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  • Full Text Search like Google

    - by Eduardo
    I would like to implement full-text-search in my off-line (android) application to search the user generated list of notes. I would like it to behave just like Google (since most people are already used to querying to Google) My initial requirements are: Fast: like Google or as fast as possible, having 100000 documents with 200 hundred words each. Searching for two words should only return documents that contain both words (not just one word) (unless the OR operator is used) Case insensitive (aka: normalization): If I have the word 'Hello' and I search for 'hello' it should match. Diacritical mark insensitive: If I have the word 'así' a search for 'asi' should match. In Spanish, many people, incorrectly, either do not put diacritical marks or fail in correctly putting them. Stop word elimination: To not have a huge index meaningless words like 'and', 'the' or 'for' should not be indexed at all. Dictionary substitution (aka: stem words): Similar words should be indexed as one. For example, instances of 'hungrily' and 'hungry' should be replaced with 'hunger'. Phrase search: If I have the text 'Hello world!' a search of '"world hello"' should not match it but a search of '"hello world"' should match. Search all fields (in multifield documents) if no field specified (not just a default field) Auto-completion in search results while typing to give popular searches. (just like Google Suggest) How may I configure a full-text-search engine to behave as much as possible as Google? (I am mostly interested in Open Source, Java and in particular Lucene)

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  • Function behaviour on shell(ksh) script

    - by footy
    Here are 2 different versions of a program: this Program: #!/usr/bin/ksh printmsg() { i=1 print "hello function :)"; } i=0; echo I printed `printmsg`; printmsg echo $i Output: # ksh e I printed hello function :) hello function :) 1 and Program: #!/usr/bin/ksh printmsg() { i=1 print "hello function :)"; } i=0; echo I printed `printmsg`; echo $i Output: # ksh e I printed hello function :) 0 The only difference between the above 2 programs is that printmsg is 2times in the above program while printmsg is called once in the below program. My Doubt arises here: To quote Be warned: Functions act almost just like external scripts... except that by default, all variables are SHARED between the same ksh process! If you change a variable name inside a function.... that variable's value will still be changed after you have left the function!! But we can clearly see in the 2nd program's output that the value of i remains unchanged. But we are sure that the function is called as the print statement gets the the output of the function and prints it. So why is the output different in both?

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  • c++, object life-time of anonymous (unnamed) variables

    - by Joe Steeve
    In the following code, the object constructed in the last line of 'main()', seems to be destroyed before the end of the expression. The destructor is called before the '<<' is executed. Is this how it is supposed to be? #include <string> #include <sstream> #include <iostream> using std::string; using std::ostringstream; using std::cout; class A : public ostringstream { public: A () {} virtual ~A () { string s; s = str(); cout << "from A: " << s << std::endl; } }; int main () { string s = "Hello"; A os; os << s; cout << os.str() << std::endl; A() << "checking this"; } This is the output: Hello from A: 0x80495f7 from A: Hello This is the gdb log: (gdb) b os.cxx : 18 Breakpoint 1 at 0x80492b1: file os.cxx, line 18. (2 locations) (gdb) r Starting program: /home/joe/sandbox/test/os Hello Breakpoint 1, ~A (this=0xbffff37c, __in_chrg=<value optimized out>, __vtt_parm=<value optimized out>) at os.cxx:18 18 cout << "from A: " << s << std::endl; (gdb) p s.c_str () $1 = 0x804b45c "0x80495f7" (gdb) p *s.c_str () $2 = 48 '0' (gdb) c Continuing. from A: 0x80495f7 Breakpoint 1, ~A (this=0xbffff2bc, __in_chrg=<value optimized out>, __vtt_parm=<value optimized out>) at os.cxx:18 18 cout << "from A: " << s << std::endl; (gdb) p s.c_str () $3 = 0x804b244 "Hello" (gdb) p *s.c_str () $4 = 72 'H' (gdb) c Continuing. from A: Hello Program exited normally. (gdb)

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  • How to inject php code from database into php script ?

    - by luxquarta
    I want to store php code inside my database and then use it into my script. class A { public function getName() { return "lux"; } } // instantiates a new A $a = new A(); Inside my database there is data like "hello {$a->getName()}, how are you ?" In my php code I load the data into a variable $string $string = load_data_from_db(); echo $string; // echoes hello {$a->getName()}, how are you ? So now $string contains "hello {$a-getName()}, how are you ?" {$a-getName()} still being un-interpretated Question: I can't find how to write the rest of the code so that {$a-getName()} gets interpretated "into hello lux, how are you". Can someone help ? $new_string = ?????? echo $new_string; //echoes hello lux, how are you ? Is there a solution with eval() ? (please no debate about evil eval ;)) Or any other solution ?

