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• A simple demo for WDF Driver development

  Access physical memory, I/O port in user mode

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• Join 2 children tables with a parent tables without duplicated

  - by user1847866

  Problem I have 3 tables: People, Phones and Emails. Each person has an UNIQUE ID, and each person can have multiple numbers or multiple emails. Simplified it looks like this: +---------+----------+ | ID | Name | +---------+----------+ | 5000003 | Amy | | 5000004 | George | | 5000005 | John | | 5000008 | Steven | | 8000009 | Ashley | +---------+----------+ +---------+-----------------+ | ID | Number | +---------+-----------------+ | 5000005 | 5551234 | | 5000005 | 5154324 | | 5000008 | 2487312 | | 8000009 | 7134584 | | 5000008 | 8451384 | +---------+-----------------+ +---------+------------------------------+ | ID | Email | +---------+------------------------------+ | 5000005 | [email protected] | | 5000005 | [email protected] | | 5000008 | [email protected] | | 5000008 | [email protected] | | 5000008 | [email protected] | | 8000009 | [email protected] | | 5000004 | [email protected] | +---------+------------------------------+ I am trying to joining them together without duplicates. It works great, when I try to join only Emails with People or only Phones with People. SELECT People.Name, People.ID, Phones.Number FROM People LEFT OUTER JOIN Phones ON People.ID=Phones.ID ORDER BY Name, ID, Number; +----------+---------+-----------------+ | Name | ID | Number | +----------+---------+-----------------+ | Steven | 5000008 | 8451384 | | Steven | 5000008 | 24887312 | | John | 5000005 | 5551234 | | John | 5000005 | 5154324 | | George | 5000004 | NULL | | Ashley | 8000009 | 7134584 | | Amy | 5000003 | NULL | +----------+---------+-----------------+ SELECT People.Name, People.ID, Emails.Email FROM People LEFT OUTER JOIN Emails ON People.ID=Emails.ID ORDER BY Name, ID, Email; +----------+---------+------------------------------+ | Name | ID | Email | +----------+---------+------------------------------+ | Steven | 5000008 | [email protected] | | Steven | 5000008 | [email protected] | | Steven | 5000008 | [email protected] | | John | 5000005 | [email protected] | | John | 5000005 | [email protected] | | George | 5000004 | [email protected] | | Ashley | 8000009 | [email protected] | | Amy | 5000003 | NULL | +----------+---------+------------------------------+ However, when I try to join Emails and Phones on People - I get this: SELECT People.Name, People.ID, Phones.Number, Emails.Email FROM People LEFT OUTER JOIN Phones ON People.ID = Phones.ID LEFT OUTER JOIN Emails ON People.ID = Emails.ID ORDER BY Name, ID, Number, Email; +----------+---------+-----------------+------------------------------+ | Name | ID | Number | Email | +----------+---------+-----------------+------------------------------+ | Steven | 5000008 | 8451384 | [email protected] | | Steven | 5000008 | 8451384 | [email protected] | | Steven | 5000008 | 8451384 | [email protected] | | Steven | 5000008 | 24887312 | [email protected] | | Steven | 5000008 | 24887312 | [email protected] | | Steven | 5000008 | 24887312 | [email protected] | | John | 5000005 | 5551234 | [email protected] | | John | 5000005 | 5551234 | [email protected] | | John | 5000005 | 5154324 | [email protected] | | John | 5000005 | 5154324 | [email protected] | | George | 5000004 | NULL | [email protected] | | Ashley | 8000009 | 7134584 | [email protected] | | Amy | 5000003 | NULL | NULL | +----------+---------+-----------------+------------------------------+ What happens is - if a Person has 2 numbers, all his emails are shown twice (They can not be sorted! which means they can not be removed by @last) What I want: Bottom line, playing with the @last, I want to end up with somethig like this, but @last won't work if I don't arrange ORDER columns in the righ way - and this seems like a big problem..Orderin the email column. Because seen from the example above: Steven has 2 phone number and 3 emails. The JOIN Emails with Numbers happens with each email - thus duplicated values that can not be sorted (SORT BY does not work on them). **THIS IS WHAT I WANT** +----------+---------+-----------------+------------------------------+ | Name | ID | Number | Email | +----------+---------+-----------------+------------------------------+ | Steven | 5000008 | 8451384 | [email protected] | | | | 24887312 | [email protected] | | | | | [email protected] | | John | 5000005 | 5551234 | [email protected] | | | | 5154324 | [email protected] | | George | 5000004 | NULL | [email protected] | | Ashley | 8000009 | 7134584 | [email protected] | | Amy | 5000003 | NULL | NULL | +----------+---------+-----------------+------------------------------+ Now I'm told that it's best to keep emails and number in separated tables because one can have many emails. So if it's such a common thing to do, what isn't there a simple solution? I'd be happy with a PHP Solution aswell. What I know how to do by now that satisfies it, but is not as pretty. If I do it with GROUP_CONTACT I geat a satisfactory result, but it doesn't look as pretty: I can't put a "Email type = work" next to it. SELECT People.Ime, GROUP_CONCAT(DISTINCT Phones.Number), GROUP_CONCAT(DISTINCT Emails.Email) FROM People LEFT OUTER JOIN Phones ON People.ID=Phones.ID LEFT OUTER JOIN Emails ON People.ID=Emails.ID GROUP BY Name; +----------+----------------------------------------------+---------------------------------------------------------------------+ | Name | GROUP_CONCAT(DISTINCT Phones.Number) | GROUP_CONCAT(DISTINCT Emails.Email) | +----------+----------------------------------------------+---------------------------------------------------------------------+ | Steven | 8451384,24887312 | [email protected],[email protected],[email protected] | | John | 5551234,5154324 | [email protected],[email protected] | | George | NULL | [email protected] | | Ashley | 7134584 | [email protected] | | Amy | NULL | NULL | +----------+----------------------------------------------+---------------------------------------------------------------------+

