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  • OS X 10.6 Snow Leopard no longer mounting an external USB drive

    - by Brant Bobby
    I have a 1TB generic external hard drive containing a single HFS partition. I originally formatted this using Disk Utility and it worked fine. Now, for some reason, it's not auto-mounting when I start up. Using mount at the command line gives the following error: $ sudo mount /dev/disk1s2 /Volumes/Test /dev/disk1s2 on /Volumes/Test: Incorrect super block. ... but if I use the mount_hfs command it works fine, mounts, and is readable. $ mount_hfs /dev/disk1s2 /Volumes/Test/ fsck gives me an error about a bad super block: $ fsck /dev/disk1 ** /dev/rdisk1 (NO WRITE) BAD SUPER BLOCK: MAGIC NUMBER WRONG ... but fsck_hfs -fn /dev/disk1s2 doesn't find any problems and reports that the volume appears to be OK. In Disk Utility, the drive appears to have a single MS-DOS partition with a curious notice about how it appears to be partitioned for Boot Camp: I have the Boot Camp HFS driver installed in WIndows 7, and that OS sees the drive/partition normally. What's wrong with my disk?

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  • UriBuilder incorrectly encoding Query Parameters value ?

    - by Fred
    Lets consider the following code sample where a path and single parameter are encoded... Parameter name: "param" Parameter value: "foo/bar?aaa=bbb&ccc=ddd" (happens to be a url with query parameters) String test = UriBuilder.fromPath("https://dummy.com"). queryParam("param", "foo/bar?aaa=bbb&ccc=ddd"). build().toURL().toString(); The encoded URL string returned is: "https://dummy.com?param=foo/bar?aaa%3Dbbb&ccc%3Dddd" Is this correct ? Should not the character "&" (and may be even "?") be encoded in the parameter value string ? Would not the URL produced be interpreted as follow: One first parameter, name="param", value = "ar?aaa%3Dbbb" followed by a second parameter, name="ccc%3Dddd", without value.

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  • Rub, regex, sentences

    - by Perello
    I'm currently building a code generator, which aim to generate boiler plate for me once i wrote the templates and/or translations, whatever the language i have to work with, and it has an educationnal part :p. So i have a problem with a regex in ruby. The regex aim to select whatever is between {{{ and }}}, so i can generae functions according to my needs. My regex is currently : /\{\{\{(([a-zA-Z]|\s)+)\}\}\}/m My test data set {{{Demande aaa}}} = {{{tagadatsouintsouin tutu}}} The results are : [["Demande aaa", "a"], ["tagadatsouintsouin tutu", "u"]] So the regex pick each time the last character twice. But, that's not exactly what i want, my need is more about this : /\{\{\{((\w|\W)+)\}\}\}/m But this as a flaw too, the results : [["Demande aaa}}} = {{{tagadatsouintsouin tutu", "u"]] Whereas, i wish to get [["Demande aaa"],["tagadatsouintsouin tutu"]] Any ideas to correct theses regex ? I could use 2 sets of delimiters, but it won't learn me anything.

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  • Libgdx actor bounds are wrong

    - by Undume
    The Actor's boundaries are not centered at the ButtonText but I used the setBounds() method. The higher the Y position is, the less centered is the boundary. The weird thing is that i only created and added to the Stage one button but the screen shows two. When i click the top button, the bottom one is the one highlighted. How can i fix that? import com.badlogic.gdx.Game; import com.badlogic.gdx.Gdx; import com.badlogic.gdx.files.FileHandle; import com.badlogic.gdx.scenes.scene2d.Stage; import com.badlogic.gdx.scenes.scene2d.ui.Skin; import com.badlogic.gdx.scenes.scene2d.ui.TextButton; public class MyGame extends Game { Stage stage; @Override public void create() { stage=new Stage(); FileHandle skinFile = new FileHandle("data/resources/uiskin/uiskin.json"); Skin skin = new Skin(skinFile); TextButton sas=new TextButton("dd",skin); sas.setBounds(0, 500, 100, 100); stage.addActor(sas); Gdx.input.setInputProcessor(stage); } @Override public void dispose() { super.dispose(); } @Override public void render() { super.render(); stage.act(Gdx.graphics.getDeltaTime()); stage.draw(); } @Override public void resize(int width, int height) { super.resize(width, height); } @Override public void pause() { super.pause(); } @Override public void resume() { super.resume(); } }

