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• Sending email in asp .net

  - by program-java

  hii every one, I m creating an project in which Admin is one user I want code or any idea how can Admin send a quotation of any type of products to his customer though email on their email ID and this quotation has to be a print option button bec. after clicking on print button quotation should be print How can i do it?

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• Python Etiquette: Importing Modules

  - by F3AR3DLEGEND

  Say I have two Python modules: module1.py: import module2 def myFunct(): print "called from module1" module2.py: def myFunct(): print "called from module2" def someFunct(): print "also called from module2" If I import module1, is it better etiquette to re-import module2, or just refer to it as module1.module2? For example (someotherfile.py): import module1 module1.myFunct() # prints "called from module1" module1.module2.myFunct() # prints "called from module2" I can also do this: module2 = module1.module2. Now, I can directly call module2.myFunct(). However, I can change module1.py to: from module2 import * def myFunct(): print "called from module1" Now, in someotherfile.py, I can do this: import module1 module1.myFunct() # prints "called from module1"; overrides module2 module1.someFunct() # prints "also called from module2" Also, by importing *, help('module1') shows all of the functions from module2. On the other hand, (assuming module1.py uses import module2), I can do: someotherfile.py: import module1, module2 module1.myFunct() # prints "called from module1" module2.myFunct() # prints "called from module2" Again, which is better etiquette and practice? To import module2 again, or to just refer to module1's importation?

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• Why jquery have problem with onbeforeprint event?

  - by Cesar Lopez

  Hi all, I have the following function. $(function() { $(".sectionHeader:gt(0)").click(function() { $(this).next(".fieldset").slideToggle("fast"); }); $("img[alt='minimize']").click(function(e) { $(this).closest("table").next(".fieldset").slideUp("fast"); e.stopPropagation(); return false; }); $("img[alt='maximize']").click(function(e) { $(this).closest("table").next(".fieldset").slideDown("fast"); e.stopPropagation(); return false; }); }); <script type="text/javascript"> window.onbeforeprint = expandAll; function expandAll(){ $(".fieldset:gt(0)").slideDown("fast"); } </script> For this html <table class="sectionHeader" ><tr ><td>Heading 1</td></tr></table> <div style="display:none;" class="fieldset">Content 1</div> <table class="sectionHeader" ><tr ><td>Heading 2</td></tr></table> <div style="display:none;" class="fieldset">Content 2</div> I have several div class="fieldset" over the page, but when I do print preview or print, I can see all divs sliding down before opening the print preview or printing but on the actual print preview or print out they are all collapse. I would appreciate if anyone comes with a solution for this. Anyone have any idea why is this or how to fix it? Thanks. PS:Using a does not work either ( I assume because jquery using toggle) and its not the kind of question I am looking for.

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• Question about multiple 'catch'

  - by chun

  Can anyone tell me why the output of this class is 'xa'? why the other exception won't be caught? public class Tree { public static void main(String... args){ try { throw new NullPointerException(new Exception().toString()); } catch (NullPointerException e) { System.out.print("x"); } catch (RuntimeException e) { System.out.print("y"); } catch (Exception e) { System.out.print("z"); } finally{System.out.println("a");} } }

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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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• variable being weirdly deleted

  - by calccrypto

  im having a weird problem with one variable: its not being recognized but its still printing. i would post my code, but it is massive. the basic idea is: # pseudocode def function(stuff): <do stuff> # These are the only 2 conditions if tag == 3: pka = <a string> if tag == 4: pka = <a string> print pka # (1) print pka # (2) <do stuff not modifying pka> print pka # (3) if pka == 'RSA': <do stuff> elif pka == 'DSA': <do stuff> my code will error at (2). however, it will print out (1), (2), and (3), all of which are the same. is there any general explanation of why this is happening? if my code is really needed, i will post it, but otherwise, i would rather not due to its size update: now the code will error at the if statement after (3), saying UnboundLocalError: local variable 'pka' referenced before assignment even though (1),(2),(3) just printed

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• issue running a batch script to kill a process

  - by user657064

  I am using the following script on a command line to kill a hypothetical notepad process (using a Korn shell in Windows XP, if that matters): kill $(tasklist | grep -i notepad.exe | awk '{print 2}') Now I take this line, and put it into a batch file c:\temp\testkill.bat, thinking that I should just as well be able to kill the process by running the batch file. However, when I run the batch file, I get the following awk error about unbalanced parentheses: C:/Temp ./testkill.bat C:\Tempkill $(tasklist | grep -i notepad.exe | awk '{print $2}') awk: unbalanced () Context is: {print $2}) <<< C:/Temp So I'm baffled as to why I am getting this error about unbalanced parentheses when I run this script via a batch file, but have no issues when I run the command directly from the command line? (Btw, I'm not necessarily tied to this way of killing a process - as a total noob to shell scripting, I am additionally wondering why if I write the following on the command line: tasklist | grep -i notepad.exe | awk '{print $2}' | kill the process ID that comes out of the tasklist/grep/awk calls doesn't seem to properly get piped to kill...)