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  • Java how to replace 2 or more spaces with single space in string and delete leading spaces only

    - by Nessa
    Looking for quick, simple way in Java to change this string " hello there " to something that looks like this "hello there" where I replace all those multiple spaces with a single space, except I also want the one or more spaces at the beginning of string to be gone. Something like this gets me partly there String mytext = " hello there "; mytext = mytext.replaceAll("( )+", " "); but not quite.

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  • Problem overridding virtual function

    - by William
    Okay, I'm writing a game that has a vector of a pairent class (enemy) that s going to be filled with children classes (goomba, koopa, boss1) and I need to make it so when I call update it calls the childclasses respective update. I have managed to create a example of my problem. #include <stdio.h> class A{ public: virtual void print(){printf("Hello from A");} }; class B : public A{ public: void print(){printf("Hello from B");} }; int main(){ A ab = B(); ab.print(); while(true){} } Output wanted: "Hello from B" Output got: "Hello from A" How do I get it to call B's print function?

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  • WPF: Change padding depending of container ?

    - by Alex Ilyin
    I have UserControl named MyUserControl, and another UserControl named MyContainer. I want MyUserControl to have padding 10 if it is placed inside MyContainer and 15 otherwise. Shortly, I want <MyContainer> <MyUserControl> Hello </MyUserControl> </MyContainer> to look like <MyContainer> <UserControl Padding="10"> Hello </UserControl> <MyContainer> and <MyUserControl> Hello </MyUserControl> to look like <UserControl Padding="15"> Hello </UserControl>

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  • html - problem with word wrapping

    - by pakita883
    Hello! I have some text in div, and I want it to wrap to fit document width (without any scrolls!). I don't want to have word-break, like div {word-wrap: break-word;} For example (this is what I want to get): hello world! today is a good day. But not: hello world! today is a good day. or: hello world! today is a go od day.

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  • Silverlight textblock binding question + MVVM

    - by AdrianDN
    Hello everyone, I'm trying to create a simple textblock control and I'm trying to insert a property from my ViewModel in the middle of the string. E.G. "Hello, My name is XX, bla, bla." (XX is a property from my ViewModel) <TextBlock Text="Hello, My name is {Binding SelectedUser.Name}, bla, bla." /> Is that possible? Regards, Adrian

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  • Replace method doesn't work properly

    - by John Smith
    Hello I have a string and when i try to use replace method in for loop it doesn't work String phrase="hello friend"; String[] wordds=phrase.split(" "); String newPhrase="sup friendhello weirdo"; for (int g=0;g<2;g++) { finalPhrase+=newPhrase.replace(wordds[g],"");} } System.out.println(finalPhrase) It prints out "sup hello weirdo" and i expect it to print "sup weirdo". What am i doing wrong?

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  • Linux termios VTIME not working?

    - by San Jacinto
    We've been bashing our heads off of this one all morning. We've got some serial lines setup between an embedded linux device and an Ubuntu box. Our reads are getting screwed up because our code usually returns two (sometimes more, sometimes exactly one) message reads instead of one message read per actual message sent. Here is the code that opens the serial port. InterCharTime is set to 4. void COMBaseClass::OpenPort() { cerr<< "openning port"<< port <<"\n"; struct termios newtio; this->fd = -1; int fdTemp; fdTemp = open( port, O_RDWR | O_NOCTTY); if (fdTemp < 0) { portOpen = 0; cerr<<"problem openning "<< port <<". Retrying"<<endl; usleep(1000000); return; } newtio.c_cflag = BaudRate | CS8 | CLOCAL | CREAD ;//| StopBits; newtio.c_iflag = IGNPAR; newtio.c_oflag = 0; /* set input mode (non-canonical, no echo,...) */ newtio.c_lflag = 0; newtio.c_cc[VTIME] = InterCharTime; /* inter-character timer in .1 secs */ newtio.c_cc[VMIN] = readBufferSize; /* blocking read until 1 char received */ tcflush(fdTemp, TCIFLUSH); tcsetattr(fdTemp,TCSANOW,&newtio); this->fd = fdTemp; portOpen = 1; } The other end is configured similarly for communication, and has one small section of particular iterest: while (1) { sprintf(out, "\r\nHello world %lu", ++ulCount); puts(out); WritePort((BYTE *)out, strlen(out)+1); sleep(2); } //while Now, when I run a read thread on the receiving machine, "hello world" is usually broken up over a couple messages. Here is some sample output: 1: Hello 2: world 1 3: Hello 4: world 2 5: Hello 6: world 3 where number followed by a colon is one message recieved. Can you see any error we are making? Thank you. Edit: For clarity, please view section 3.2 of this resource href="http://www.faqs.org/docs/Linux-HOWTO/Serial-Programming-HOWTO.html. To my understanding, with a VTIME of a couple seconds (meaning vtime is set anywhere between 10 and 50, trial-and-error), and a VMIN of 1, there should be no reason that the message is broken up over two separate messages.

