reading and writing file in javafx program
I'm writing a program writes and read the file. it has 50 numbers and if users enter number more than 50 it should ask the users to enter it again. my question is for the error it should show the message box but again return to the text field for input. Also one more question, for example, user1 is entering 1 and user2 is also entering 1, it should tell the user2 "please enter another number it's already entered." how can I do it in writing and reading files.
this is my code:
public class JavaFXApplication24 extends Application
private int seatInput;
@Override
public void start(Stage primaryStage)
StackPane pane = new StackPane();
HBox hbox = new HBox();
Label num = new Label("Please enter the number from 1 to 50: ");
TextField numInput = new TextField();
File file = new File("NumberBook.txt");
Button ok = new Button("ok");
ok.setOnAction((new EventHandler<ActionEvent>()
public void handle(final ActionEvent event ) NumberFormatException e)
try
PrintWriter output = new PrintWriter(file);
catch (IOException ex)
System.out.printf("error", ex);
try
Scanner input1 = new Scanner(file);
System.out.printf("Entered: %s", seatInput);
catch (FileNotFoundException ex)
System.out.printf("error", ex);
));
hbox.getChildren().addAll(num,numInput,ok);
pane.getChildren().addAll(hbox);
Scene secondScene = new Scene(pane, 600, 400);
// New window (Stage)
Stage newWindow = new Stage();
newWindow.setTitle("Booking and Timing");
newWindow.setScene(secondScene);
newWindow.show();
javafx
add a comment |
I'm writing a program writes and read the file. it has 50 numbers and if users enter number more than 50 it should ask the users to enter it again. my question is for the error it should show the message box but again return to the text field for input. Also one more question, for example, user1 is entering 1 and user2 is also entering 1, it should tell the user2 "please enter another number it's already entered." how can I do it in writing and reading files.
this is my code:
public class JavaFXApplication24 extends Application
private int seatInput;
@Override
public void start(Stage primaryStage)
StackPane pane = new StackPane();
HBox hbox = new HBox();
Label num = new Label("Please enter the number from 1 to 50: ");
TextField numInput = new TextField();
File file = new File("NumberBook.txt");
Button ok = new Button("ok");
ok.setOnAction((new EventHandler<ActionEvent>()
public void handle(final ActionEvent event ) NumberFormatException e)
try
PrintWriter output = new PrintWriter(file);
catch (IOException ex)
System.out.printf("error", ex);
try
Scanner input1 = new Scanner(file);
System.out.printf("Entered: %s", seatInput);
catch (FileNotFoundException ex)
System.out.printf("error", ex);
));
hbox.getChildren().addAll(num,numInput,ok);
pane.getChildren().addAll(hbox);
Scene secondScene = new Scene(pane, 600, 400);
// New window (Stage)
Stage newWindow = new Stage();
newWindow.setTitle("Booking and Timing");
newWindow.setScene(secondScene);
newWindow.show();
javafx
This is probably not the main issue, but: NeithernumInput.getText().equals(1)
nornumInput.getText().equals(50)
yieldtrue
regardless of the input in theTextField
. ComparingString
toInteger
usingequals
always yieldsfalse
.
– fabian
Nov 12 '18 at 19:21
add a comment |
I'm writing a program writes and read the file. it has 50 numbers and if users enter number more than 50 it should ask the users to enter it again. my question is for the error it should show the message box but again return to the text field for input. Also one more question, for example, user1 is entering 1 and user2 is also entering 1, it should tell the user2 "please enter another number it's already entered." how can I do it in writing and reading files.
