Planetarion Forums

Androme · 5 Dec 2006, 17:57

This is procedural before you ask. We're starting object orientated next term.

The problem I'm having with this is that when you answer yes to "do you have a headache?" and then yes to the next "do you have a rash"- you get given "ring 999" message. But then it goes back and asks "do you have a rash?" - i don't want it to ask that, I just want it to go back to the start of the program.

I've tried adding breaks in two of the questions but this doesn't seem to work.

Any help is greatly appreciated - thanks.

Here is the coding:

Code:

problem fixed :D

SpaceMonkey · 6 Dec 2006, 00:19

The trouble is the condition on your loops in the questions:

while (!answer[1].equalsIgnoreCase("yes") || !answer[1].equalsIgnoreCase("no"))

it should be && because:
!answer[1].equalsIgnoreCase("yes") == false
!answer[1].equalsIgnoreCase("no") == true
false || true == true
but
false && true == false

alternatively you could use:
while (! (answer[1].equalsIgnoreCase("yes") || answer[1].equalsIgnoreCase("no")) )

This might help or just be confusing, I don't know, but it's because of De Morgans Law:
!(a || b) === !a && !b

Androme · 6 Dec 2006, 18:23

Cheers SpaceMonkey - that helped solve the problem!!!

However, I noticed something else just during testing. When I type in "exit" as players name to stop the program going through all the questions it adds "exit" to the names arraylist - I don't want exit to appear in the arraylist. I realise I have to 'remove' exit from the arraylist but I'm not sure how to search an arraylist for an item and then remove it.

Do I have to go through "i<names.length" and do "if names[i] = "exit",
names.remove[i]" kind of thing?

Also, this doesn't seem to work as I get these 2 errors:

>> "cannot find symbol variable length"
>> "array required, but java.util.ArrayList found"

Code:

for (int i=1; i<names.length; i++)
{
    System.out.println("Name: " + names[i] + "\nAnswer: " + ans);
}

Does this mean I cannot look at the length of the arraylist because in theory it is infinate? Does that mean .length can only be used on arrays? :|

Androme · 6 Dec 2006, 23:33

Also, how can I link 2 arraylists? Is a LinkedList the solution?

I have a names arraylist and an answers arraylist (one answer per person). I'm going to run an algorithm using the comparator to sort the names arraylist into alphabetical order but that means the arraylist will no longer be in the same position as the players name.

Wandows · 7 Dec 2006, 00:10

Although i have no idea what your actual code looks like and it has been a while since i fooled around in Java, i'll give it a go...

Quote:

Originally Posted by Androme2

Cheers SpaceMonkey - that helped solve the problem!!!

However, I noticed something else just during testing. When I type in "exit" as players name to stop the program going through all the questions it adds "exit" to the names arraylist - I don't want exit to appear in the arraylist. I realise I have to 'remove' exit from the arraylist but I'm not sure how to search an arraylist for an item and then remove it.

Do I have to go through "i<names.length" and do "if names[i] = "exit",
names.remove[i]" kind of thing?

This doesn't make a whole lot of sense in a program. How about instead of adding every player name submitted to the names list by default, you first check for the values you don't want to be added and simply don't allow them to be added at all (and immediatly start the desired procedure on the exception instead of doing 'extra' work).

i.e.

Code:

if( entered_name == 'exit' ) {
  this.startExitProcedure();
} else {
  namesList.Add( New name());
}

Quote:

Also, this doesn't seem to work as I get these 2 errors:

>> "cannot find symbol variable length"
>> "array required, but java.util.ArrayList found"

Code:

for (int i=1; i<names.length; i++)
{
    System.out.println("Name: " + names[i] + "\nAnswer: " + ans);
}

Does this mean I cannot look at the length of the arraylist because in theory it is infinate? Does that mean .length can only be used on arrays? :|

In Java there is a difference between Array and Arraylist. Going by those error messages your names list is an Arraylist, which is resizable array/list of objects rather than a normal static array of values. Therefor you have to use a different technique to access the values. Meaning you have to use names.size() to get the amount of elements in the nameslist and names.get(i) to get the object from a specific index and then use.

Quote:

I have a names arraylist and an answers arraylist (one answer per person). I'm going to run an algorithm using the comparator to sort the names arraylist into alphabetical order but that means the arraylist will no longer be in the same position as the players name.