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• Phone-book Database Help - Python

  - by IDOntWantThat

  I'm new to programming and have an assignment I've been working at for awhile. I understand defining functions and a lot of the basics but I'm kind of running into a brick wall at this point. I'm trying to figure this one out and don't really understand how the 'class' feature works yet. I'd appreciate any help with this one; also any help with some python resources that have can dummy down how/why classes are used. You've been going to work on a database project at work for sometime now. Your boss encourages you to program the database in Python. You disagree, arguing that Python is not a database language but your boss persists by providing the source code below for a sample telephone database. He asks you to do two things: Evaluate the existing source code and extend it to make it useful for managers in the firm. (You do not need a GUI interface, just work on the database aspects: data entry and retrieval - of course you must get the program to run or properly work He wants you to critically evaluate Python as a database tool. Import the sample code below into the Python IDLE and enhance it, run it and debug it. Add features to make this a more realistic database tool by providing for easy data entry and retrieval. import shelve import string UNKNOWN = 0 HOME = 1 WORK = 2 FAX = 3 CELL = 4 class phoneentry: def __init__(self, name = 'Unknown', number = 'Unknown', type = UNKNOWN): self.name = name self.number = number self.type = type # create string representation def __repr__(self): return('%s:%d' % ( self.name, self.type )) # fuzzy compare or two items def __cmp__(self, that): this = string.lower(str(self)) that = string.lower(that) if string.find(this, that) >= 0: return(0) return(cmp(this, that)) def showtype(self): if self.type == UNKNOWN: return('Unknown') if self.type == HOME: return('Home') if self.type == WORK: return('Work') if self.type == FAX: return('Fax') if self.type == CELL: return('Cellular') class phonedb: def __init__(self, dbname = 'phonedata'): self.dbname = dbname; self.shelve = shelve.open(self.dbname); def __del__(self): self.shelve.close() self.shelve = None def add(self, name, number, type = HOME): e = phoneentry(name, number, type) self.shelve[str(e)] = e def lookup(self, string): list = [] for key in self.shelve.keys(): e = self.shelve[key] if cmp(e, string) == 0: list.append(e) return(list) # if not being loaded as a module, run a small test if __name__ == '__main__': foo = phonedb() foo.add('Sean Reifschneider', '970-555-1111', HOME) foo.add('Sean Reifschneider', '970-555-2222', CELL) foo.add('Evelyn Mitchell', '970-555-1111', HOME) print 'First lookup:' for entry in foo.lookup('reifsch'): print '%-40s %s (%s)' % ( entry.name, entry.number, entry.showtype() ) print print 'Second lookup:' for entry in foo.lookup('e'): print '%-40s %s (%s)' % ( entry.name, entry.number, entry.showtype() ) I'm not sure if I'm on the right track but here is what I have so far: def openPB(): foo = phonedb() print 'Please select an option:' print '1 - Lookup' print '2 - Add' print '3 - Delete' print '4 - Quit' entry=int(raw_input('>> ')) if entry==1: namelookup=raw_input('Please enter a name: ') for entry in foo.lookup(namelookup): print '%-40s %s (%s)' % (entry.name, entry.number, entry.showtype() ) elif entry==2: name=raw_input('Name: ') number=raw_input('Number: ') showtype=input('Type (UNKNOWN, HOME, WORK, FAX, CELL): \n>> ') for entry in foo.add(name, number, showtype): #Trying to figure out this part print '%-40s %s (%s)'% (entry.name, entry.number, entry.showtype() ) elif entry==3: delname=raw_input('Please enter a name to delete: ') # #Trying to figure out this part print "Contact '%s' has been deleted" (delname) elif entry==4: print "Phone book is now closed" quit else: print "Your entry was not recognized." openPB() openPB()

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• C Number to Text problem with ones and tens..