this is my code:
public class JavaFXApplication24 extends Application
private int seatInput;
@Override
public void start(Stage primaryStage)
StackPane pane = new StackPane();
HBox hbox = new HBox();
Label num = new Label("Please enter the number from 1 to 50: ");
TextField numInput = new TextField();
File file = new File("NumberBook.txt");
Button ok = new Button("ok");
ok.setOnAction((new EventHandler<ActionEvent>()
public void handle(final ActionEvent event ) NumberFormatException e)
try
PrintWriter output = new PrintWriter(file);
catch (IOException ex)
System.out.printf("error", ex);
try
Scanner input1 = new Scanner(file);
System.out.printf("Entered: %s", seatInput);
catch (FileNotFoundException ex)
System.out.printf("error", ex);
));
hbox.getChildren().addAll(num,numInput,ok);
pane.getChildren().addAll(hbox);
Scene secondScene = new Scene(pane, 600, 400);
// New window (Stage)
Stage newWindow = new Stage();
newWindow.setTitle("Booking and Timing");
newWindow.setScene(secondScene);
newWindow.show();
javafx
I'm writing a program writes and read the file. it has 50 numbers and if users enter number more than 50 it should ask the users to enter it again. my question is for the error it should show the message box but again return to the text field for input. Also one more question, for example, user1 is entering 1 and user2 is also entering 1, it should tell the user2 "please enter another number it's already entered." how can I do it in writing and reading files.
this is my code:
public class JavaFXApplication24 extends Application
private int seatInput;
@Override
public void start(Stage primaryStage)
StackPane pane = new StackPane();
HBox hbox = new HBox();
Label num = new Label("Please enter the number from 1 to 50: ");
TextField numInput = new TextField();
File file = new File("NumberBook.txt");
Button ok = new Button("ok");
ok.setOnAction((new EventHandler<ActionEvent>()
public void handle(final ActionEvent event ) NumberFormatException e)
try
PrintWriter output = new PrintWriter(file);
catch (IOException ex)
System.out.printf("error", ex);
try
Scanner input1 = new Scanner(file);
System.out.printf("Entered: %s", seatInput);
catch (FileNotFoundException ex)
System.out.printf("error", ex);
));
hbox.getChildren().addAll(num,numInput,ok);
pane.getChildren().addAll(hbox);
Scene secondScene = new Scene(pane, 600, 400);
// New window (Stage)
Stage newWindow = new Stage();
newWindow.setTitle("Booking and Timing");
newWindow.setScene(secondScene);
newWindow.show();
javafx
javafx
asked Nov 12 '18 at 19:02
MjahMjah
65
65
This is probably not the main issue, but: NeithernumInput.getText().equals(1)
nornumInput.getText().equals(50)
yieldtrue
regardless of the input in theTextField
. ComparingString
toInteger
usingequals
always yieldsfalse
.
– fabian
Nov 12 '18 at 19:21
add a comment |
This is probably not the main issue, but: NeithernumInput.getText().equals(1)
nornumInput.getText().equals(50)
yieldtrue
regardless of the input in theTextField
. ComparingString
toInteger
usingequals
always yieldsfalse
.
– fabian
Nov 12 '18 at 19:21
This is probably not the main issue, but: Neither
numInput.getText().equals(1)
nor numInput.getText().equals(50)
yield true
regardless of the input in the TextField
. Comparing String
to Integer
using equals
always yields false
.– fabian
Nov 12 '18 at 19:21
This is probably not the main issue, but: Neither
numInput.getText().equals(1)
nor numInput.getText().equals(50)
yield true
regardless of the input in the TextField
. Comparing String
to Integer
using equals
always yields false
.– fabian
Nov 12 '18 at 19:21
add a comment |
1 Answer
1
active
oldest
votes
From a quick glance here are some problems with your code:
1) if ((numInput.getText().equals(1)&& numInput.getText().equals(50)))
. There are 2 problems here. Firstly, as fabian pointed out, this statement will always evaluate to false. Secondly, from your question I think you want the user to enter a number from 1 to 50, but here you're checking if they enter 1 and 50 (which also means it's always false).
2) JOptionPane.showInputDialog("Please enter again")
Because your application is JavaFX, you should use the built in Dialog instead.