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  • cross domain access in iframe from parent to child

    - by Aparna
    Hi, Could someone please help me with this- I've 2 applications AAA and BBB. The homepage of AAA contains an iFrame which displays the application BBB. When I login to AAA, the same login details should be used to login to BBB(single signon) and on pageload of the homepage of AAA, homepage of BBB should also be loaded in the iFrame. I tried to use javascript to access the form elements of login page of BBB to enter the login data and submit. But the browser gives me a 'Access is denied' error. I did a little reading and came to know that cross- domain communication is not allowed by the browser. Could someone tell me how i can go about achieving this?

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  • Compiz command plugin won't register keyboard shortcuts

    - by David Moles
    Per this discussion I've enabled the Compiz commands plugin in order to try to bind some keyboard shortcuts to wmctrl actions. CCSM captures my keystrokes just fine, but no matter what keystroke I try or what command I bind it to (everything from my original intention of binding Super-1, Super-2 etc. to wmctrl -o 0,0, wmctrl -o 2560,0, etc., to binding Ctrl-Alt-Shift-L to gnome-terminal). Basic compiz shortcuts for window switching and so on -- even custom ones -- seem to work fine, but the command plugin doesn't seem to be working at all. I also notice the following symptom: when I open the keyboard shortcut tab in CCSM, the keyboard shortcuts often at first appear blank, though if you click on the blank button, the correct value is still there. Also possibly related, I've noticed that gnome-terminal doesn't seem to notice the Super key, though other apps (e.g. CCSM, Emacs) register it fine. Anyway, it seems like something's eating my keystrokes. Any ideas?

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  • How to run some code only once in view

    - by Freewind
    I have a partial view called '_comment.erb', and it may be called by parent many times(e.g. in a loop). The '_comment.erb' looks like: <script> function aaa() {} </script> <%= comment.content %> <%=link_to_function 'Do', 'aaa()' %> You can see if the '_comment.erb' be called many times, that the javascript function 'aaa' will be re-defined many times. I hope it can be define only once, but I don't want to move it to parent view. I hope there is a method, say 'run_once', and I can use it like this: <%= run_once do %> <script> function aaa() {} </script> <% end %> <%= comment.content %> <%=link_to_function 'Do', 'aaa()' %> No matter how many time I call the '_comment.erb', the code inside 'run_once' will be run only once. What shall I do?

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  • Synergy on Mac OS X 10.7

    - by matt
    Unfortunately, I can't seem to get Synergy to work between two Mac clients both using 10.7. One is an older 21" iMac and the other is a new 2011 Macbook Air. Here's my synergy.conf, symlinked to my home directory as ~/.synergy.conf section: screens foo: bar: super = alt alt = super end section: links foo: right = bar bar: left = foo end That is, I thought the super = alt trick was mandatory in order to have alt work on Mac but unfortunately, nothing really works. Both the Command and Control keys do NOT work on bar but work fine on foo given that the keyboard is paired with that screen. The keyboard modifier keys are the same on both computers and the making the mouse go between screens works as well. I was wondering if anyone else had any success or problems running into this issue on 10.7 and was hoping there was a possible fix.

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  • Possible to load an Enum based on a string name?

    - by Cooter
    OK, I don't think the title says it right... but here goes: I have a class with about 40 Enums in it. i.e: Class Hoohoo { public enum aaa : short { a = 0, b = 3 } public enum bbb : short { a = 0, b = 3 } public enum ccc : short { a = 0, b = 3 } } Now say I have a Dictionary of strings and values, and each string is the name of above mentioned enums: Dictionary<string,short>{"aaa":0,"bbb":3,"ccc":0} I need to change "aaa" into HooBoo.aaa to look up 0. Can't seem to find a way to do this since the enum is static. Otherwise I'll have to write a method for each enum to tie the string to it. I can do that but thats mucho code to write. Thanks, Cooter

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  • Restore default keyboard shortcut for Workspace Switcher/Show Desktop

    - by To Do
    I tried setting the default keyboard shortcut to Hide normal windows (Show desktop) to Super + S. It didn't work and now whenever I press Super + S, I get the workspace switcher. I tried setting Hide normal windows back to Ctrl + Super + S, but it doesn't work. I'm still getting the Workspace switcher. How can I reset these two settings? I use the Show Desktop quite a lot and it is quite annoying not being able to do it.