  - by Joegabb

  #include<stdio.h> #include<conio.h> main() { int ones,tens,ventoteen, myloop = 0; long num2,cents2,centeens,cents1,thousands,hundreds; double num; do{ printf("Enter a number: "); scanf("%lf",&num); if(num<=10000 || num>=0) { if (num==0) { printf("\t\tZero"); } num=(num*100); num2= (long)num; thousands=num2/100000; num2=num2%100000; hundreds=num2/10000; num2=num2%10000; if ((num2>=1100) || (num2<=1900)) { tens=0; ones=0; ventoteen=num2%1000; } else { tens=num2/1000; num2=num2%1000; ones=num2/100; num2=num2%100; } if((num2>=11) && (num2<=19)) { cents1=0; cents2=0; centeens=num2%10; } else { cents1=num2/10; num2=num2%10; cents2=num2/1; } if (thousands == 1) printf("One thousand "); else if (thousands == 2) printf("Two thousand "); else if (thousands == 3) printf("Three Thousand "); else if (thousands == 4) printf("Four thousand "); else if (thousands == 5) printf("Five Thousand "); else if (thousands == 6) printf("Six thousand "); else if (thousands == 7) printf("Seven Thousand "); else if (thousands == 8) printf("Eight thousand "); else if (thousands == 9) printf("Nine Thousand "); else {} if (hundreds == 1) printf("one hundred "); else if (hundreds == 2) printf("two hundred "); else if (hundreds == 3) printf("three hundred "); else if (hundreds == 4) printf("four hundred "); else if (hundreds == 5) printf("five hundred "); else if (hundreds == 6) printf("six hundred "); else if (hundreds == 7) printf("seven hundred "); else if (hundreds == 8) printf("eight hundred "); else if (hundreds == 9) printf("nine hundred "); else {} switch(ventoteen) { case 1: printf("eleven ");break; case 2: printf("twelve ");break; case 3: printf("thirteen ");break; case 4: printf("fourteen ");break; case 5: printf("fifteen ");break; case 6: printf("sixteen ");break; case 7: printf("seventeen ");break; case 8: printf("eighteen ");break; case 9: printf("nineteen ");break; } switch(tens) { case 1: printf("ten ");break; case 2: printf("twenty ");break; case 3: printf("thirty ");break; case 4: printf("forty ");break; case 5: printf("fifty ");break; case 6: printf("sixty ");break; case 7: printf("seventy ");break; case 8: printf("eighty ");break; case 9: printf("ninety ");break; } switch(ones) { case 1: printf("one ");break; case 2: printf("two ");break; case 3: printf("three ");break; case 4: printf("four ");break; case 5: printf("five ");break; case 6: printf("six ");break; case 7: printf("seven ");break; case 8: printf("eight ");break; case 9: printf("nine ");break; } switch(cents1) { case 1: printf("and ten centavos ");break; case 2: printf("and twenty centavos ");break; case 3: printf("and thirty centavos ");break; case 4: printf("and fourty centavos ");break; case 5: printf("and fifty centavos ");break; case 6: printf("and sixty centavos ");break; case 7: printf("and seventy centavos ");break; case 8: printf("and eighty centavos ");break; case 9: printf("and ninety centavos ");break; } switch(centeens) { case 1: printf("and eleven centavos ");break; case 2: printf("and twelve centavos ");break; case 3: printf("and thirteen centavos ");break; case 4: printf("and fourteen centavos ");break; case 5: printf("and fifteen centavos ");break; case 6: printf("and sixteen centavos ");break; case 7: printf("and seventeen centavos ");break; case 8: printf("and eighteen centavos ");break; case 9: printf("and nineteen centavos ");break; } switch(cents2) { case 1: printf("and one centavos ");break; case 2: printf("and two centavos ");break; case 3: printf("and three centavos ");break; case 4: printf("and four centavos ");break; case 5: printf("and five centavos ");break; case 6: printf("and six centavos ");break; case 7: printf("and seven centavos ");break; case 8: printf("and eight centavos ");break; case 9: printf("and nine centavos ");break; } } getch(); }while(myloop == 0); return 0; } my code is working fine but the problem is when i input 1 - 90 nothing appears but when i input 100 the output would be fine and that is "One Hundred" and so as 1000 the output would be "One Thousand". thanks for the help..

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• Simple perl program failing to execute

  - by yves Baumes

  Here is a sample that fails: #!/usr/bin/perl -w # client.pl #---------------- use strict; use Socket; # initialize host and port my $host = shift || 'localhost'; my $port = shift || 55555; my $server = "10.126.142.22"; # create the socket, connect to the port socket(SOCKET,PF_INET,SOCK_STREAM,(getprotobyname('tcp'))[2]) or die "Can't create a socket $!\n"; connect( SOCKET, pack( 'Sn4x8', AF_INET, $port, $server )) or die "Can't connect to port $port! \n"; my $line; while ($line = <SOCKET>) { print "$line\n"; } close SOCKET or die "close: $!"; with the error: Argument "10.126.142.22" isn't numeric in pack at D:\send.pl line 16. Can't connect to port 55555! I am using this version of Perl: This is perl, v5.10.1 built for MSWin32-x86-multi-thread (with 2 registered patches, see perl -V for more detail) Copyright 1987-2009, Larry Wall Binary build 1006 [291086] provided by ActiveState http://www.ActiveState.com Built Aug 24 2009 13:48:26 Perl may be copied only under the terms of either the Artistic License or the GNU General Public License, which may be found in the Perl 5 source kit. Complete documentation for Perl, including FAQ lists, should be found on this system using "man perl" or "perldoc perl". If you have access to the Internet, point your browser at http://www.perl.org/, the Perl Home Page. While I am running the netcat command on the server side. Telnet does work.