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• Which faces technology for use with GlassFish 2.1 and NetBeans 6.7?

  - by SteJav

  I'm running GlassFish 2.1 and using NetBeans 6.7. I'd like to create a web interface to my data using JSF 1.2. Trouble is, I'm not sure which 'faces' technology to learn (that includes some good documentation). JBoss/RichFaces seem pretty good on documentation, but I'm using GlassFish. Any thoughts? The choices appear overwhelming: Tomahawk Tobago Trinidad ICEfaces RCFaces Netadvantage WebGalileoFaces QuipuKit BluePrints Woodstock JBoss RichFaces Ajax4jsf ILOG Oracle ADF G4JSF Simplica Backbase jenia4faces VisualWebPack DynaFaces IBM Impl Dinamica Mojarra PrimeFaces jQuery OpenFaces ZK ExtJS Anybody had any experience with any of the above and found the documentation to be clear to a beginner? Being a JSF/Web beginner, I tried some ICEFaces, Mojarra tutorials and had a go at getting RichFaces working with NBeans and GlassFish, but no luck. Lots of XML complaints. I'm clearly missing some huge chunks of configuration, but I can't find any documentation to help me. Any suggestions would be much appreciated :-)

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• Why will this for loop not return one field from list rather than the list?

  - by Dick Eshelman

  import csv """sample row = 10/6/2010,73.42,74.43,72.9,74.15,2993500""" filename_in = 'c:/python27/scripts/fiverows.csv' reader = csv.reader(open(filename_in, "rb"), dialect="excel", delimiter="\t", quoting =csv.QUOTE_MINIMAL) for row in reader: for item in row: print 'row = ',row print 'item = ', item When you run this script and print the row you get the sample row returned in [] as a list. When you print the item you get the sample row as an unquoted string. Why do I not get each field ie, (10/6/2010), (73.42), etc. returned as an item? How do I return a single item?

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• What's wrong with this statement in perl?

  - by benjamin button

  print "$_", join(',',sort keys %$h),"\n"; It's giving me an error below: Use of uninitialized value in string at missing_months.pl line 36. 1,10,11,12 this print statement is present in a for loop as below: foreach my $num ( sort keys %hash ) { my $h = $hash{$num}; print "$_", join(',',sort keys %$h),"\n"; }

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• It says i have an indented block when i dont?

  - by user3728373

  def cave(): global key global response print(''' You find yourself standing infront of a cave. You venture into the cave to find a large door blocking your path. (insert key, turn around''') response = input("Enter a command: ") while response != 'insert key' or response != 'turn around': if response =='insert key' or response == 'turn around': break print('Choose one of the options: ") response = input() if response == 'insert key': if key == 1: win() else: print('''You don't have a key. Get One!!''') elif response == 'turn around' : home()

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• python multiprocessing.Process.Manager not producing consistent results?

  - by COpython

  I've written the following code to illustrate the problem I'm seeing. I'm trying to use a Process.Manager.list() to keep track of a list and increment random indices of that list. Each time there are 100 processes spawned, and each process increments a random index of the list by 1. Therefore, one would expect the SUM of the resulting list to be the same each time, correct? I get something between 203 and 205. from multiprocessing import Process, Manager import random class MyProc(Process): def __init__(self, A): Process.__init__(self) self.A = A def run(self): i = random.randint(0, len(self.A)-1) self.A[i] = self.A[i] + 1 if __name__ == '__main__': procs = [] M = Manager() a = M.list(range(15)) print('A: {0}'.format(a)) print('sum(A) = {0}'.format(sum(a))) for i in range(100): procs.append(MyProc(a)) map(lambda x: x.start(), procs) map(lambda x: x.join(), procs) print('A: {0}'.format(a)) print('sum(A) = {0}'.format(sum(a)))

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• R matrix handling expressions aren't evaluated in a function or a "for" loop - Column extract doesn't seem to work