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  • GROUP BY as a way to pick the first row from a group of similar rows, is this correct, is there any

    - by FipS
    I have a table which stores test results like this: user | score | time -----+-------+------ aaa | 90% | 10:30 bbb | 50% | 9:15 *** aaa | 85% | 10:15 aaa | 90% | 11:00 *** ... What I need is to get the top 10 users: user | score | time -----+-------+------ aaa | 90% | 11:00 bbb | 50% | 9:15 ... I've come up with the following SELECT: SELECT * FROM (SELECT user, score, time FROM tests_score ORDER BY user, score DESC, time DESC) t1 GROUP BY user ORDER BY score DESC, time LIMIT 10 It works fine but I'm not quite sure if my use of ORDER BY is the right way to pick the first row of each group of sorted records. Is there any better practice to achieve the same result? (I use MySQL 5)

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  • Passing Strings by Ref

    - by SGWellens
    Humbled yet again…DOH! No matter how much experience you acquire, no matter how smart you may be, no matter how hard you study, it is impossible to keep fully up to date on all the nuances of the technology we are exposed to. There will always be gaps in our knowledge: Little 'dead zones' of uncertainty. For me, this time, it was about passing string parameters to functions. I thought I knew this stuff cold. First, a little review... Value Types and Ref Integers and structs are value types (as opposed to reference types). When declared locally, their memory storage is on the stack; not on the heap. When passed to a function, the function gets a copy of the data and works on the copy. If a function needs to change a value type, you need to use the ref keyword.  Here's an example:     // ---- declaration -----------------     public struct MyStruct    {        public string StrTag;    }     // ---- functions -----------------------     void SetMyStruct(MyStruct myStruct)     // pass by value    {        myStruct.StrTag = "BBB";    }     void SetMyStruct(ref MyStruct myStruct)  // pass by ref    {        myStruct.StrTag = "CCC";    }     // ---- Usage -----------------------     protected void Button1_Click(object sender, EventArgs e)    {        MyStruct Data;        Data.StrTag = "AAA";         SetMyStruct(Data);        // Data.StrTag is still "AAA"         SetMyStruct(ref Data);        // Data.StrTag is now "CCC"    } No surprises here. All value types like ints, floats, datetimes, enums, structs, etc. work the same way. And now on to... Class Types and Ref     // ---- Declaration -----------------------------     public class MyClass    {        public string StrTag;    }     // ---- Functions ----------------------------     void SetMyClass(MyClass myClass)  // pass by 'value'    {        myClass.StrTag = "BBB";    }     void SetMyClass(ref MyClass myClass)   // pass by ref    {        myClass.StrTag = "CCC";    }     // ---- Usage ---------------------------------------     protected void Button2_Click(object sender, EventArgs e)    {        MyClass Data = new MyClass();        Data.StrTag = "AAA";         SetMyClass(Data);          // Data.StrTag is now "BBB"         SetMyClass(ref Data);        // Data.StrTag is now "CCC"    }  No surprises here either. Since Classes are reference types, you do not need the ref keyword to modify an object. What may seem a little strange is that with or without the ref keyword, the results are the same: The compiler knows what to do. So, why would you need to use the ref keyword when passing an object to a function? Because then you can change the reference itself…ie you can make it refer to a completely different object. Inside the function you can do: myClass = new MyClass() and the old object will be garbage collected and the new object will be returned to the caller. That ends the review. Now let's look at passing strings as parameters. The String Type and Ref Strings are reference types. So when you pass a String to a function, you do not need the ref keyword to change the string. Right? Wrong. Wrong, wrong, wrong. When I saw this, I was so surprised that I fell out of my chair. Getting up, I bumped my head on my desk (which really hurt). My bumping the desk caused a large speaker to fall off of a bookshelf and land squarely on my big toe. I was screaming in pain and hopping on one foot when I lost my balance and fell. I struck my head on the side of the desk (once again) and knocked myself out cold. When I woke up, I was in the hospital where due to a database error (thanks Oracle) the doctors had put casts on both my hands. I'm typing this ever so slowly with just my ton..tong ..tongu…tongue. But I digress. Okay, the only true part of that story is that I was a bit surprised. Here is what happens passing a String to a function.     // ---- Functions ----------------------------     void SetMyString(String myString)   // pass by 'value'    {        myString = "BBB";    }     void SetMyString(ref String myString)  // pass by ref    {        myString = "CCC";    }     // ---- Usage ---------------------------------     protected void Button3_Click(object sender, EventArgs e)    {        String MyString = "AAA";         SetMyString(MyString);        // MyString is still "AAA"  What!!!!         SetMyString(ref MyString);        // MyString is now "CCC"    } What the heck. We should not have to use the ref keyword when passing a String because Strings are reference types. Why didn't the string change? What is going on?   I spent hours unssuccessfully researching this anomaly until finally, I had a Eureka moment: This code: String MyString = "AAA"; Is semantically equivalent to this code (note this code doesn't actually compile): String MyString = new String(); MyString = "AAA"; Key Point: In the function, the copy of the reference is pointed to a new object and THAT object is modified. The original reference and what it points to is unchanged. You can simulate this behavior by modifying the class example code to look like this:      void SetMyClass(MyClass myClass)  // call by 'value'    {        //myClass.StrTag = "BBB";        myClass = new MyClass();        myClass.StrTag = "BBB";    } Now when you call the SetMyClass function without using ref, the parameter is unchanged...just like the string example.  I hope someone finds this useful. Steve Wellens