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  • Ruby proc vs lambda in initialize()

    - by Jimmy Chu
    I found out this morning that proc.new works in a class initialize method, but not lambda. Concretely, I mean: class TestClass attr_reader :proc, :lambda def initialize @proc = Proc.new {puts "Hello from Proc"} @lambda = lambda {puts "Hello from lambda"} end end c = TestClass.new c.proc.call c.lambda.call In the above case, the result will be: Hello from Proc test.rb:14:in `<main>': undefined method `call' for nil:NilClass (NoMethodError) Why is that? Thanks!

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  • Scala return type for tuple-functions

    - by Felix
    Hello Guys, I want to make a scala function which returns a scala tuple. I can do a function like this: def foo = (1,"hello","world") and this will work fine, but now I want to tell the compiler what I expect to be returned from the function instead of using the built in type inference (after all, I have no idea what a (1,"hello","world") is) I thought I remembered the classname being something like Tuple3[Int,String,String] but that doesnt work for me. Suggestions? :D (ps: I love stack overflow!)

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  • Why cant Git merge file changes with a modified parent/master.

    - by Andy
    I have a file with one line in it. I create a branch and add a second line to the same file. Save and commit to the branch. I switch back to the master. And add a different, second line to the file. Save and commit to the master. So there's now 3 unique lines in total. If I now try and merge the branch back to the master, it suffers a merge conflict. Why cant Git simple merge each line, one after the other? My attempt at merge behaves something like this: PS D:\dev\testing\test1> git merge newbranch Auto-merging hello.txt CONFLICT (content): Merge conflict in hello.txt Automatic merge failed; fix conflicts and then commit the result. PS D:\dev\testing\test1> git diff diff --cc hello.txt index 726eeaf,e48d31a..0000000 --- a/hello.txt +++ b/hello.txt @@@ -1,2 -1,2 +1,6 @@@ This is the first line. - New line added by master. -Added a line in newbranch. ++<<<<<<< HEAD ++New line added by master. ++======= ++Added a line in newbranch. ++>>>>>>> newbranch Is there a way to make it slot lines in automatically, one after the other?

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  • Get the common prefix substring through Regex

    - by Dreampuf
    like this text = " \t hello there\n \t how are you?\n \t HHHH" hello there how are you? HHHH Could I get the common prefix substring through regex? I try to In [36]: re.findall(r"(?m)(?:(^[ \t]+).+[\n\r]+\1)", " \t hello there\n \t how are you?\n \t HHHH") Out[36]: [' \t '] But apparently that common prefix substring is ' \t ' I want use for dedent function like python textwrap module.

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  • C++ game loop example

    - by David
    Can someone write up a source for a program that just has a "game loop", which just keeps looping until you press Esc, and the program shows a basic image. Heres the source I have right now but I have to use SDL_Delay(2000); to keep the program alive for 2 seconds, during which the program is frozen. #include "SDL.h" int main(int argc, char* args[]) { SDL_Surface* hello = NULL; SDL_Surface* screen = NULL; SDL_Init(SDL_INIT_EVERYTHING); screen = SDL_SetVideoMode(640, 480, 32, SDL_SWSURFACE); hello = SDL_LoadBMP("hello.bmp"); SDL_BlitSurface(hello, NULL, screen, NULL); SDL_Flip(screen); SDL_Delay(2000); SDL_FreeSurface(hello); SDL_Quit(); return 0; } I just want the program to be open until I press Esc. I know how the loop works, I just don't know if I implement inside the main() function, or outside of it. I've tried both, and both times it failed. If you could help me out that would be great :P

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  • multiple inheritance

    - by hitech
    when we say "a member declated as protected is accessible to any class imediately derived from it" what does this mean. in the follwing example get_number function can be accessible by the result class , as per the statement it sould only be accessile to test class. class student { protected: int roll_number; public: void get_number(int){ cout<< "hello"; } void put_number(void) {cout<< "hello"; } }; class test : public student { protected : float sub1; float sub2; public: void get_marks(float, float) {cout<< "hello"; roll_number = 10; } void put_marks(void) {cout<< "hello"; cout << "roll_number = " << roll_number ; } }; class result :public test { float total; public: void display(){cout<< "hello"; roll_number = 10; } }; int main() { result student; student.get_marks(2.2, 2.2); student.put_marks(); return 0; } i changed the code as per the first statement the protected variable roll_number not be accessible upto the result class ?

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  • How can I have a newline in a string in sh?

    - by juannavarroperez
    This STR="Hello\nWorld" echo $STR produces as output Hello\nWorld instead of Hello World What should I do to have a newline in a string? I'm aware of echo -e, but I'm no sending the string to echo, the string will be used as an argument by another command that doesn't know how to interpret \n as a newline.

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