  - by Sal Leggio

  I have an R matrix named ddd. When I enter this, everything works fine: i <- 1 shapiro.test(ddd[,y]) ad.test(ddd[,y]) stem(ddd[,y]) print(y) The calls to Shapiro Wilk, Anderson Darling, and stem all work, and extract the same column. If I put this code in a "for" loop, the calls to Shapiro Wilk, and Anderson Darling stop working, while the the stem & leaf call and the print call continue to work. for (y in 7:10) { shapiro.test(ddd[,y]) ad.test(ddd[,y]) stem(ddd[,y]) print(y) } The decimal point is 1 digit(s) to the right of the | 0 | 0 0 | 899999 1 | 0 [1] 7 The same thing happens if I try and write a function. SW & AD do not work. The other calls do. D <- function (y) { + shapiro.test(ddd[,y]) + ad.test(ddd[,y]) + stem(ddd[,y]) + print(y) } D(9) The decimal point is at the | 9 | 000 9 | 10 | 00000 [1] 9 Why don't all the calls behave the same way?

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• python - returns incorrect positive #

  - by tekknolagi

  what i'm trying to do is write a quadratic equation solver but when the solution should be -1, as in quadratic(2, 4, 2) it returns 1 what am i doing wrong? #!/usr/bin/python import math def quadratic(a, b, c): #a = raw_input("What\'s your `a` value?\t") #b = raw_input("What\'s your `b` value?\t") #c = raw_input("What\'s your `c` value?\t") a, b, c = float(a), float(b), float(c) disc = (b*b)-(4*a*c) print "Discriminant is:\n" + str(disc) if disc = 0: root = math.sqrt(disc) top1 = b + root top2 = b - root sol1 = top1/(2*a) sol2 = top2/(2*a) if sol1 != sol2: print "Solution 1:\n" + str(sol1) + "\nSolution 2:\n" + str(sol2) if sol1 == sol2: print "One solution:\n" + str(sol1) else: print "No solution!" EDIT: it returns the following... import mathmodules mathmodules.quadratic(2, 4, 2) Discriminant is: 0.0 One solution: 1.0

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• For improving the join of two wave files

  - by kaki

  i want to get the values of the last 30 frames of the first wav file and first thirty frames of the second wave file in integer format and stored in a list or array. i have written the code for joining but during this manupalation i am getting in byte format and tried to convert it to integer but couldn't. as told before i want to get the frame detail of 1st 30 and last 30 in integer format,and by performing other operations join can be more successful looking for your help in this,please... thanking you, import wave m=['C:/begpython/S0001_0002.wav', 'C:/begpython/S0001_0001.wav'] i=1 a=m[i] infiles = [a] outfile = "C:/begpython/S0001_00367.wav" data= [] data1=[] for infile in infiles: w = wave.open(infile, 'rb') data1=[w.getnframes] #print w.readframes(100) data.append( [w.getparams(), w.readframes(w.getnframes())] ) #print w.readframes(1) #data1 = [ord(character) for character in data1] #print data1 #data1 = ''.join(chr(character) for character in data1) w.close() print data output = wave.open(outfile, 'wb') output.setparams(data[0][0]) output.writeframes(data[0][1]) output.writeframes(data[1][1]) output.writeframes(data[2][1]) output.close()

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• Is there a difference between `==` and `is` in python?

  - by Bernard

  My Google-fu has failed me. In Python, are these: n = 5 # Test one. if n == 5: print 'Yay!' # Test two. if n is 5: print 'Yay!' two tests for equality equivalent (ha!)? Does this hold true for objects where you would be comparing instances (a list say)? Okay, so this kind of answers my question: l = list() l.append(1) if l == [1]: print 'Yay!' # Holds true, but... if l is [1]: print 'Yay!' # Doesn't. So == tests value where is tests to see if they are the same object?

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• How to lazy load a data structure (python)

  - by Anton Geraschenko

  I have some way of building a data structure (out of some file contents, say): def loadfile(FILE): return # some data structure created from the contents of FILE So I can do things like puppies = loadfile("puppies.csv") # wait for loadfile to work kitties = loadfile("kitties.csv") # wait some more print len(puppies) print puppies[32] In the above example, I wasted a bunch of time actually reading kitties.csv and creating a data structure that I never used. I'd like to avoid that waste without constantly checking if not kitties whenever I want to do something. I'd like to be able to do puppies = lazyload("puppies.csv") # instant kitties = lazyload("kitties.csv") # instant print len(puppies) # wait for loadfile print puppies[32] So if I don't ever try to do anything with kitties, loadfile("kitties.csv") never gets called. Is there some standard way to do this? After playing around with it for a bit, I produced the following solution, which appears to work correctly and is quite brief. Are there some alternatives? Are there drawbacks to using this approach that I should keep in mind? class lazyload: def __init__(self,FILE): self.FILE = FILE self.F = None def __getattr__(self,name): if not self.F: print "loading %s" % self.FILE self.F = loadfile(self.FILE) return object.__getattribute__(self.F, name) What might be even better is if something like this worked: class lazyload: def __init__(self,FILE): self.FILE = FILE def __getattr__(self,name): self = loadfile(self.FILE) # this never gets called again # since self is no longer a # lazyload instance return object.__getattribute__(self, name) But this doesn't work because self is local. It actually ends up calling loadfile every time you do anything.

