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  • WPF MediaElement source in C# 2010

    - by Alex Farber
    The sample from the book is compiled and executed successfully in VS2008. Project contains file Bear.wmw in the project directory. XAML: MediaElement Source="Bear.wmv" Build Action = Content, Copy to output directory = Copy always In C# 2008 this file is shown in the window. In C# 2010 Express file is not shown. Output directory contains this file. How can this be fixed?

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  • Set caret position in contenteditable div layer in Chrome/webkit

    - by Alex
    Hello, I'm trying to set the caret position in a contenteditable div layer, and after a bit of searching the web and experimenting I got it to work in firefox using this: function set(element,position){ element.focus(); var range= window.getSelection().getRangeAt(0); range.setStart(element.firstChild,position); range.setEnd(element.firstChild,position); } [...] set(document.getElementById("test"),3); But in Chrome/webkit it selects all of the content in the div. Is this a bug with webkit or am I doing something wrong? Thank you in advance.

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  • WPF zoom canvas and maintain scroll position

    - by Alex McBride
    I have a Canvas element, contained within a ScrollViewer, which I'm zooming using ScaleTransform. However, I want to be able to keep the scroll position of the viewer focused on the same part of the canvas after the zoom operation has finished. Currently when I zoom the canvas the scroll position of the viewer stays where it was and the place the user was viewing is lost. I'm still learning WPF, and I've been going backwards and forwards a bit on this, but I can't figure out a nice XAML based way to accomplish what I want. Any help in this matter would be greatly appreciated and would aid me in my learning process. Here is the kind of code I'm using... <Grid> <ScrollViewer Name="TrackScrollViewer" HorizontalScrollBarVisibility="Auto" VerticalScrollBarVisibility="Auto"> <Canvas Width="2560" Height="2560" Name="TrackCanvas"> <Canvas.LayoutTransform> <ScaleTransform ScaleX="{Binding ElementName=ZoomSlider, Path=Value}" ScaleY="{Binding ElementName=ZoomSlider, Path=Value}"/> </Canvas.LayoutTransform> <!-- Some complex geometry describing a motor racing circuit --> </Canvas> </ScrollViewer> <StackPanel Orientation="Horizontal" Margin="8" VerticalAlignment="Top" HorizontalAlignment="Left"> <Slider Name="ZoomSlider" Width="80" Minimum="0.1" Maximum="10" Value="1"/> <TextBlock Margin="4,0,0,0" VerticalAlignment="Center" Text="{Binding ElementName=ZoomSlider, Path=Value, StringFormat=F1}"/> </StackPanel> </Grid>

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  • How to manually check a YUI radio "button".

    - by alex
    <script type="text/javascript"> (function () { var ButtonGroup = YAHOO.widget.ButtonGroup; var onCheckedButtonChange = function (p_oEvent) { }; YAHOO.util.Event.onContentReady("mediaFilterButtonsFieldset", function () { var oButtonGroup = new ButtonGroup("mediaFilterButtons"); oButtonGroup.on("checkedButtonChange", onCheckedButtonChange); }); }()); </script> <div id="resultInfo"> <form id="button-example-form" name="button-example-form" method="post"> <fieldset id="mediaFilterButtonsFieldset"> <div id="mediaFilterButtons" class="yui-buttongroup ie7filter" style="z-index:11;"> <div id="mediaFilterLabel">Go to</div> <input id="radio1" class="filter_but" type="radio" name="0" value="First" checked rel="0" > <input id="radio2" class="filter_but" type="radio" name="2" value="Second" rel="2"> <input id="radio3" class="filter_but" type="radio" name="1" value="Third" rel="1"> </div> </fieldset> </form> </div> These are my YUI buttons. They're just 3 radio buttons turned into "buttons"--literally. My question is this: After people click the third button, I cannot manually check the first button anymore. How can I manually check "radio1"? I tried JQuery: $("#radio1").attr("checked",true); But this didn't work. The third button still remained pressed down.