Now to answer your questions, you can call requestFocus on the text field right after the error message is displayed. Your second question seems like you want some networking functionality in the application, which adds a whole lot of complexity to it. You might need to do some research on how to do it in Java.
figured out the dialog but still not able if statement
– Mjah
Nov 12 '18 at 22:55
if ( (numInput.getText().length() > 50) && (numInput.getText().length() < 1)) {
– Mjah
Nov 12 '18 at 22:55
please help with the if statement
– Mjah
Nov 12 '18 at 22:56
Convert the text input to int with something like Integer.parseInt then compare to 1 and 50.
– Gnas
Nov 13 '18 at 9:08
if ((Integer.parseInt(numInput.getText()) > 50) || ((Integer.parseInt(numInput.getText()) < 1)))
– Mjah
Nov 14 '18 at 0:15
|
show 2 more comments
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1 Answer
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oldest
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1 Answer
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active
oldest
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oldest
votes
active
oldest
votes
From a quick glance here are some problems with your code:
1) if ((numInput.getText().equals(1)&& numInput.getText().equals(50)))
. There are 2 problems here. Firstly, as fabian pointed out, this statement will always evaluate to false. Secondly, from your question I think you want the user to enter a number from 1 to 50, but here you're checking if they enter 1 and 50 (which also means it's always false).
2) JOptionPane.showInputDialog("Please enter again")
Because your application is JavaFX, you should use the built in Dialog instead.
Now to answer your questions, you can call requestFocus on the text field right after the error message is displayed. Your second question seems like you want some networking functionality in the application, which adds a whole lot of complexity to it. You might need to do some research on how to do it in Java.
figured out the dialog but still not able if statement
– Mjah
Nov 12 '18 at 22:55
if ( (numInput.getText().length() > 50) && (numInput.getText().length() < 1)) {
– Mjah
Nov 12 '18 at 22:55
please help with the if statement
– Mjah
Nov 12 '18 at 22:56
Convert the text input to int with something like Integer.parseInt then compare to 1 and 50.
– Gnas
Nov 13 '18 at 9:08
if ((Integer.parseInt(numInput.getText()) > 50) || ((Integer.parseInt(numInput.getText()) < 1)))
– Mjah
Nov 14 '18 at 0:15
|
show 2 more comments
From a quick glance here are some problems with your code:
1) if ((numInput.getText().equals(1)&& numInput.getText().equals(50)))
. There are 2 problems here. Firstly, as fabian pointed out, this statement will always evaluate to false. Secondly, from your question I think you want the user to enter a number from 1 to 50, but here you're checking if they enter 1 and 50 (which also means it's always false).
2) JOptionPane.showInputDialog("Please enter again")
Because your application is JavaFX, you should use the built in Dialog instead.
Now to answer your questions, you can call requestFocus on the text field right after the error message is displayed. Your second question seems like you want some networking functionality in the application, which adds a whole lot of complexity to it. You might need to do some research on how to do it in Java.
figured out the dialog but still not able if statement
– Mjah
Nov 12 '18 at 22:55
if ( (numInput.getText().length() > 50) && (numInput.getText().length() < 1)) {
– Mjah
Nov 12 '18 at 22:55
please help with the if statement
– Mjah
Nov 12 '18 at 22:56
Convert the text input to int with something like Integer.parseInt then compare to 1 and 50.
– Gnas
Nov 13 '18 at 9:08
if ((Integer.parseInt(numInput.getText()) > 50) || ((Integer.parseInt(numInput.getText()) < 1)))
– Mjah
Nov 14 '18 at 0:15
|
show 2 more comments
From a quick glance here are some problems with your code:
1) if ((numInput.getText().equals(1)&& numInput.getText().equals(50)))
. There are 2 problems here. Firstly, as fabian pointed out, this statement will always evaluate to false. Secondly, from your question I think you want the user to enter a number from 1 to 50, but here you're checking if they enter 1 and 50 (which also means it's always false).