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• Python Socket Getting Connection Reset

  - by Ian

  I created a threaded socket listener that stores newly accepted connections in a queue. The socket threads then read from the queue and respond. For some reason, when doing benchmarking with 'ab' (apache benchmark) using a concurrency of 2 or more, I always get a connection reset before it's able to complete the benchmark (this is taking place locally, so there's no external connection issue). class server: _ip = '' _port = 8888 def __init__(self, ip=None, port=None): if ip is not None: self._ip = ip if port is not None: self._port = port self.server_listener(self._ip, self._port) def now(self): return time.ctime(time.time()) def http_responder(self, conn, addr): httpobj = http_builder() httpobj.header('HTTP/1.1 200 OK') httpobj.header('Content-Type: text/html; charset=UTF-8') httpobj.header('Connection: close') httpobj.body("Everything looks good") data = httpobj.generate() sent = conn.sendall(data) def http_thread(self, id): self.log("THREAD %d: Starting Up..." % id) while True: conn, addr = self.q.get() ip, port = addr self.log("THREAD %d: responding to request: %s:%s - %s" % (id, ip, port, self.now())) self.http_responder(conn, addr) self.q.task_done() conn.close() def server_listener(self, host, port): self.q = Queue.Queue(0) sock = socket.socket(socket.AF_INET, socket.SOCK_STREAM) sock.bind( (host, port) ) sock.listen(5) for i in xrange(4): #thread count thread.start_new(self.http_thread, (i+1, )) while True: self.q.put(sock.accept()) sock.close() server('', 9999) When running the benchmark, I get totally random numbers of good requests before it errors out, usually between 4 and 500. Edit: Took me a while to figure it out, but the problem was in sock.listen(5). Because I was using apache benchmark with a higher concurrency (5 and up) it was causing the backlog of connections to pile up, at which point the connections started getting dropped by the socket.

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• Is it possible to defer member initialization to the constructor body?

  - by Kjir

  I have a class with an object as a member which doesn't have a default constructor. I'd like to initialize this member in the constructor, but it seems that in C++ I can't do that. Here is the class: #include <boost/asio.hpp> #include <boost/array.hpp> using boost::asio::ip::udp; template<class T> class udp_sock { public: udp_sock(std::string host, unsigned short port); private: boost::asio::io_service _io_service; udp::socket _sock; boost::array<T,256> _buf; }; template<class T> udp_sock<T>::udp_sock(std::string host = "localhost", unsigned short port = 50000) { udp::resolver res(_io_service); udp::resolver::query query(udp::v4(), host, "spec"); udp::endpoint ep = *res.resolve(query); ep.port(port); _sock(_io_service, ep); } The compiler tells me basically that it can't find a default constructor for udp::socket and by my research I understood that C++ implicitly initializes every member before calling the constructor. Is there any way to do it the way I wanted to do it, or is it too "Java-oriented" and not feasible in C++? I worked around the problem by defining my constructor like this: template<class T> udp_sock<T>::udp_sock(std::string host = "localhost", unsigned short port = 50000) : _sock(_io_service) { udp::resolver res(_io_service); udp::resolver::query query(udp::v4(), host, "spec"); udp::endpoint ep = *res.resolve(query); ep.port(port); _sock.bind(ep); } So my question is more out of curiosity and to better understand OOP in C++

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• Error when loading YAML config files in Rails

  - by ZelluX

  I am configuring Rails with MongoDB, and find a strange problem when paring config/mongo.yml file. config/mongo.yml is generated by executing script/rails generate mongo_mapper:config, and it looks like following: defaults: &defaults host: 127.0.0.1 port: 27017 development: <<: *defaults database: tc_web_development test: <<: *defaults database: tc_web_test From the config file we can see the objects development and test should both have a database field. But when it is parsed and loaded in config/initializers/mongo.db, config = YAML::load(File.read(Rails.root.join('config/mongo.yml'))) puts config.inspect MongoMapper.setup(config, Rails.env) the strange thing comes: the output of puts config.inspect is {"defaults"=>{"host"=>"127.0.0.1", "port"=>27017}, "development"=>{"host"=>"127.0.0.1", "port"=>27017}, "test"=>{"host"=>"127.0.0.1", "port"=>27017}} which does not contain database attribute. But when I execute the same statements in a plain ruby console, instead of using rails console, mongo.yml is parsed in a right way. {"defaults"=>{"host"=>"127.0.0.1", "port"=>27017}, "development"=>{"host"=>"127.0.0.1", "port"=>27017, "database"=>"tc_web_development"}, "test"=>{"host"=>"127.0.0.1", "port"=>27017, "database"=>"tc_web_test"}} I am wondering what may be the cause of this problem. Any ideas? Thanks.

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• Spring: Bean fails to read off values from external Properties file when using @Value annotation

  - by daydreamer

  XML Configuration <beans xmlns="http://www.springframework.org/schema/beans" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:util="http://www.springframework.org/schema/util" xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.0.xsd http://www.springframework.org/schema/util http://www.springframework.org/schema/util/spring-util.xsd"> <util:properties id="mongoProperties" location="file:///storage/local.properties" /> <bean id="mongoService" class="com.business.persist.MongoService"></bean> </beans> and MongoService looks like @Service public class MongoService { @Value("#{mongoProperties[host]}") private String host; @Value("#{mongoProperties[port]}") private int port; @Value("#{mongoProperties[database]}") private String database; private Mongo mongo; private static final Logger LOGGER = LoggerFactory.getLogger(MongoService.class); public MongoService() throws UnknownHostException { LOGGER.info("host=" + host + ", port=" + port + ", database=" + database); mongo = new Mongo(host, port); } public void putDocument(@Nonnull final DBObject document) { LOGGER.info("inserting document - " + document.toString()); mongo.getDB(database).getCollection(getCollectionName(document)).insert(document, WriteConcern.SAFE); } I write my MongoServiceTest as public class MongoServiceTest { @Autowired private MongoService mongoService; public MongoServiceTest() throws UnknownHostException { mongoRule = new MongoRule(); } @Test public void testMongoService() { final DBObject document = DBContract.getUniqueQuery("001"); document.put(DBContract.RVARIABLES, "values"); document.put(DBContract.PVARIABLES, "values"); mongoService.putDocument(document); } and I see failures in tests as 12:37:25.224 [main] INFO c.s.business.persist.MongoService - host=null, port=0, database=null java.lang.NullPointerException at com.business.persist.MongoServiceTest.testMongoService(MongoServiceTest.java:40) Which means bean was not able to read the values from local.properties local.properties ### === MongoDB interaction === ### host="127.0.0.1" port=27017 database=contract How do I fix this? update It doesn't seem to read off the values even after creating setters/getters for the fields. I am really clueless now. How can I even debug this issue? Thanks much!