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  • c#Repository pattern: One repository per subclass?

    - by Alex
    I am wondering if you would create a repository for each subclass of a domain model. There are two classes for example: public class Person { public virtual String GivenName { set; get; } public virtual String FamilyName { set; get; } public virtual String EMailAdress { set; get; } } public class Customer : Person { public virtual DateTime RegistrationDate { get; set; } public virtual String Password { get; set; } } Would you create both a PersonRepository and a CustomerRepository or just the PersonRepository which would also be able to execute Customer related queries?

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  • VSS 2005 and VS 2005

    - by Alex
    I use VS 2005 and VSS 2005. Every time I close VS I get error: ss.ini not found. But except this VSS works fine, no problems when I open VS and do check in and check out. ss.ini is present and VSS repository specified as network path. I just worry that I can have problems later.

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  • How to change image and disable UIBarButtonItem

    - by Alex
    I have a NavigationBar app with two views: a parent and a sub view. In the sub view I'm adding a button to the right corner as follows: - (void)viewDidLoad { UIBarButtonItem *tempButton = [[UIBarButtonItem alloc] initWithImage:[UIImage imageNamed:@"lock-unlocked.png"] style:UIBarButtonItemStylePlain target:self action:@selector(lockScreen)]; self.navigationItem.rightBarButtonItem = tempButton; [tempButton release]; } When that button is clicked I want to change the image of this rightBarButtonItem and disable the leftBarButtonItem (which was added automatically by the controller). Basically have two states of a button, locked and unlocked. Question 1: The only way I can find how to change the image is to create a new UIButtonItem with a new image and replace rightBarButtonItem with that new one. But I'm wondering if there's a way to just change the image without creating a new UIBarButtonItem. Am I creating a memory leak if I keep creating new UIBarButtonItem? Question 2: How can I get a hold of self.navigationItem.leftBarButtonItem and disable/enable it? I don't create that one manually, it's created automatically for me by the controller. I don't see any method/property on UIBarButtonItem to enable/disable user interaction with it.

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  • SAT Thread and Process output capture in c#

    - by alex
    Hi: This is a strange problem I encountered. I have an window application written in c# to do testing. It has a MDI parent form that is hosting a few children forms. One of the forms launch test cripts by creating processes and capture the scripts output to a text box. Another form open serial port and monitoring the status of the device I am working on(like a shell). If I ran both of them together, the output of the script seems only appear in the text box after the test is done. However, If I don't open the serial port form, the output of the script is captured in real time. Does anyone knows what's causing the problem? I notice the onDataReceived even handler for serial port form has a [SAThread] header to it. Will this cause the serial port thread having higher priority than other processes? Thanks in advance.

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  • Maven freezes while perfoming scm:update on VSS

    - by Alex
    Im trying to get source files from VSS using Maven. But when I execute command mvn -DvssDirectory="C:\Program Files\VisualSourceSafe\win32" scm:update the log goes to [INFO] [scm:update {execution: default-cli}] [INFO] Executing: cmd.exe /X /C ""C:\Program Files\VisualSourceSafe\win32\ss" Get $/TEST -R -I- -GWS" [INFO] Working directory: C:\temp\test and then nothing happens. No error, no success. Can someone advice how to proceed with downloading sources from VSS?

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  • SQLite file locking and DropBox

