path of a mounted disk image with just its mount point. I want to write a "simple" Finder service that ejects the disk image and trashes the accompanying .dmg. The ejecting is trivial, but I'm at a loss as to how to figure out the path of the .dmg, given just the mount point. diskutil doesn't seem to know or isn't saying. It's for a script, so AppleScript- or shell-based suggestions are preferred.

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  • How to convert a string into a Point in C#

    - by NateD
    I have a list of strings of the format "x,y". I would like to make them all into Points. The best Point constructor I can find takes two ints. What is the best way in C# to turn "14,42" into new Point(14,42);? I know the Regex for doing that is /(\d+),(\d+)/, but I'm having a hard time turning those two match groups into ints in C#. any help you could offer would be appreciated.

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  • Calculate pixels within a polygon

    - by DoomStone
    In an assignment for school do we need to do some image recognizing, where we have to find a path for a robot. So far have we been able to find all the polygons in the image, but now we need to generate a pixel map, that be used for an astar algorithm later. We have found a way to do this, show below, but the problem is that is very slow, as we go though each pixel and test if it is inside the polygon. So my question is, are there a way that we can generate this pixel map faster? We have a list of cordinates for the polygon private List<IntPoint> hull; The fuction "getMap" is called to get the pixel map public Point[] getMap() { List<Point> points = new List<Point>(); lock (hull) { Rectangle rect = getRectangle(); for (int x = rect.X; x <= rect.X + rect.Width; x++) { for (int y = rect.Y; y <= rect.Y + rect.Height; y++) { if (inPoly(x, y)) points.Add(new Point(x, y)); } } } return points.ToArray(); } Get Rectangle is used to limit the search, se we don't have to go thoug the whole image public Rectangle getRectangle() { int x = -1, y = -1, width = -1, height = -1; foreach (IntPoint item in hull) { if (item.X < x || x == -1) x = item.X; if (item.Y < y || y == -1) y = item.Y; if (item.X > width || width == -1) width = item.X; if (item.Y > height || height == -1) height = item.Y; } return new Rectangle(x, y, width-x, height-y); } And atlast this is how we check to see if a pixel is inside the polygon public bool inPoly(int x, int y) { int i, j = hull.Count - 1; bool oddNodes = false; for (i = 0; i < hull.Count; i++) { if (hull[i].Y < y && hull[j].Y >= y || hull[j].Y < y && hull[i].Y >= y) { try { if (hull[i].X + (y - hull[i].X) / (hull[j].X - hull[i].X) * (hull[j].X - hull[i].X) < x) { oddNodes = !oddNodes; } } catch (DivideByZeroException e) { if (0 < x) { oddNodes = !oddNodes; } } } j = i; } return oddNodes; }

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