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• C: incompatible types in assignment

  - by The.Anti.9

  I'm writing a program to check to see if a port is open in C. One line in particular copies one of the arguments to a char array. However, when I try to compile, it says: error: incompatible types in assignment Heres the code. The error is on the assignment of addr #include <sys/socket.h> #include <sys/time.h> #include <sys/types.h> #include <arpa/inet.h> #include <netinet/in.h> #include <errno.h> #include <fcntl.h> #include <stdio.h> #include <netdb.h> #include <stdlib.h> #include <string.h> #include <unistd.h> int main(int argc, char **argv) { u_short port; /* user specified port number */ char addr[1023]; /* will be a copy of the address entered by u */ struct sockaddr_in address; /* the libc network address data structure */ short int sock = -1; /* file descriptor for the network socket */ port = atoi(argv[1]); addr = strncpy(addr, argv[2], 1023); bzero((char *)&address, sizeof(address)); /* init addr struct */ address.sin_addr.s_addr = inet_addr(addr); /* assign the address */ address.sin_port = htons(port); /* translate int2port num */ sock = socket(AF_INET, SOCK_STREAM, 0); if (connect(sock,(struct sockaddr *)&address,sizeof(address)) == 0) { printf("%i is open\n", port); } if (errno == 113) { fprintf(stderr, "Port not open!\n"); } close(sock); return 0; } I'm new to C, so I'm not sure why it would do this.

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• Unit testing a method with many possible outcomes

  - by Cthulhu

  I've built a simple~ish method that constructs an URL out of approximately 5 parts: base address, port, path, 'action', and a set of parameters. Out of these, only the address part is mandatory, the other parts are all optional. A valid URL has to come out of the method for each permutation of input parameters, such as: address address port address port path address path address action address path action address port action address port path action address action params address path action params address port action params address port path action params andsoforth. The basic approach for this is to write one unit test for each of these possible outcomes, each unit test passing the address and any of the optional parameters to the method, and testing the outcome against the expected output. However, I wonder, is there a Better (tm) way to handle a case like this? Are there any (good) unit test patterns for this? (rant) I only now realize that I've learned to write unit tests a few years ago, but never really (feel like) I've advanced in the area, and that every unit test is a repeat of building parameters, expected outcome, filling mock objects, calling a method and testing the outcome against the expected outcome. I'm pretty sure this is the way to go in unit testing, but it gets kinda tedious, yanno. Advice on that matter is always welcome. (/rant) (note) christmas weekend approaching, probably won't reply to suggestions until next week. (/note)

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• PHP Infine Loope Problem

  - by Ashwin

  function httpGet( $url, $followRedirects=true ) { global $final_url; $url_parsed = parse_url($url); if ( empty($url_parsed['scheme']) ) { $url_parsed = parse_url('http://'.$url); } $final_url = $url_parsed; $port = $url_parsed["port"]; if ( !$port ) { $port = 80; } $rtn['url']['port'] = $port; $path = $url_parsed["path"]; if ( empty($path) ) { $path="/"; } if ( !empty($url_parsed["query"]) ) { $path .= "?".$url_parsed["query"]; } $rtn['url']['path'] = $path; $host = $url_parsed["host"]; $foundBody = false; $out = "GET $path HTTP/1.0\r\n"; $out .= "Host: $host\r\n"; $out .= "User-Agent: Mozilla/5.0 (Windows; U; Windows NT 5.1; en-US; rv:1.8.1) Gecko/20061010 Firefox/2.0\r\n"; $out .= "Connection: Close\r\n\r\n"; if ( !$fp = @fsockopen($host, $port, $errno, $errstr, 30) ) { $rtn['errornumber'] = $errno; $rtn['errorstring'] = $errstr; } fwrite($fp, $out); while (!@feof($fp)) { $s = @fgets($fp, 128); if ( $s == "\r\n" ) { $foundBody = true; continue; } if ( $foundBody ) { $body .= $s; } else { if ( ($followRedirects) && (stristr($s, "location:") != false) ) { $redirect = preg_replace("/location:/i", "", $s); return httpGet( trim($redirect) ); } $header .= $s; } } fclose($fp); return(trim($body)); } This code sometimes go infinite loop. What's wrong here?