    - by Alex Jenter
    I'm developing an app in Visual C++ that uses an SQLite3 DB for storing data. Usually it sits in the tray most of the time. I also would like to enable putting my app in a DropBox folder to share it across several PCs. It worked really well up until DropBox has recently updated itself. And now it says that it "can't sync the file in use". The SQLite file is open in my app, but the lock is shared. There are some prepared statements, but all are reset immediately after using step. Is there any way to enable synchronizing of an open SQLite database file? Thanks! Here is the simple wrapper that I use just for testing (no error handling), in case this helps: class Statement { private: Statement(sqlite3* db, const std::wstring& sql) : db(db) { sqlite3_prepare16_v2(db, sql.c_str(), sql.length() * sizeof(wchar_t), &stmt, NULL); } public: ~Statement() { sqlite3_finalize(stmt); } public: void reset() { sqlite3_reset(stmt); } int step() { return sqlite3_step(stmt); } int getInt(int i) const { return sqlite3_column_int(stmt, i); } tstring getText(int i) const { const wchar_t* v = (const wchar_t*)sqlite3_column_text16(stmt, i); int sz = sqlite3_column_bytes16(stmt, i) / sizeof(wchar_t); return std::wstring(v, v + sz); } private: friend class Database; sqlite3* db; sqlite3_stmt* stmt; }; class Database { public: Database(const std::wstring& filename = L"")) : db(NULL) { sqlite3_open16(filename.c_str(), &db); } ~Database() { sqlite3_close(db); } void exec(const std::wstring& sql) { auto_ptr<Statement> st(prepare(sql)); st->step(); } auto_ptr<Statement> prepare(const tstring& sql) const { return auto_ptr<Statement>(new Statement(db, sql)); } private: sqlite3* db; };

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  • Intelligencia URL ReWriter mapping with regex

    - by alex
    I am using the Intelligencia URL rewriter in my asp.net web application. I use the web.config mappings I'm trying to map the following url: www.mydomain.com/product-deals/manufacturer-model_PRODUCTId.aspx To: www.mydomain.com/ProductInfo.aspx?productID=xxx obviously in the above example, xxx is replaced from the "productId" from the "friendly" url. In my web.config, I've got so far: <rewrite url="~/contract-deals/([\w-_]+)/_(.+).aspx" to="~/ProductInfo.aspx?productId=$1"/> This isn't working however. I need the correct regex to use for my requirements (regex really isn't my strong point!!)

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  • object of type 'closure' is not subsettable - contradiction?

    - by Alex
    I'm writing a function to produce time series plots of stock prices. However, I'm getting the following error "Error in df[, 7] : object of type 'closure' is not subsettable" Here's an example of the function: plot.prices <- function(df) { require(ggplot2) g <- ggplot(df, aes(x= as.Date(Date, format= "%Y-%m-%d"), y= df[, 7])) + geom_point(size=1) # ... code not shown... g } And example data: spy <- read.csv(file= 'http://ichart.finance.yahoo.com/table.csv?s=SPY&d=11&e=1&f=2012&g=d&a=0&b=29&c=1993&ignore=.csv', header= T) plot.prices(spy) # produces error ggplot(spy, aes(x= as.Date(Date, format= "%Y-%m-%d"), y= spy[, 7])) + geom_point(size=1) ## does not produce error As you can see, the code is identical. I get an error if the call to ggplot() is INSIDE the function but not if the call to ggplot() is OUTSIDE the function. Anyone have any idea why the seeming contradiction?

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  • jax-ws on glassfish3 init method

    - by Alex
    Hi all, I've created simple jax-ws (anotated Java 6 class to web service) service and deploied it on glassfish v3. The web.xml <?xml version="1.0" encoding="ISO-8859-1"?> <web-app> <servlet> <servlet-name>MyServiceName</servlet-name> <description>Blablabla</description> <servlet-class>com.foo-bar.somepackage.TheService</servlet-class> <load-on-startup>1</load-on-startup> </servlet> <servlet-mapping> <servlet-name>MyServiceName</servlet-name> <url-pattern>/MyServiceName</url-pattern> </servlet-mapping> <session-config> <session-timeout>30</session-timeout> </session-config> </web-app> There is no sun-jaxws.xml in the war. The service works fine but I have 2 issues: I'm using apache common configuration package to read my configuration, so i have init function that calls configuration stuff. 1. How can I configure init method for jaxws service (like i can do for the servlets for example) 2. the load on startup parameter is not affecting the service, I see that for every request init function called again (and c-tor). How can I set scope for my service? Thanks a lot,

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  • How do you construct an array suitable for numpy sorting?