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  • Finding spelling of an element in an ignore case dictionary

    - by Andrew White
    Hi, Weird question, but I have a dictionary created with StringComparer.OrdinalIgnoreCase, looks something like this AaA, 10 aAB, 20 AAC, 12 I then use myDictionary["AAA"] to find the value associated with the key, but what I also need to know is what the actual spelling of the key is in myDictionary, e.g. in this case I want it to return AaA. Any way to do this without a loop? Thx.

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  • Create a group indicator (SQL)

    - by user1723699
    I am looking to create a group indicator for a query using SQL (Oracle specifically). Basically, I am looking for duplicate entries for certain columns and while I can find those what I also want is some kind of indicator to say what rows the duplicates are from. Below is an example of what I am looking to do (looking for duplicates on Name, Zip, Phone). The rows with Name = aaa are all in the same group, bb are not, and c are. Is there even a way to do this? I was thinking something with OVER (PARTITION BY ... but I can't think of a way to only increment for each group. +----------+---------+-----------+------------+-----------+-----------+ | Name | Zip | Phone | Amount | Duplicate | Group | +----------+---------+-----------+------------+-----------+-----------+ | aaa | 1234 | 5555555 | 500 | X | 1 | | aaa | 1234 | 5555555 | 285 | X | 1 | | bb | 545 | 6666666 | 358 | | 2 | | bb | 686 | 7777777 | 898 | | 3 | | aaa | 1234 | 5555555 | 550 | X | 1 | | c | 5555 | 8888888 | 234 | X | 4 | | c | 5555 | 8888888 | 999 | X | 4 | | c | 5555 | 8888888 | 230 | X | 4 | +----------+---------+-----------+------------+-----------+-----------+

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  • A elusive multi level accordion