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• Python client / server question

  - by AustinM

  I'm working on a bit of a project in python. I have a client and a server. The server listens for connections and once a connection is received it waits for input from the client. The idea is that the client can connect to the server and execute system commands such as ls and cat. This is my server code: import sys, os, socket host = '' port = 50105 s = socket.socket(socket.AF_INET, socket.SOCK_STREAM) s.bind((host, port)) print("Server started on port: ", port) s.listen(5) print("Server listening\n") conn, addr = s.accept() print 'New connection from ', addr while (1): rc = conn.recv(5) pipe = os.popen(rc) rl = pipe.readlines() file = conn.makefile('w', 0) file.writelines(rl[:-1]) file.close() conn.close() And this is my client code: import sys, socket s = socket.socket(socket.AF_INET, socket.SOCK_STREAM) host = 'localhost' port = input('Port: ') s.connect((host, port)) cmd = raw_input('$ ') s.send(cmd) file = s.makefile('r', 0) sys.stdout.writelines(file.readlines()) When I start the server I get the right output, saying the server is listening. But when I connect with my client and type a command the server exits with this error: Traceback (most recent call last): File "server.py", line 21, in <module> rc = conn.recv(2) File "/usr/lib/python2.6/socket.py", line 165, in _dummy raise error(EBADF, 'Bad file descriptor') socket.error: [Errno 9] Bad file descriptor On the client side, I get the output of ls but the server gets screwed up.

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• Python server open all ports

  - by user1670178

  I am trying to open all ports using this code, why can I not create a loop to perform this function? http://www.kellbot.com/2010/02/tutorial-writing-a-tcp-server-in-python/ #!/usr/bin/python # This is server.py file ##server.py from socket import * #import the socket library n=1025 while n<1050: ##let's set up some constants HOST = '' #we are the host PORT = n #arbitrary port not currently in use ADDR = (HOST,PORT) #we need a tuple for the address BUFSIZE = 4096 #reasonably sized buffer for data ## now we create a new socket object (serv) ## see the python docs for more information on the socket types/flags serv = socket( AF_INET,SOCK_STREAM) ##bind our socket to the address serv.bind((ADDR)) #the double parens are to create a tuple with one element serv.listen(5) #5 is the maximum number of queued connections we'll allow serv = socket( AF_INET,SOCK_STREAM) ##bind our socket to the address serv.bind((ADDR)) #the double parens are to create a tuple with one element serv.listen(5) #5 is the maximum number of queued connections we'll allow print 'listening...' n=n+1 conn,addr = serv.accept() #accept the connection print '...connected!' conn.send('TEST') conn.close() How do I make this work so that I can specify input range and have the server open all ports up to 65535? #!/usr/bin/python # This is server.py file from socket import * #import the socket library startingPort=input("\nPlease enter starting port: ") startingPort=int(startingPort) #print startingPort def connection(): ## let's set up some constants HOST = '' #we are the host PORT = startingPort #arbitrary port not currently in use ADDR = (HOST,PORT) #we need a tuple for the address BUFSIZE = 4096 #reasonably sized buffer for data def socketObject(): ## now we create a new socket object (serv) serv = socket( AF_INET,SOCK_STREAM) def bind(): ## bind our socket to the address serv = socket( AF_INET,SOCK_STREAM) serv.bind((ADDR)) #the double parens are to create a tuple with one element serv.listen(5) #5 is the maximum number of queued connections we'll allow serv = socket( AF_INET,SOCK_STREAM) print 'listening...' def accept(): conn,addr = serv.accept() #accept the connection print '...connected!' conn.send('TEST') def close(): conn.close() ## Main while startingPort<65535: connection() socketObject() bind() accept() startingPort=startingPort+1

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• PHP Infine Loop Problem

  - by Ashwin

  function httpGet( $url, $followRedirects=true ) { global $final_url; $url_parsed = parse_url($url); if ( empty($url_parsed['scheme']) ) { $url_parsed = parse_url('http://'.$url); } $final_url = $url_parsed; $port = $url_parsed["port"]; if ( !$port ) { $port = 80; } $rtn['url']['port'] = $port; $path = $url_parsed["path"]; if ( empty($path) ) { $path="/"; } if ( !empty($url_parsed["query"]) ) { $path .= "?".$url_parsed["query"]; } $rtn['url']['path'] = $path; $host = $url_parsed["host"]; $foundBody = false; $out = "GET $path HTTP/1.0\r\n"; $out .= "Host: $host\r\n"; $out .= "User-Agent: Mozilla/5.0 (Windows; U; Windows NT 5.1; en-US; rv:1.8.1) Gecko/20061010 Firefox/2.0\r\n"; $out .= "Connection: Close\r\n\r\n"; if ( !$fp = @fsockopen($host, $port, $errno, $errstr, 30) ) { $rtn['errornumber'] = $errno; $rtn['errorstring'] = $errstr; } fwrite($fp, $out); while (!@feof($fp)) { $s = @fgets($fp, 128); if ( $s == "\r\n" ) { $foundBody = true; continue; } if ( $foundBody ) { $body .= $s; } else { if ( ($followRedirects) && (stristr($s, "location:") != false) ) { $redirect = preg_replace("/location:/i", "", $s); return httpGet( trim($redirect) ); } $header .= $s; } } fclose($fp); return(trim($body)); } This code sometimes go infinite loop. What's wrong here?

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• NSUserDefaults always returns 0, in Simulator and Device

  - by joec

  I have a settings bundle in my iPhone app, with the following PList: Type: PSMultiValueSpecifier Title: Within (Kilometres) DefaultValue: Number: 1 Titles: Array: Item 1: Number: 1 Item 2: Number: 2 Values: Array: Item 1: Number: 1 Item 2: Number: 2 Key: String: km Then in my code i have: NSUserDefaults *defaults = [NSUserDefaults standardUserDefaults]; NSInteger choiceOne = [defaults integerForKey:@"km"]; choiceOne is always 0. What am i doing wrong? Thanks

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• adding zeros in objective-c string formats

  - by Chris

  Hello- Quick question: I am trying to fill in empty spaces with a specific number of zeroes in an NSString stringWithFormat formatting string. For example, I want: @"The number is %d", 5 // I want this to output 'the number is 05' @"the number is %d", 10 // I want this to output 'the number is 10' I know how to do this in Java, but I cant seem to find the same function in objective-c. Any help would be great.