    - by Alex
    I need to sort two arrays simultaneously, or rather I need to sort one of the arrays and bring the corresponding element of its associated array with it as I sort. That is if the array is [(5, 33), (4, 44), (3, 55)] and I sort by the first axis (labeled below dtype='alpha') then I want: [(3.0, 55.0) (4.0, 44.0) (5.0, 33.0)]. These are really big data sets and I need to sort first ( for nlog(n) speed ) before I do some other operations. I don't know how to merge my two separate arrays though in the proper manner to get the sort algorithm working. I think my problem is rather simple. I tried three different methods: import numpy x=numpy.asarray([5,4,3]) y=numpy.asarray([33,44,55]) dtype=[('alpha',float), ('beta',float)] values=numpy.array([(x),(y)]) values=numpy.rollaxis(values,1) #values = numpy.array(values, dtype=dtype) #a=numpy.array(values,dtype=dtype) #q=numpy.sort(a,order='alpha') print "Try 1:\n", values values=numpy.empty((len(x),2)) for n in range (len(x)): values[n][0]=y[n] values[n][1]=x[n] print "Try 2:\n", values #values = numpy.array(values, dtype=dtype) #a=numpy.array(values,dtype=dtype) #q=numpy.sort(a,order='alpha') ### values = [(x[0], y[0]), (x[1],y[1]) , (x[2],y[2])] print "Try 3:\n", values values = numpy.array(values, dtype=dtype) a=numpy.array(values,dtype=dtype) q=numpy.sort(a,order='alpha') print "Result:\n",q I commented out the first and second trys because they create errors, I knew the third one would work because that was mirroring what I saw when I was RTFM. Given the arrays x and y (which are very large, just examples shown) how do I construct the array (called values) that can be called by numpy.sort properly? *** Zip works great, thanks. Bonus question: How can I later unzip the sorted data into two arrays again?

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  • Help with a compiler warning: Initialization from distinct Objective-C type when types match

    - by Alex Gosselin
    Here is the function where I get the compiler warning, I can't seem to figure out what is causing it. Any help is appreciated. -(void)displaySelector{ //warning on the following line: InstanceSelectorViewController *controller = [[InstanceSelectorViewController alloc] initWithCreator:self]; [self.navController pushViewController:controller animated:YES]; [controller release]; } Interface and implementation for the initWithCreator: method -(InstanceSelectorViewController*)initWithCreator:(InstanceCreator*)creator; -(InstanceSelectorViewController*)initWithCreator:(InstanceCreator*)crt{ if (self = [self initWithNibName:@"InstanceSelectorViewController" bundle:nil]) { creator = crt; } return self; }

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  • Can Spring.Net function as PostSharp?

    - by Alex K
    A few months back I've discovered PostSharp, and for a time, it was good. But then legal came back with an answer saying that they don't like the licence of the old versions. Then the department told me that 2.0's price was unacceptably high (for the number of seats we need)... I was extremely disapponted, but not disheartened. Can't be the only such framework, I thought. I kept looking for a replacement, but most of it was either dead, ill maintained (especially in documentation department), for academic use, or all of the above (I'm looking at you Aspect.Net) Then I've discovered Spring.Net, and for a time, it was good. I've been reading the documentation and it went on to paint what seemed to be a supperior picture of an AOP nirvana. No longer was I locked to attributes to mark where I wanted code interception to take place, but it could be configured in XML and changes to it didn't require re-compile. Great. Then I looked at the samples and saw the following, in every single usage scenario: // Create AOP proxy using Spring.NET IoC container. IApplicationContext ctx = ContextRegistry.GetContext(); ICommand command = (ICommand)ctx["myServiceCommand"]; command.Execute(); if (command.IsUndoCapable) { command.UnExecute(); } Why must the first two lines of code exist? It ruins everything. This means I cannot simply provide a user with a set of aspects and attributes or XML configs that they can use by sticking appropriate attributes on the appropriate methods/classes/etc or editing the match pattern in XML. They have to modify their program logic to make this work! Is there a way to make Spring.Net behave as PostSharp in this case? (i.e. user only needs to add attributes/XML config, not edit content of any methods. Alternatively, is there a worthy and functioning alternative to PostSharp? I've seen a few question titled like this on SO, but none of them were actually looking to replace PostSharp, they only wanted to supplement its functionality. I need full replacement.