    - by juanpablo
    Hi, I try make a multi level accordion with jquery: <div class="acordion"> <a href="#"><h1>aaa</h1></a> <div class="acoTitulo"> <h1>aaa</h1> <div class="acoSubTitulo"> <a href="#"><h2>bbb</h2></a> <div class="acoSubSubTitulo"> <h2>bbb</h2> <div class="acoLink"> <a href="#">ccc</a><br> <a href="#">ccc</a> </div> </div> </div> <!-- acoSubTitulo --> </div> <!-- end acoTitulo --> <div class="acoTitulo"> <h1>aaa</h1> <div class="acoSubTitulo"> <a href="#"><h2>bbb</h2></a> <div class="acoSubSubTitulo"> <h2>bbb</h2> <div class="acoLink"> <a href="#">ccc</a><br> <a href="#">ccc</a> </div> </div> </div> <!-- acoSubTitulo --> </div> <!-- end acoTitulo --> <div class="acoTitulo"> <h1>aaa</h1> <div class="acoSubTitulo"> <a href="#"><h2>bbb</h2></a> <a href="#"><h2>bbb</h2></a> </div> <!-- acoSubTitulo --> </div> <!-- end acoTitulo --> <a href="#"><h1>aaa</h1></a> <a href="#"><h1>aaa</h1></a> <div class="acoTitulo"> <h1>aaa</h1> <div class="acoSubTitulo"> <a href="#"><h2>bbb</h2></a> <a href="#"><h2>bbb</h2></a> </div> <!-- acoSubTitulo --> </div> <!-- end acoTitulo --> <div class="acoTitulo"> <h1>aaa</h1> <div class="acoSubTitulo"> <a href="#"><h2>bbb</h2></a> <a href="#"><h2>bbb</h2></a> </div> <!-- acoSubTitulo --> </div> <!-- end acoTitulo --> </div> <!-- end acordion --> .acordion h1, .acordion h2, .acoLink { font-size: 11px; padding: 8px; } .acoTitulo h2 { margin: 0px; } .acordion * a, .acordion a { text-decoration:none; color: #fff; } .acordion { width: 160px; color: #fff; background-color: #06f; border: 1px solid #06f; } .acoSubTitulo { background-color: #09f; } $(document).ready(function(){ $(".acoSubTitulo").hide(); $(".acoLink").hide(); $(".acoTitulo").hover( function(){ $(this).find(".acoSubTitulo").slideToggle("slow"); },function(){ $(this).find(".acoSubTitulo").slideToggle("slow"); } ); $(".acoSubSubTitulo").hover( function(){ $(this).find(".acoLink").slideToggle(); },function(){ $(this).find(".acoLink").slideToggle(); } ); }); but, this accordion is very elusive with the mouse. any idea? many thanks.

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  • GRUB2 not working after installing xubuntu 14.04

    - by h3bm
    I have a vaio laptop and it used to have installed windows 8 and Xubuntu 13.04 in dual boot, everything was working fine. I decided to update my version of xubuntu 14.04 LTS mainly because the support for 13.04 is finished and LTS version have 3 years of support. What I did was to format the partition where xubuntu 13.04 was installed and install 14.04 in that formated partition. When I restarted my computer willing to start using my new system I got the following message: error: symbol 'grub_term_highlight_color' not found and I was not able to enter any OS. I tried boot-repair from live USB more than two times and it did not fix the problem. I tried to enter to my computer using super GRUB2 disk, however it does not apperar to work with UEFI active (besides super grub2 disk says it can) I only get the message "no operating system found". If I boot super grub2 disk with UEFI disabled, super grub2 disk can not detect any OS,I also tried Rescatux distro, however, as of super grub2 disk, rescatux cannot enter when UEFI is active. I tried boot-repair with the option of "restore EFI backups", after that I was able to boot on windows, but no grub menu appeared. I ran boot-repair again with no improving results Here is the last Bootinfo report I got: http://paste.ubuntu.com/7609801/ Do you have any idea of what is happening? I really appreciate your help, Best regards,

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  • Issue about Tomcat 5.5 "ROOT" directory.

    - by Captain Kidd
    Now I specify a directory name for "appBase" attribute on "{TomcatHome}/config/server.xml". <host appBase="d:/aaa"> <Context docBase="d:/aaa/bbb"> </Context> </host> When I navigate URL to "http://localhost:8080", TOMCAT think "d:/aaa/ROOT" as application directory. I want to know how can I modify this mechanism to make TOMCAT auto search my specified directory. For example: When I input "http://localhost:8080", TOMCAT would search "d:/aaa/{SpecifiedDirectoryName}/"

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  • How to display image in second layer in Cocos2d