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• SFINAE and detecting if a C++ function object returns void.

  - by Tom Swirly

  I've read the various authorities on this, include Dewhurst and yet haven't managed to get anywhere with this seemingly simple question. What I want to do is to call a C++ function object, (basically, anything you can call, a pure function or a class with ()), and return its value, if that is not void, or "true" otherwise. #include <stdio.h> struct Foo { void operator()() {} }; struct Bar { bool operator()() { return false; } }; Foo foo; Bar bar; bool baz() { return false; } void bang() {} const char* print(bool b) { printf(b ? "true, " : "false, "); } template <typename Functor> bool magicCallFunction(Functor f) { return true; // lots of template magic occurs here... } int main(int argc, char** argv) { print(magicCallFunction(foo)); print(magicCallFunction(bar)); print(magicCallFunction(baz)); print(magicCallFunction(bang)); printf("\n"); }

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• Error compiling basic java code

  - by Michael Younani

  New to java. Practicing coding by following a book. Heres my code: class Motorcycle { //Three instance variables - make and color are strings. while a boolean refers to TRUE OR FLASE(in this case off or on) String make; String color; boolean engineState; void startEngine() { if (engineState == true) System.out.print("The engine is already on."); else { engineState = true; System.out.print("The engine is now on."); } void showAtts() { System.out.print("This motorcycle is a " + color + " " + make); if (engineState ==true) System.out.print("The engine is on."); else System.out.print("The engine is off."); } } } When I compile I get 2 errors: 1) illegal start of expression 2) ; expected I can't pin point the problem. If anyone can direct me or hint me please do.

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• regular expression not behaving as expected - Python

  - by philippe

  I have the following function which is supposed to read a .html file and search for <input> tags, and inject a <input type='hidden' > tag into the string to be shown into the page. However, that condition is never met:( e.g the if statement is never executed. ) What's wrong with my regex? def print_choose( params, name ): filename = path + name f = open( filename, 'r' ) records = f.readlines() print "Content-Type: text/html" print page = "" flag = True for record in records: if re.match( '<input*', str(record) ) != None: print record page += record page += "<input type='hidden' name='pagename' value='psychology' />" else: page += record print page Thank you

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• Archive tar files to a different location inperl

  - by user314261

  Hi, I am a newbee in Perl. I am reading a directory having some archive files and uncompressing the archive files one by one. Everything seems well however the files are getting uncompressed in the folder which has the main perl code module which is running the sub modules. I want the archive to be generated in the folder I specify. This is my code: sub ExtractFile { #Read the folder which was copied in the output path recursively and extract if any file is compressed my $dirpath = $_[0]; opendir(D, "$dirpath") || die "Can't open dir $dirpath: $!\n"; my @list = readdir(D); closedir(D); foreach my $f (@list) { print " \$f = $f"; if(-f $dirpath."/$f") { #print " File in directory $dirpath \n ";#is \$f = $f\n"; my($file_name, $file_dirname,$filetype)= fileparse($f,qr{\..*}); #print " \nThe file extension is $filetype"; #print " \nThe file name is is $file_name"; # If compressed file then extract the file if($filetype eq ".tar" or $filetype eq ".tzr.gz") { my $arch_file = $dirpath."/$f"; print "\n file to be extracted is $arch_file"; my $tar = Archive::Tar->new($arch_file); #$tar->extract() or die ("Cannot extract file $arch_file"); #mkdir($dirpath."/$file_name"); $tar->extract_file($arch_file,$dirpath."/$file_name" ) or die ("Cannot extract file $arch_file"); } } if(-d $dirpath."/$f") { if($f eq "." or $f eq "..") { next; } print " Directory\n";# is $f"; ExtractFile($dirpath."/$f"); } } } The method ExtractFile is called recursively to loop all the archives. When using $tar-extract() it uncompresses in the folder which calls this metohd. when I use $tar-extract_file($arch_file,$dirpath."/$file_name" ) I get an error : No such file in archive: '/home/fsang/dante/workspace/output/s.tar' at /home/fsang/dante/lib/Extraction.pm line 80 Please help I have checked that path and input output there is no issue with it. Seems some usage problem I am not aware of for $tar-extract_file(). Many thanks for anyone resolving this issue. Regards, Sakshi