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  • WPF TextBox SpellChecking Problem

    - by Alex
    How can I change the spellchecking language of a WPF textbox to french using XAML? I tried this but it doesn't work. <TextBox AcceptsReturn="true" SpellCheck.IsEnabled="true" FontSize="12" BorderBrush="Blue" Height="100" Language="fr-fr" /> French is supposed to be one of the 4 supported languages for spellchecking in WPF so I don't understand why it doesn't work. I also tried fr-CA but it still won't work.

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  • How do you pass user credentials from WebClient to a WCF REST service?

    - by Alex
    I am trying to expose a WCT REST service and only users with valid username and password would be able to access it. The username and password are stored in a SQL database. Here is the service contract: public interface IDataService { [OperationContract] [WebGet(ResponseFormat = WebMessageFormat.Json)] byte[] GetData(double startTime, double endTime); } Here is the WCF configuration: <bindings> <webHttpBinding> <binding name="SecureBinding"> <security mode="Transport"> <transport clientCredentialType="Basic"/> </security> </binding> </webHttpBinding> </bindings> <behaviors> <serviceBehaviors> <behavior name="DataServiceBehavior"> <serviceMetadata httpGetEnabled="true"/> <serviceCredentials> <userNameAuthentication userNamePasswordValidationMode="Custom" customUserNamePasswordValidatorType= "CustomValidator, WCFHost" /> </serviceCredentials> </behavior> </serviceBehaviors> </behaviors> <services> <service behaviorConfiguration="DataServiceBehavior" name="DataService"> <endpoint address="" binding="webHttpBinding" bindingConfiguration="SecureBinding" contract="IDataService" /> </service> </services> I am accessing the service via the WebClient class within a Silverlight application. However, I have not been able to figure out how to pass the user credentials to the service. I tried various values for client.Credentials but none of them seems to trigger the code in my custom validator. I am getting the following error: The underlying connection was closed: An unexpected error occurred on a send. Here is some sample code I have tried: WebClient client = new WebClient(); client.Credentials = new NetworkCredential("name", "password", "domain"); client.OpenReadCompleted += new OpenReadCompletedEventHandler(GetData); client.OpenReadAsync(new Uri(uriString)); If I set the security mode to None, the whole thing works. I also tried other clientCredentialType values and none of them worked. I also self-hosted the WCF service to eliminate the issues related to IIS trying to authenticate a user before the service gets a chance. Any comment on what the underlying issues may be would be much appreciated. Thanks. Update: Thanks to Mehmet's excellent suggestions. Here is the tracing configuration I had: <system.diagnostics> <sources> <source name="System.ServiceModel" switchValue="Information, ActivityTracing" propagateActivity="true"> <listeners> <add name="xml" /> </listeners> </source> <source name="System.IdentityModel" switchValue="Information, ActivityTracing" propagateActivity="true"> <listeners> <add name="xml" /> </listeners> </source> </sources> <sharedListeners> <add name="xml" type="System.Diagnostics.XmlWriterTraceListener" initializeData="c:\Traces.svclog" /> </sharedListeners> </system.diagnostics> But I did not see any message coming from my Silverlight client. As for https vs http, I used https as follows: string baseAddress = "https://localhost:6600/"; _webServiceHost = new WebServiceHost(typeof(DataServices), new Uri(baseAddress)); _webServiceHost.Open(); However, I did not configure any SSL certificate. Is this the problem?