    - by PeterK
    I am very new at Cocos2d and is testing to displaying an image over the "Hello World" text on a second layer and need help to get it work. I guess it is some basic stuff here and appreciate any tips etc. with this. I know that if i put the display-code (myLayer1) in the "init" it work or do the call [self goHere] from the "init" in myLayer1 it works but i want to call the "goHere" directly. I have the following code: HelloWorld.m: #import "HelloWorldLayer.h" #import "myLayer1.h" // HelloWorldLayer implementation @implementation HelloWorldLayer +(CCScene *) scene { // 'scene' is an autorelease object. CCScene *scene = [CCScene node]; // 'layer' is an autorelease object. HelloWorldLayer *layer = [HelloWorldLayer node]; myLayer1 *layer1 = [myLayer1 node]; // add layer as a child to scene [scene addChild: layer]; [scene addChild: layer1]; // return the scene return scene; } // on "init" you need to initialize your instance -(id) init { // always call "super" init // Apple recommends to re-assign "self" with the "super" return value if( (self=[super init])) { // create and initialize a Label CCLabelTTF *label = [CCLabelTTF labelWithString:@"Hello World" fontName:@"Marker Felt" fontSize:64]; // ask director the the window size CGSize size = [[CCDirector sharedDirector] winSize]; // position the label on the center of the screen label.position = ccp( size.width /2 , size.height/2 ); // add the label as a child to this Layer [self addChild: label]; myLayer1 *a1 = [myLayer1 new]; [a1 goHere]; [myLayer1 release]; } return self; } myLayer1.m: #import "myLayer1.h" @implementation myLayer1 -(void)goHere { NSLog(@">>>>goHere<<<<"); CGSize size = [[CCDirector sharedDirector] winSize]; CCSprite *vv = [CCSprite spriteWithFile:@"hand.png"]; vv.position = ccp( size.width /2 , size.height/2 ); [self addChild:vv z:3]; } -(id) init { // always call "super" init // Apple recommends to re-assign "self" with the "super" return value if( (self=[super init])) { } return self; } @end

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  • how to code multiple button navigation with java activities [migrated]

    - by user1738212
    Question 1: I have 2 activities. I was wondering how to optimize it. I can either create 2 activities with multiple listeners. Or create multiple java files for each button(onclick listener) Question 2: I have tried to create multiple listeners in one java but can only get one button to work. What is the syntax for multiple listeners in one java file? Here is my *updated code: now the issue is no matter what button is clicked on it leads to the same page. package install.fineline; import android.app.Activity; import android.content.Context; import android.content.Intent; import android.os.Bundle; import android.widget.Button; import android.view.View; import android.view.View.OnClickListener; public class Activity1 extends Activity2 { Button Button1; Button Button2; Button Button3; Button Button4; Button Button5; Button Button6; public void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); setContentView(R.layout.fineline); addListenerOnButton(); } public void addListenerOnButton() { final Context context = this; Button1 = (Button) findViewById(R.id.autobody); Button1.setOnClickListener(new OnClickListener() { public void onClick(View arg0) { Intent intent = new Intent(context, Activity1.class); startActivity(intent); } }); Button2 = (Button) findViewById(R.id.glass); Button2.setOnClickListener(new OnClickListener() { @Override public void onClick(View arg0) { Intent intent = new Intent(context, Activity1.class); startActivity(intent); } }); Button3 = (Button) findViewById(R.id.wheels); Button3.setOnClickListener(new OnClickListener() { @Override public void onClick(View arg0) { Intent intent = new Intent(context, Activity1.class); startActivity(intent); } }); Button4 = (Button) findViewById(R.id.speedy); Button4.setOnClickListener(new OnClickListener() { @Override public void onClick(View arg0) { Intent intent = new Intent(context, Activity1.class); startActivity(intent); } }); Button5 = (Button) findViewById(R.id.sevan); Button5.setOnClickListener(new OnClickListener() { @Override public void onClick(View arg0) { Intent intent = new Intent(context, Activity1.class); startActivity(intent); } }); Button6 = (Button) findViewById(R.id.towing); Button6.setOnClickListener(new OnClickListener() { @Override public void onClick(View arg0) { Intent intent = new Intent(context, Activity1.class); startActivity(intent); } }); }} activity2.java package install.fineline; import android.app.Activity; import android.os.Bundle; import android.widget.Button; public class Activity2 extends Activity { Button Button1; public void onCreate1(Bundle savedInstanceState) { super.onCreate(savedInstanceState); setContentView(R.layout.autobody); } Button Button2; public void onCreate2(Bundle savedInstanceState) { super.onCreate(savedInstanceState); setContentView(R.layout.glass); } Button Button3; public void onCreate3(Bundle savedInstanceState) { super.onCreate(savedInstanceState); setContentView(R.layout.wheels); } Button button4; public void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); setContentView(R.layout.speedy); } Button Button5; public void onCreate5(Bundle savedInstanceState) { super.onCreate(savedInstanceState); setContentView(R.layout.sevan); } Button Button6; public void onCreate6(Bundle savedInstanceState) { super.onCreate(savedInstanceState); setContentView(R.layout.towing); }}

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  • How to quickly open an application for the 2nd time via Dash?

