how we can exit from program by hitting any key
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From: "beginner_in->" on 15 May 2010 02:13 hello there, I want to exit from main class by pressing any key.... the code is in this way:- public class mainClass { public static void main(String[]ar) { //...some code here... Scanner in = new Scanner(System.in); int choice = in.nextInt(); //collecting user input here if(choice == 1) { td1.C2F(); } else if (choice == 2) { td2.F2C(); } else //if any key press program should be terminate. { System.out.println("<- You select to exit - >"); //results in error java.util.InputMismatch exception //due to Strongly Type characteristic } }//end of main }//end of public class so please, suggest me what should I do to exit from the class by pressing any key. i m using jdk1.5.0_22 -Thanks
From: Alan Malloy on 15 May 2010 03:51 beginner_in-> wrote: > hello there, > > I want to exit from main class by pressing any key.... the code is in > this way:- > > > public class mainClass > { > public static void main(String[]ar) > { > //...some code here... > > Scanner in = new Scanner(System.in); > int choice = in.nextInt(); //collecting user > input here > if(choice == 1) > { > td1.C2F(); > } > else if (choice == 2) > { > td2.F2C(); > } > else //if any key press program should be > terminate. > { > > System.out.println("<- You select to exit - >> "); > //results in error java.util.InputMismatch exception > //due to Strongly Type characteristic > } > }//end of main > }//end of public class > > > so please, suggest me what should I do to exit from the class by > pressing any key. i m using jdk1.5.0_22 > > -Thanks The scanner is complaining that if you type, say, "J", it can't turn that into an integer: true enough. The simplest way to solve this is read the user's input as a string, and check to see whether it is a number. If so, you can convert it to a number with Java's built-in methods and then process it; if not, you can exit the program. -- Cheers, Alan (San Jose, California, USA)
From: "beginner_in->" on 15 May 2010 13:14 On May 15, 12:51 pm, Alan Malloy <alan.NO.S...(a)malloys.org> wrote: > beginner_in-> wrote: > > hello there, > > > I want to exit from main class by pressing any key.... the code is in > > this way:- > > > public class mainClass > > { > > public static void main(String[]ar) > > { > > //...some code here... > > > Scanner in = new Scanner(System.in); > > int choice = in.nextInt(); //collecting user > > input here > > if(choice == 1) > > { > > td1.C2F(); > > } > > else if (choice == 2) > > { > > td2.F2C(); > > } > > else //if any key press program should be > > terminate. > > { > > > System.out.println("<- You select to exit - > >> "); > > //results in error java.util.InputMismatch exception > > //due to Strongly Type characteristic > > } > > }//end of main > > }//end of public class > > > so please, suggest me what should I do to exit from the class by > > pressing any key. i m using jdk1.5.0_22 > > > -Thanks > > The scanner is complaining that if you type, say, "J", it can't turn > that into an integer: true enough. The simplest way to solve this is > read the user's input as a string, and check to see whether it is a > number. If so, you can convert it to a number with Java's built-in > methods and then process it; if not, you can exit the program. > > -- > Cheers, > Alan (San Jose, California, USA) i have follow your suggestion ... but i wanna show you .. int ch = Integer.parseInt(choice); System.out.println("String has converted: " +ch); if(ch == 1 || ch == 2) { //inner if block if(ch == 1) { td1.getVal(37.4); res = td1.C2F(); System.out.println("Resulted Temperature: "+res); } // start of else-if block else if(ch == 2) { td1.getVal(104.2); res = td1.F2C(); System.out.println("Resulted Temperature: "+res); } }//end of outer if and starting of else else { System.out.println("<- You select to termination of program ->"); System.exit(0); } is that what you envisage about.... now i m not able to exit from any key input but on pressing numeric key. Sir Should I again convert this into String to exit from the menu. Thanks! -Niks
From: Lew on 15 May 2010 19:17 Alan Malloy wrote: >> The scanner is complaining that if you type, say, "J", it can't turn >> that into an integer: true enough. The simplest way to solve this is >> read the user's input as a string, and check to see whether it is a >> number. If so, you can convert it to a number with Java's built-in >> methods and then process it; if not, you can exit the program. beginner_in-> wrote: >> -- >> Cheers, >> Alan (San Jose, California, USA) Please don't quote sigs. > i [sic] have follow your suggestion ... but i [sic] wanna show you .. What happened to the part of the code where you invoke the Scanner method? We can't tell if you took Alan's advice not to use 'Scanner#nextInt()'. You should provide a complete example as explained at <http://sscce.org/> because these code fragments don't tell enough of the story for us to help you. Now for some comments not related directly to your question. Disregard them if you will, for now, but in the long run they'll help your coding. > int ch = Integer.parseInt(choice); > System.out.println("String has converted: " +ch); > if(ch == 1 || ch == 2) > { //inner if block You want to lighten up on that indentation a little, there, sport? Four spaces is about the maximum per indent level in a Usenet post if you wish to keep it readable, and thus keep those interested who might help you. Your code would be simpler and more maintainable if you used a 'switch' instead of these complicated 'if' blocks: switch ( ch ) { case 1: whateverOne(); break; case 2: whateverTwo(); break; } > if(ch == 1) > { There are two popular conventions for brace placement in Java. The official one places the opening brace at the end of the conditional introducing a block, e.g, if ( ch == 1 ) { The more readable one puts the opening brace at the same level as the introductory conditional. if ( ch == 1 ) { Both put the closing brace at the same level as the conditional. if ( ch == 1 ) { ... } or if ( ch == 1 ) { ... } > td1.getVal(37.4); Magic number. What's 37.4? > res = td1.C2F(); The Java coding conventions call for method names (that /is/ a method call, right?) and non-constant variable names to begin with a lower-case letter. <http://java.sun.com/docs/codeconv/index.html> > System.out.println("Resulted Temperature: "+res); > } // start of else-if block > else if(ch == 2) > { > td1.getVal(104.2); Magic number. What's 104.2? > res = td1.F2C(); > System.out.println("Resulted Temperature: "+res); > } > > }//end of outer if and starting of else Comments are supposed to add to understanding of the code. A closing brace on an 'if' followed by an 'else' already informs the maintainer that you are at the end of an 'if' and starting an 'else'. > else > { > > System.out.println("<- You select to termination of program ->"); > System.exit(0); > } > > is that what you envisage about.... > > now i [sic] m not able to exit from any key input but on pressing numeric > key. Sir Should I again convert this into String to exit from the > menu. Back to your main question: Provide a short, self-contained compilable example (SSCCE) as instructed at <http://sscce.org/> and we will find the part of the code that you have not shown here that is the source of your trouble - or perhaps you'll find it for yourself as you construct the example. I suspect that you are still using the 'Scanner#nextInt()' method and that is the problem, but I don't know. -- Lew
From: "beginner_in->" on 16 May 2010 07:25
On May 16, 4:17 am, Lew <no...(a)lewscanon.com> wrote: > Alan Malloy wrote: > >> The scanner is complaining that if you type, say, "J", it can't turn > >> that into an integer: true enough. The simplest way to solve this is > >> read the user's input as a string, and check to see whether it is a > >> number. If so, you can convert it to a number with Java's built-in > >> methods and then process it; if not, you can exit the program. > beginner_in-> wrote: > >> -- > >> Cheers, > >> Alan (San Jose, California, USA) > > Please don't quote sigs. > > > i [sic] have follow your suggestion ... but i [sic] wanna show you .. > > What happened to the part of the code where you invoke the Scanner method? We > can't tell if you took Alan's advice not to use 'Scanner#nextInt()'. > > You should provide a complete example as explained at > <http://sscce.org/> > because these code fragments don't tell enough of the story for us to help you. > > Now for some comments not related directly to your question. Disregard them > if you will, for now, but in the long run they'll help your coding. > > > int ch = Integer.parseInt(choice); > > System.out.println("String has converted: " +ch); > > if(ch == 1 || ch == 2) > > { //inner if block > > You want to lighten up on that indentation a little, there, sport? > > Four spaces is about the maximum per indent level in a Usenet post if you wish > to keep it readable, and thus keep those interested who might help you. > > Your code would be simpler and more maintainable if you used a 'switch' > instead of these complicated 'if' blocks: > > switch ( ch ) > { > case 1: > whateverOne(); > break; > case 2: > whateverTwo(); > break; > } > > > if(ch == 1) > > { > > There are two popular conventions for brace placement in Java. The official > one places the opening brace at the end of the conditional introducing a > block, e.g, > > if ( ch == 1 ) { > > The more readable one puts the opening brace at the same level as the > introductory conditional. > > if ( ch == 1 ) > { > > Both put the closing brace at the same level as the conditional. > > if ( ch == 1 ) { > ... > } > > or > if ( ch == 1 ) > { > ... > } > > > td1.getVal(37.4); > > Magic number. What's 37.4? > > > res = td1.C2F(); > > The Java coding conventions call for method names (that /is/ a method call, > right?) and non-constant variable names to begin with a lower-case letter.. > > <http://java.sun.com/docs/codeconv/index.html> > > > System.out.println("Resulted Temperature: "+res); > > } // start of else-if block > > else if(ch == 2) > > { > > td1.getVal(104.2); > > Magic number. What's 104.2? > > > res = td1.F2C(); > > System.out.println("Resulted Temperature: "+res); > > } > > > }//end of outer if and starting of else > > Comments are supposed to add to understanding of the code. A closing brace on > an 'if' followed by an 'else' already informs the maintainer that you are at > the end of an 'if' and starting an 'else'. > > > else > > { > > > System.out.println("<- You select to termination of program ->"); > > System.exit(0); > > } > > > is that what you envisage about.... > > > now i [sic] m not able to exit from any key input but on pressing numeric > > key. Sir Should I again convert this into String to exit from the > > menu. > > Back to your main question: > > Provide a short, self-contained compilable example (SSCCE) as instructed at > <http://sscce.org/> > and we will find the part of the code that you have not shown here that is the > source of your trouble - or perhaps you'll find it for yourself as you > construct the example. > > I suspect that you are still using the 'Scanner#nextInt()' method and that is > the problem, but I don't know. > > -- > Lew Respected Sir, thanks for your suggestion to go through http://sscce.org/ i have got the concept of sscce. Sir, i want to tell you that i am accessing this Usenet group from google group service... so i could not able to guess the out look of my given code snippet at usenet group platform. I would like to draw your attention the word [sic] that you mostly use before each 'i' in quoted words. I still don't know the meaning of [sic] as I m possibly not the countryman of your place, however, your each useful advise I would like to follow to remove objections and errors to learn the concepts effectively. This is the thing that matters to me and all other things come on second priority so please avoid those words which are of no business, here. About the erroneous code, i have solved the problem using hasNextInt() and nextInt(). I also want to know why you are not in favour to use Scanner#nextInt() method. -Niks |