Java has been a while for me this is quite hard for me to remember. But, since Java is a OO language another solution would be to make a Player class which holds the name of the player (string value) and the list of questions for that player. When a new name is entered a new Player class is initiated with a new Question list, the player object is added to the names arraylist and you can access the question list through the selected player object (advantage being that in such a solution the question lists can 'never' be detached from the player/name or be mixed up with a list from another player/name).

Androme · 7 Dec 2006, 00:30

cheers for the help dude

why is it this works:

Code:

for (int i=0; i<nameslist.size(); i++)
{
   System.out.println("Name: " + nameslist.get(i) + "\nAnswer: " + anslist.get(i) + "\n");
}

but the following will not and gives me this error: "cannot find symbol method showMessageDialog(java.lang.String)"

Code:

for (int i=0; i<nameslist.size(); i++)
{
   JOptionPane.showMessageDialog("Name: " + nameslist.get(i) + "\nAnswer: " + anslist.get(i) + "\n");
}

I want to display the 2 lists in a message box but it will not allow it - seems to only let me do so in the command line.

Phil^ · 7 Dec 2006, 02:37

dont forget the null before the message in showMessageDialog.
JOptionPane.showMessageDialog(null,message); etc

Androme · 7 Dec 2006, 02:53

Quote:

Originally Posted by Phil^

dont forget the null before the message in showMessageDialog.
JOptionPane.showMessageDialog(null,message); etc

that gives me this error message for the coding i gave above your post

cannot find symbol method println(<nulltype>,java.lang.String)

thanks though!

Phil^ · 7 Dec 2006, 03:01

You only want the null param in the JOptionpane line, Not the system.out.println line.
This is becase the showmessagedialog function can take a componant as a parameter. Since you have no others - substituting it with null gets around it

Androme · 7 Dec 2006, 03:34

LOL

my stupid! Thanks.

Just an update on the coding btw - I'm combining the two arraylists into a new arraylist called "cbal". I haven't touched multiple classes and object orientated stuff yet - I don't want to confuse myself right now heh.

First part combines the two arraylists into one. Second part displays a message. ** PROBLEM** - it displays each element of my new combined arraylist with a seperate message box - i.e.

say I have the elements mark, paul, jack - it'll display mark in one message dialog, paul in another etc. - I want them to be displayed in one - in a list - any idea how?

Code:

for (int i=0; i<nameslist.size(); i++)
{
   cbal.add(nameslist.get(i) + " : " + anslist.get(i));
}
	   
for (int i=0; i<cbal.size(); i++)
{
   JOptionPane.showMessageDialog(null, cbal.get(i));
}

Phil^ · 7 Dec 2006, 04:50

put them all into a single string first., then use joptionpane to display the string

Androme · 7 Dec 2006, 11:22

thanks

annoying questions should now stop

Phil^ · 7 Dec 2006, 11:39

a little trick you can use if you want it to be 'pretty' when it appears in the joptionpane - use html coding for it

Code:

String out = "<html><ul>";
for (int i=0; i<cbal.size(); i++)
{
   out += "<li>" + cbal.get(i) + "</li>";
}
out += "</ul></html>";
JOptionPane.showMessageDialog(null,out);

Now, assuming ive got the html tags correct that should put it in a nice bullet point'd list.

Androme · 7 Dec 2006, 11:49

Wow! But it gives the same result as what I've done:

Code:

for (int i = 0; i<xd.size(); i++) 
{ stringans += xd.get(i); }
JOptionPane.showMessageDialog(null, xd);

Thanks though

**UPDATE**

Why is it combining example elements of two arraylists into one to give something like ---> "Mark - Has The Flu" but then when these 2 arraylists are combined into one AND THEN displayed in a list give an output like this?

[Mark - Has the flu
, paul - not ill
, test - diahorrea
]

My coding for combining the two lists is below and my coding for displaying the combined list is at the top of thisg msg

Code:

for (int i=0; i<nameslist.size(); i++)
{  
  xd.add(nameslist.get(i) + " --- " + anslist.get(i) + "\n");
}

Phil^ · 7 Dec 2006, 11:59

Its displaying the arraylist like that because thats how its tostring method formats the data inside of it. When you use
JOptionPane.showMessageDialog(null, xd); , you are indirectly calling the tostring method for the arraylist object. I suspect you meant to use xd.get(i) , or stringans instead

Androme · 7 Dec 2006, 12:49

Thanks for all those who have helped me put my project together.

Especially for that book Phil, nice algorithms in it. I might not understand that book yet but it doesn't seem out of reach which is nice. Thanks