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  • [Ruby] [gem] A GEM error shown during run the commend: gem update --system

    - by Alex
    I’m a freshman on Ruby and now trying to install ruby on my machine according to the Tutorial on http://wiki.openqa.org/display/WTR/Tutorial However, after I installed the ruby186-26, and run the command “gem update --system”, the following error occurred: C:\Documents and Settings\e482090\Desktopgem update --system c:/ruby/lib/ruby/site_ruby/1.8/rubygems/config_file.rb:51:in initialize': Inval id argument - <Not Set>/.gemrc (Errno::EINVAL) from c:/ruby/lib/ruby/site_ruby/1.8/rubygems/config_file.rb:51:inopen' from c:/ruby/lib/ruby/site_ruby/1.8/rubygems/config_file.rb:51:in initi alize' from c:/ruby/lib/ruby/site_ruby/1.8/rubygems/gem_runner.rb:36:innew' from c:/ruby/lib/ruby/site_ruby/1.8/rubygems/gem_runner.rb:36:in do_con figuration' from c:/ruby/lib/ruby/site_ruby/1.8/rubygems/gem_runner.rb:25:inrun' from c:/ruby/bin/gem:23 C:\Documents and Settings\e482090\Desktopgem install watir c:/ruby/lib/ruby/site_ruby/1.8/rubygems/config_file.rb:51:in initialize': Inval id argument - <Not Set>/.gemrc (Errno::EINVAL) from c:/ruby/lib/ruby/site_ruby/1.8/rubygems/config_file.rb:51:inopen' from c:/ruby/lib/ruby/site_ruby/1.8/rubygems/config_file.rb:51:in initi alize' from c:/ruby/lib/ruby/site_ruby/1.8/rubygems/gem_runner.rb:36:innew' from c:/ruby/lib/ruby/site_ruby/1.8/rubygems/gem_runner.rb:36:in do_con figuration' from c:/ruby/lib/ruby/site_ruby/1.8/rubygems/gem_runner.rb:25:inrun' from c:/ruby/bin/gem:23 Meanwhile, we have tried this on other machines and the result turned out ok. Thus, my question is why the error happened on my pc? Have you met this kind of error before?

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  • Why is it so difficult to get a working IDE for Scala?

    - by Alex R
    I recently gave up trying to use Scala in Eclipse (basic stuff like completion doesn't work). So now I'm trying IntelliJ. I'm not getting very far. This was the original error. See below for update: Scala signature Predef has wrong version Expected 5.0 found: 4.1 in .... scala-library.jar I tried both versions 2.7.6 and 2.8 RC1 of scala-*.jar, the result was the same. JDK is 1.6.u20. UPDATE Today I uninstalled IntelliJ 9.0.1, and installed 9.0.2 Early Availability, with the 4/14 stable version of the Scala plug-in. Then I setup a project from scratch through the wizards: new project from scratch JDK is 1.6.u20 accept the default (project) instead of global / module accept the download of Scala 2.8.0beta1 into project's lib folder Created a new class: object hello { def main(args: Array[String]) { println("hello: " + args); } } For my efforts, I now have a brand-new error :) Here it is: Scalac internal error: class java.lang.ClassNotFoundException [java.net.URLClassLoader$1.run(URLClassLoader.java:202), java.security.AccessController.doPrivileged(Native Method), java.net.URLClassLoader.findClass(URLClassLoader.java:190), java.lang.ClassLoader.loadClass(ClassLoader.java:307), sun.misc.Launcher$AppClassLoader.loadClass(Launcher.java:301), java.lang.ClassLoader.loadClass(ClassLoader.java:248), java.lang.Class.forName0(Native Method), java.lang.Class.forName(Class.java:169), org.jetbrains.plugins.scala.compiler.rt.ScalacRunner.main(ScalacRunner.java:72)] Thanks

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  • "Launch Failed. Binary Not Found." Snow Leopard and Eclipse C/C++ IDE issue.