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• What is the best algorithm for this problem?

  - by mark

  What is the most efficient algorithm to solve the following problem? Given 6 arrays, D1,D2,D3,D4,D5 and D6 each containing 6 numbers like: D1[0] = number D2[0] = number ...... D6[0] = number D1[1] = another number D2[1] = another number .... ..... .... ...... .... D1[5] = yet another number .... ...... .... Given a second array ST1, containing 1 number: ST1[0] = 6 Given a third array ans, containing 6 numbers: ans[0] = 3, ans[1] = 4, ans[2] = 5, ......ans[5] = 8 Using as index for the arrays D1,D2,D3,D4,D5 and D6, the number that goes from 0, to the number stored in ST1[0] minus one, in this example 6, so from 0 to 6-1, compare each res array against each D array My algorithm so far is: I tried to keep everything unlooped as much as possible. EML := ST1[0] //number contained in ST1[0] EML1 := 0 //start index for the arrays D While EML1 < EML if D1[ELM1] = ans[0] goto two if D2[ELM1] = ans[0] goto two if D3[ELM1] = ans[0] goto two if D4[ELM1] = ans[0] goto two if D5[ELM1] = ans[0] goto two if D6[ELM1] = ans[0] goto two ELM1 = ELM1 + 1 return 0 //bad row of numbers, if while ends two: EML1 := 0 start index for arrays Ds While EML1 < EML if D1[ELM1] = ans[1] goto two if D2[ELM1] = ans[1] goto two if D3[ELM1] = ans[1] goto two if D4[ELM1] = ans[1] goto two if D5[ELM1] = ans[1] goto two if D6[ELM1] = ans[1] goto two ELM1 = ELM1 + 1 return 0 three: EML1 := 0 start index for arrays Ds While EML1 < EML if D1[ELM1] = ans[2] goto two if D2[ELM1] = ans[2] goto two if D3[ELM1] = ans[2] goto two if D4[ELM1] = ans[2] goto two if D5[ELM1] = ans[2] goto two if D6[ELM1] = ans[2] goto two ELM1 = ELM1 + 1 return 0 four: EML1 := 0 start index for arrays Ds While EML1 < EML if D1[ELM1] = ans[3] goto two if D2[ELM1] = ans[3] goto two if D3[ELM1] = ans[3] goto two if D4[ELM1] = ans[3] goto two if D5[ELM1] = ans[3] goto two if D6[ELM1] = ans[3] goto two ELM1 = ELM1 + 1 return 0 five: EML1 := 0 start index for arrays Ds While EML1 < EML if D1[ELM1] = ans[4] goto two if D2[ELM1] = ans[4] goto two if D3[ELM1] = ans[4] goto two if D4[ELM1] = ans[4] goto two if D5[ELM1] = ans[4] goto two if D6[ELM1] = ans[4] goto two ELM1 = ELM1 + 1 return 0 six: EML1 := 0 start index for arrays Ds While EML1 < EML if D1[ELM1] = ans[0] return 1 //good row of numbers if D2[ELM1] = ans[0] return 1 if D3[ELM1] = ans[0] return 1 if D4[ELM1] = ans[0] return 1 if D5[ELM1] = ans[0] return 1 if D6[ELM1] = ans[0] return 1 ELM1 = ELM1 + 1 return 0 As language of choice, it would be pure c

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• Code-Golf: Modulus Divide

  - by thyrgle

  Challenge: Without using the modulus divide operator provided already by your language, write a program that will take two integer inputs from a user and then displays the result of the first number modulus divided number by the second number. Example: Input of first number:2 Input of second number:2 Result:0 Who wins: In case you don't know how Code Golf the winner is the person who writes this program in the least amount of characters.

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• Find a repeated numbers out of 3 boxes

  - by james1

  I have 3 boxes, each box contain 10 piece of numbered paper (1 - 10) but there is a number the same in all 3 boxes eg: box1 has number 4 and box2 has number 4 and box3 also has number 4. How to find that repeated number in java with an efficient/fastest way possible?

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• Max value amongst 4 columns in a row.

  - by KandadaBoggu

  I have test_scores table with following fields: Table schema: id (number) score1 (number) score2 (number) score3 (number) score4 (number) Sample data: id score1 score2 score3 score4 1 10 05 30 50 2 05 15 10 00 3 25 10 05 15 Expected result set: id col_name col_value 1 score4 50 2 score2 15 3 score1 25 What is a good SQL for this?(I am using MySQL.)

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• TinyMCE: Looking for line count solution.