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• Find all occurrences of a substring in Python

  - by cru3l

  Python has string.find() and string.rfind() to get the index of a substring in string. I wonder, maybe there is something like string.find_all() which can return all founded indexes (not only first from beginning or first from end)? For example: string = "test test test test" print string.find('test') # 0 print string.rfind('test') # 15 #that's the goal print string.find_all('test') # [0,5,10,15]

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• How to write curiously recurring templates with more than 2 layers of inheritance?

  - by Kyle

  All the material I've read on Curiously Recurring Template Pattern seems to one layer of inheritance, ie Base and Derived : Base<Derived>. What if I want to take it one step further? #include <iostream> using std::cout; template<typename LowestDerivedClass> class A { public: LowestDerivedClass& get() { return *static_cast<LowestDerivedClass*>(this); } void print() { cout << "A\n"; } }; template<typename LowestDerivedClass> class B : public A<LowestDerivedClass> { public: void print() { cout << "B\n"; } }; class C : public B<C> { public: void print() { cout << "C\n"; } }; int main() { C c; c.get().print(); // B b; // Intentionally bad syntax, // b.get().print(); // to demonstrate what I'm trying to accomplish return 0; } How can I rewrite this code to compile without errors (and output "C\nB\n")?

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• Login with Google OAuth api and return url with variable

  - by user2833721

  I am using the Google API for login in my site and I appending my variable with URL and I want that variable in the return URL of OAuth API because of update purpose can I return back that variable For example: <a href="<?php echo($authUrl); ?>&kicker"> I append the kicker in $authUrl and when I return back from Oauth api print $me['displayName']; print $user['email']; print $me['gender']; print $me['id']; with this output I also want my variable "kicker" how can I get it

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• Concatenate, sort and swap array in Java

  - by sblck

  I am trying to concatenate two arrays into new array, sort in order, and swap two values of index. I'm kind of new to java and only use C before so having a hard time handling an Object. In main method it declares two object arrays IntVector vector = new IntVector(3); and IntVector vector2 = new IntVector(3); I can only do this if the types are int[], but I want to use as an object How should I code the concat, sort, and swap method? public class IntVector { private int[] items_; private int itemCount_; private IntVector(int[] data, int n) { items_ = data.clone(); itemCount_ = n; } public IntVector(int itemSize) { itemCount_ =0; if(itemSize<1) itemSize =10; items_ = new int[itemSize]; } public void push(int value) { if(itemCount_ + 1 >= items_.length) overflow(); items_[itemCount_++] = value; } public void log() { for (int i=0 ; i<itemCount_; ++i) { System.out.print(items_[i]); if(i<itemCount_ -1) System.out.println(); } } public void overflow() { int[] newItems = new int[items_.length * 2]; for(int i=0 ; i<itemCount_; ++i) { newItems[i] = items_[i]; } items_=newItems; } public int getValue(int index) { if(index < 0 || index >= itemCount_) { System.out.println("[error][IntVector][setValue] Incorrect index=" + index); return 0; } return items_[index]; } public void setValue(int index, int value) { if(index < 0 || index >= itemCount_) { System.out.println("[error][IntVector][setValue] Incorrect index=" + index); return ; } items_[index] = value; } public IntVector clone() { return new IntVector(items_, itemCount_); } public IntVector concat() { return null; } public IntVector sort() { return null; } public IntVector swap() { return null; } public static void main(String[] args) { IntVector vector = new IntVector(3); IntVector vector2 = new IntVector(3); vector.push(8); vector.push(200); vector.push(3); vector.push(41); IntVector cloneVector = vector.clone(); vector2.push(110); vector2.push(12); vector2.push(7); vector2.push(141); vector2.push(-32); IntVector concatResult = vector.concat(vector2); IntVector sortResult = concatResult.sort(); IntVector swapResult = sortResult.clone(); //swapResult.swap(1,5); System.out.print("vector : "); vector.log(); System.out.print("\n\ncloneVector : "); cloneVector.log(); System.out.print("\n\nvector2 : "); vector2.log(); System.out.print("\n\nconcatvector : "); concatResult.log(); System.out.print("vector : "); vector.log(); System.out.print("vector : "); vector.log(); } }

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