    - by Alex
    Not a question, I've just scoured the internet in search of a solution for this problem and thought I'd share it with the good folks of SO. I'll put it in plain terms so that it's accessible to newbs. :) (Apologies if this is the wrong place -- just trying to be helpful.) This issue occurs with almost any user OS X Snow Leopard who tries to use the Eclipse C/C++ IDE, but is particularly annoying for the people (like me) who were using the Eclipse C/C++ IDE in Leopard, and were unable to work with Eclipse anymore when they upgraded. The issue occurs When users go to build/compile/link their software. They get the following error: Launch Failed. Binary Not Found. Further, the "binaries" branch in the project window on the left is simply nonexistent. THE PROBLEM: is that GCC 4.2 (the GNU Compiler Collection) that comes with Snow Leopard compiles binaries in 64-bit by default. Unfortunately, the linker that Eclipse uses does not understand 64-bit binaries; it reads 32-bit binaries. There may be other issues here, but in short, they culminate in no binary being generated, at least not one that Eclipse can read, which translates into Eclipse not finding the binaries. Hence the error. One solution is to add an -arch i686 flag when making the file, but manually making the file every time is annoying. Luckily for us, Snow Leopard also comes with GCC 4.0, which compiles in 32 bits by default. So one solution is merely to link this as the default compiler. This is the way I did it. THE SOLUTION: The GCCs are in /usr/bin, which is normally a hidden folder, so you can't see it in the Finder unless you explicitly tell the system that you want to see hidden folders. Anyway, what you want to do is go to the /usr/bin folder and delete the path that links the GCC command with GCC 4.2 and add a path that links the GCC command with GCC 4.0. In other words, when you or Eclipse try to access GCC, we want the command to go to the compiler that builds in 32 bits by default, so that the linker can read the files; we do not want it to go to the compiler that compiles in 64 bits. The best way to do this is to go to Applications/Utilities, and select the app called Terminal. A text prompt should come up. It should say something like "(Computer Name):~ (Username)$ " (with a space for you user input at the end). The way to accomplish the tasks above is to enter the following commands, entering each one in sequence VERBATIM, and pressing enter after each individual line. cd /usr/bin rm cc gcc c++ g++ ln -s gcc-4.0 cc ln -s gcc-4.0 gcc ln -s c++-4.0 c++ ln -s g++-4.0 g++ Like me, you will probably get an error that tells you you don't have permission to access these files. If so, try the following commands instead: cd /usr/bin sudo rm cc gcc c++ g++ sudo ln -s gcc-4.0 cc sudo ln -s gcc-4.0 gcc sudo ln -s c++-4.0 c++ sudo ln -s g++-4.0 g++ Sudo may prompt you for a password. If you've never used sudo before, try just pressing enter. If that doesn't work, try the password for your main admin account. OTHER POSSIBLE SOLUTIONS You may be able to enter build variables into Eclipse. I tried this, but I don't know enough about it. If you want to feel it out, the flag you will probably need is -arch i686. In earnest, GCC-4.0 worked for me all this time, and I don't see any reason to switch now. There may be a way to alter the default for the compiler itself, but once again, I don't know enough about it. Hope this has been helpful and informative. Good coding!

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  • Is this so bad when using MySQL queries in PHP?

    - by alex
    I need to update a lot of rows, per a user request. It is a site with products. I could... Delete all old rows for that product, then loop through string building a new INSERT query. This however will lose all data if the INSERT fails. Perform an UPDATE through each loop. This loop currently iterates over 8 items, but in the future it may get up to 15. This many UPDATEs doesn't sound like too good an idea. Change DB Schema, and add an auto_increment Id to the rows. Then first do a SELECT, get all old rows ids in a variable, perform one INSERT, and then a DELETE WHERE IN SET. What is the usual practice here? Thanks

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