  - by Thariama

  I am looking for a plugin or code with a functionality like the "line count" or "line number" in common editors. The line number usually is shown on the left border of the editors content. Anyone got an idea how to do it with TinyMCE? Example: line number | content line number one number two 3. one line skipped (three is empty) 5. contents end

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• Else without if

  - by user2808951

  I'm trying to write a code for my computer programming class for a project due Monday, and I'm pretty new to Java, but I'm trying to write a program that will first determine if a number the user inputs is even or odd and then determine if the number is prime or not. I'm not sure if I did the algorithm right or not, so if anyone has any corrections on the program to my algorithm or anything else please say so, but my real issue is that the program is refusing to compile. Every time I try, it says it's having an else without if problem. Here's a link to my command box: http://s1341.photobucket.com/user/Emi_Nightshade/media/Capture_zps45f9a2ea.png.html Here's my code: import java.io.*; import java.util.*; public class Lesson9p1_ThuotteEmily { public static void main(String args[]) { Scanner kbReader0=new Scanner(System.in); System.out.print("\n\nPlease enter an integer. An integer is whole number, and it can be either negative or positive. Please enter your number: "); long num=kbReader0.nextLong(); if(num%2==0) //if and else with braces { System.out.println("Your integer " + num + " is even."); } else { System.out.println("Your integer " + num + " is odd."); } Scanner kbReader1=new Scanner(System.in); System.out.print("\n\nWould you like to know if your number is prime? Please enter yes or no: "); String yn=kbReader1.nextLine(); if(yn.equals.IgnoreCase("Yes")) { System.out.println("Okay. Give me a moment."); { if(num%2==0) { System.out.println("Your number isn't prime."); } else if(num==2) { System.out.println("Your number is 2, which is the only even prime number in existence. Cool, right?"); } for(int i=3;i*i<=n;i+=2) { if(n%1==0) { System.out.println("Your number isn't prime."); } } else { System.out.println("Your number is prime!"); } } } if(yn.equals.IgnoreCase("No")) { System.out.println("Okay."); } } } If anyone could help me out with this and also any problems I may have made elsewhere in the program, I'd be very grateful! Thanks.

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• Any reliable polygon normal calculation code?

  - by Jenko

  I'm currently calculating the normal vector of a polygon using this code, but for some faces here and there it calculates a wrong normal. I don't really know what's going on or where it fails but its not reliable. Do you have any polygon normal calculation that's tested and found to be reliable? // calculate normal of a polygon using all points var n:int = points.length; var x:Number = 0; var y:Number = 0; var z:Number = 0 // ensure all points above 0 var minx:Number = 0, miny:Number = 0, minz:Number = 0; for (var p:int = 0, pl:int = points.length; p < pl; p++) { var po:_Point3D = points[p] = points[p].clone(); if (po.x < minx) { minx = po.x; } if (po.y < miny) { miny = po.y; } if (po.z < minz) { minz = po.z; } } for (p = 0; p < pl; p++) { po = points[p]; po.x -= minx; po.y -= miny; po.z -= minz; } var cur:int = 1, prev:int = 0, next:int = 2; for (var i:int = 1; i <= n; i++) { // using Newell method x += points[cur].y * (points[next].z - points[prev].z); y += points[cur].z * (points[next].x - points[prev].x); z += points[cur].x * (points[next].y - points[prev].y); cur = (cur+1) % n; next = (next+1) % n; prev = (prev+1) % n; } // length of the normal var length:Number = Math.sqrt(x * x + y * y + z * z); // turn large values into a unit vector if (length != 0){ x = x / length; y = y / length; z = z / length; }else { throw new Error("Cannot calculate normal since triangle has an area of 0"); }

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• Perl syntax error [closed]

  - by Linny

  I am a beginner taking a Perl programming course. We are trying to write a basic program for counting nucleotides in a DNA string. I'm getting syntax errors on the lines that have a single bracket on lines 28 & 70 and don't know why. It also reads that I have compilation errors. I have no idea where to start figuring that out. # The purpose of this program is to count the number of nucleotides in a strand. Each protein is counted separately # print "/n NOTE: Nucleotide counting /n"; # use strict; # enforce variable declarations use warnings; # enable compiler warnings # Display number of A,a,T,t,G,g,C,c, nucleotides in a word or sequence of letters. # my ($base) = ''; # an extracted letter from a string my ($nuceotide_count) = 0 ; # the current position within the word my ($position) = 0 ; # number of vowels in user-supplied word my ($word) = ''; # word to be processed my ($A_count) = 0 ; # of A nucleotides in the user-supplied sequence my ($a_count) = 0 ; # of A nucleotides in the user-supplied sequence my ($C_count) = 0 ; # of C nucleotides in the user-supplied sequence my ($c_count) = 0 ; # of C nucleotides in the user-supplied sequence my ($G_count) = 0 ; # of G nucleotides in the user-supplied sequence my ($g_count) = 0 ; # of G nucleotides in the user-supplied sequence my ($T_count) = 0 ; # of T nucleotides in the user-supplied sequence my ($t_count) = 0 ; # of T nucleotides in the user-supplied sequence word = (STDIN) for ($position = 0);($position if (($base eq 'a') or ($base eq 'A')) { ++$A_count; } # end if ++$position; if (($base eq 'T') or ($base eq 't')) { ++$T_count; } end if ++$position; if (($base eq 'G') or ($base eq 'g')) { ++$G_count; } # end if ++$position; if (($base eq 'C') or ($base eq 'c')) { ++$C_count; } # end if ++$position; } # end for # Display final results. # print " \n The number of A or a neucleotides is: $A_count"; print " \n The number of T or t neucleotides is: $T_count"; print " \n The number of G or g neucleotides is: $G_count"; print " \n The number of C or c neucleotides is: $C_count"; print " \n\n Program completed successfully. \n" ; exit ;

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