2) JOptionPane.showInputDialog("Please enter again")
Because your application is JavaFX, you should use the built in Dialog instead.
Now to answer your questions, you can call requestFocus on the text field right after the error message is displayed. Your second question seems like you want some networking functionality in the application, which adds a whole lot of complexity to it. You might need to do some research on how to do it in Java.
From a quick glance here are some problems with your code:
1) if ((numInput.getText().equals(1)&& numInput.getText().equals(50)))
. There are 2 problems here. Firstly, as fabian pointed out, this statement will always evaluate to false. Secondly, from your question I think you want the user to enter a number from 1 to 50, but here you're checking if they enter 1 and 50 (which also means it's always false).
2) JOptionPane.showInputDialog("Please enter again")
Because your application is JavaFX, you should use the built in Dialog instead.
Now to answer your questions, you can call requestFocus on the text field right after the error message is displayed. Your second question seems like you want some networking functionality in the application, which adds a whole lot of complexity to it. You might need to do some research on how to do it in Java.
edited Nov 13 '18 at 9:01
answered Nov 12 '18 at 21:12
GnasGnas
50329
50329
figured out the dialog but still not able if statement
– Mjah
Nov 12 '18 at 22:55
if ( (numInput.getText().length() > 50) && (numInput.getText().length() < 1)) {
– Mjah
Nov 12 '18 at 22:55
please help with the if statement
– Mjah
Nov 12 '18 at 22:56
Convert the text input to int with something like Integer.parseInt then compare to 1 and 50.
– Gnas
Nov 13 '18 at 9:08
if ((Integer.parseInt(numInput.getText()) > 50) || ((Integer.parseInt(numInput.getText()) < 1)))
– Mjah
Nov 14 '18 at 0:15
|
show 2 more comments
figured out the dialog but still not able if statement
– Mjah
Nov 12 '18 at 22:55
if ( (numInput.getText().length() > 50) && (numInput.getText().length() < 1)) {
– Mjah
Nov 12 '18 at 22:55
please help with the if statement
– Mjah
Nov 12 '18 at 22:56
Convert the text input to int with something like Integer.parseInt then compare to 1 and 50.
– Gnas
Nov 13 '18 at 9:08
if ((Integer.parseInt(numInput.getText()) > 50) || ((Integer.parseInt(numInput.getText()) < 1)))
– Mjah
Nov 14 '18 at 0:15
figured out the dialog but still not able if statement
– Mjah
Nov 12 '18 at 22:55
figured out the dialog but still not able if statement
– Mjah
Nov 12 '18 at 22:55
if ( (numInput.getText().length() > 50) && (numInput.getText().length() < 1)) {
– Mjah
Nov 12 '18 at 22:55
if ( (numInput.getText().length() > 50) && (numInput.getText().length() < 1)) {
– Mjah
Nov 12 '18 at 22:55
please help with the if statement
– Mjah
Nov 12 '18 at 22:56
please help with the if statement
– Mjah
Nov 12 '18 at 22:56
Convert the text input to int with something like Integer.parseInt then compare to 1 and 50.
– Gnas
Nov 13 '18 at 9:08
Convert the text input to int with something like Integer.parseInt then compare to 1 and 50.
– Gnas
Nov 13 '18 at 9:08
if ((Integer.parseInt(numInput.getText()) > 50) || ((Integer.parseInt(numInput.getText()) < 1)))
– Mjah
Nov 14 '18 at 0:15
if ((Integer.parseInt(numInput.getText()) > 50) || ((Integer.parseInt(numInput.getText()) < 1)))
– Mjah
Nov 14 '18 at 0:15
|
show 2 more comments
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This is probably not the main issue, but: Neither
numInput.getText().equals(1)
nornumInput.getText().equals(50)
yieldtrue
regardless of the input in theTextField
. ComparingString
toInteger
usingequals
always yieldsfalse
.– fabian
Nov 12 '18 at 19:21