    - by Andre
    When I want to open an application via Dash, I just hit Super, type the first letters, and hit Enter. For instance: Super, "drop", Enter to start Dropbox. However, if I want to start an application again, Dash remembers it, but I cannot start it by hitting ENTER although "drop" is still in there, and Dropbox is in the first position. Why? How can I (without using the mouse) start an application again? UPDATE: better example (hopefully): Super ... type "ged" ... Enter to start Gedit close Gedit Super ... and now? "ged" is remembered, Gedit is still in pole position ready to be started. However, hitting Enter does not work. How can I start an application again? - Without using the mouse or retyping? If I have to retype, it makes no sense that Dash remembers the application and my typed letters. I assume there is a way to open the application again by just: Super + Enter (or something similar). Thanks!

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  • Disable Dash Home Shortcut in 12.04

    - by kunigami
    I'm using Ubuntu 12.04 in VirtualBox on a MacBook. The current shortcut for the dash home launcher is the Command key (Super). I'm trying to disable it. First, this option is not present in the shortcuts preferences. I have already tried: Changing it through gconf-editor Changing it through ccsm Even though when I press Super or Super+F it still launches the dash home. Any ideas on what I can still try?

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  • Automatically create valid links

    - by Marcos Placona
    Hi, I'm new to python, so please bare with me :) I was wondering if there's any built-in way in python to append variables to URL's regardless of it's structure. I would like to have a URL variable (test=1) added to an URL which could have any of the following structures http://www.aaa.com (would simply add "/?test=1") to the end http://www.aaa.com/home (like the one above, would simply add "/?test=1") to the end http://www.aaa.com/?location=home (would figure out there's already a ? being used, and would add &test=1 to the end) http://www.aaa.com/?location=home&page=1 (like the one above, would figure out there's already a ? being used, and would add &test=1 to the end) I'd be happy to write domething to do it myself, but if python can already do it somehow, I'd me more than happy to use any built-in functionality that would save me some time ;-) Thanks in advance

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  • Restore the Ctrl + Alt + Num Pad windows positioning commands?

    - by holocronweaver
    Using Unity in Ubuntu 12.04, the Ctrl + Alt + Num Pad combination for positioning windows has been fragmented by Ctrl + Alt + 4 (move window to left half of screen) being changed to Super + Left Arrow. A similar change moved Ctrl + Alt + 6 to Super + Right Arrow. Thus one moves windows to corners using Ctrl + Alt combos, but Super combos are needed to move to the left or right. This is more than a convenience problem since the new windows positioning provided by the super key combos seems to give different sizes than the Ctrl + Alt combos they replaced, leading to distracting gaps between windows when you combine the two methods to position three or more windows on one screen. Is there a way to restore the previous behavior so that I can use Ctrl + Alt + Num Pad for all windows positioning?

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  • Porting Python algorithm to C++ - different solution

    - by cb0
    Hello, I have written a little brute string generation script in python to generate all possible combinations of an alphabet within a given length. It works quite nice, but for the reason I wan't it to be faster I try to port it to C++. The problem is that my C++ Code is creating far too much combination for one word. Heres my example in python: ./test.py gives me aaa aab aac aad aa aba .... while ./test (the c++ programm gives me) aaa aaa aaa aaa aa Here I also get all possible combinations, but I get them twice ore more often. Here is the Code for both programms: #!/usr/bin/env python import sys #Brute String Generator #Start it with ./brutestringer.py 4 6 "abcdefghijklmnopqrstuvwxyz1234567890" "" #will produce all strings with length 4 to 6 and chars from a to z and numbers 0 to 9 def rec(w, p, baseString): for c in "abcd": if (p<w - 1): rec(w, p + 1, baseString + "%c" % c) print baseString for b in range(3,4): rec(b, 0, "") And here the C++ Code #include <iostream> using namespace std; string chars="abcd"; void rec(int w,int b,string p){ unsigned int i; for(i=0;i<chars.size();i++){ if(b < (w-1)){ rec(w, (b+1), p+chars[i]); } cout << p << "\n"; } } int main () { int a=3, b=0; rec (a+1,b, ""); return 0; } Does anybody see my fault ? I don't have much experience with C++. Thanks indeed

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