问题
As I said before, how do I test if the entered character is one of the parameters? I've written this code, but it doesn't seem to run very well(or at all), no errors, however. Also, I need to use the basic code I've used here. Its for school and we lose points if we use things they haven't taught us (darn school).
class doody
{
public static void main(String[] args)
{ char i;
char input='D';
for(i='A';i<='Z';i++)//check if uppercase
{
if(input==i){
System.out.println("Uppercase");
switch(input){
case 'A':
case 'E':
case 'I':
case 'O':
case 'U':
System.out.println("Vowel"); break;
default: System.out.println("Not a vowel"); break;}
}
for(i='a';i<='z';i++)//check if lowercase
{
if(input==i){
System.out.println("Lowercase");
switch(input){
case 'a':
case 'e':
case 'i':
case 'o':
case 'u':
System.out.println("Vowel"); break;
default: System.out.println("Not a vowel"); break;
}}
for(i='0';i<='9';i++)//check if number
{
if(input==i)
System.out.println("Number");
}
}
}}}
Edit: Here is some code I threw together today. Much simpler. I don't know why this didn't occur to me earlier. It was probably because I was groggy, it was late.
class doody
{
public static void main(String[] args)
{
char input='$';//input here.
boolean lorn=false;
if(input>='a'&&input<='z')
{System.out.println("Lowercase");
lorn=true;
if(input=='a')System.out.println("Vowel.");
if(input=='e')System.out.println("Vowel.");
if(input=='i')System.out.println("Vowel.");
if(input=='o')System.out.println("Vowel.");
if(input=='u')System.out.println("Vowel.");
}
if(input>='A'&&input<='Z')
{System.out.println("Uppercase");
lorn=true;
if(input=='A')System.out.println("Vowel.");
if(input=='E')System.out.println("Vowel.");
if(input=='I')System.out.println("Vowel.");
if(input=='O')System.out.println("Vowel.");
if(input=='U')System.out.println("Vowel.");
}
if(input>='0'&&input<='9')
{
lorn=true;
System.out.println("Number");
}
if(lorn==false)System.out.println("It is a special character");
}
}
回答1:
If it weren't a homework, you could use existing methods such as Character.isDigit(char), Character.isUpperCase(char) and Character.isLowerCase(char) which are a bit "smarter", because they don't operate only in ASCII, but also in various charsets.
static final char[] VOWELS = { 'a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U' };
static boolean isVowel(char ch) {
for (char vowel : VOWELS) {
if (vowel == ch) {
return true;
}
}
return false;
}
static boolean isDigit(char ch) {
return ch >= '0' && ch <= '9';
}
static boolean isLowerCase(char ch) {
return ch >= 'a' && ch <= 'z';
}
static boolean isUpperCase(char ch) {
return ch >= 'A' && ch <= 'Z';
}
回答2:
You don't need a for loop in your code.
Here is how you can re implement your method
- If input is between 'A' and 'Z' its uppercase
- If input is between 'a' and 'z' its lowercase
- If input is one of 'a,e,i,o,u,A,E,I,O,U' its Vowel
- Else Consonant
Edit:
Here is hint for you to proceed, Following code snippet gives int values for chars
System.out.println("a="+(int)'a');
System.out.println("z="+(int)'z');
System.out.println("A="+(int)'A');
System.out.println("Z="+(int)'Z');
Output
a=97
z=122
A=65
Z=90
Here is how you can check if a number x exists between two numbers say a and b
// x greater than or equal to a and x less than or equal to b
if ( x >= a && x <= b )
During comparisons chars can be treated as numbers
If you can combine these hints, you should be able to find what you want ;)
回答3:
In Java : Character class has static method called isLowerCase(Char ch) ans isUpperCase(Char ch) , Character.isDigit(Char ch)gives you Boolean value, base on that you can easily achieve your task
example:
String abc = "HomePage";
char ch = abc.charAt(i); // here i= 1,2,3......
if(Character.isLowerCase(ch))
{
// do something : ch is in lower case
}
if(Character.isUpperCase(ch))
{
// do something : ch is in Upper case
}
if(Character.isDigit(ch))
{
// do something : ch is in Number / Digit
}
回答4:
Some comments on your code
- why would you want to have 2 for loops like
for(i='A';i<='Z';i++), if you can check this with a simpleifstatement ... you loop over a whole range while you can simply check whether it is contained in that range - even when you found your answer (for example when input is
Ayou will have your result the first time you enter the first loop) you still loop over all the rest - your
System.out.println("Lowercase");statement (and the uppercase statement) are placed in the wrong loop - If you are allowed to use it, I suggest to look at the
Characterclass which has for example niceisUpperCaseandisLowerCasemethods
I leave the rest up to you since it is homework
回答5:
You appear to have upper and lowercase back to front. A-Z would be upper, a-z would be lower. While not exactly efficient - with the inverted case exception, I think it should output correctly.
回答6:
This may not be what you are looking for but I thought you oughta know the real way to do this. You can use the java.lang.Character class's isUpperCase() to find aout about the case of the character. You can use isDigit() to differentiate between the numbers and letters(This is just FYI :) ). You can then do a toUpperCase() and then do the switch for vowels. This will improve your code quality.
回答7:
This is your homework, so I assume you NEED to use the loops and switch statments. That's O.K, but why are all your loops inner to the privious ones ?
Just take them out to the same "level" and your code is just fine! (part to the low/up mixup).
A tip: Pressing extra 'Enter' & 'Space' is free! (That's the first thing I did to your code, and the problem became very trivial)
回答8:
Char input;
if (input.matches("^[a-zA-Z]+$"))
{
if (Character.isLowerCase(input))
{
// lowercase
}
else
{
// uppercase
}
if (input.matches("[^aeiouAEIOU]"))
{
// vowel
}
else
{
// consonant
}
}
else if (input.matches("^(0|[1-9][0-9]*)$"))
{
// number
}
else
{
// invalid
}
来源：https://stackoverflow.com/questions/8962025/java-program-to-test-if-a-character-is-uppercase-lowercase-number-vowel