[vickyvvv44]

String st = "abcd";
int i=0;
while(i<st.length())
{
char c = ?........(st.substring(i,i+1));
System.out.Println(c);
i++;
}

[Locke]

I wouldn't use a while loop. And you don't have to use the substring() method.

String st = "abcd";

// edit the loop a bit

for (int i = 0; i < st.length(); i++)
{
    char ch = st.charAt(i); // use the charAt() method to access characters

    System.out.println(ch);
}

Hope this helps!

This post has been edited by Locke: 20 January 2010 - 09:16 AM

[vickyvvv44]

String  st ="condition";
int len = st.length();
int y= 0;
while(len>y)
{
String c = st.Substring(y,y+1);
char ch = c;
if(ch>='A'&& ch<='Z' || ch>='a' && ch<= 'z' )
{y++;
}
else{break;}
}
if(i==len){System.out.Println("valid");}
else{System.out.Println("unvalid ");}

Edited by Locke. Please, in the future -->

This post has been edited by Locke: 20 January 2010 - 09:48 AM

[SwiftStriker00]

please put [ code] [ /code] tags around your code
& Please tell us what your issue is

oh and its "check" not "cheq"

This post has been edited by SwiftStriker00: 20 January 2010 - 09:47 AM

[Locke]

You don't have a variable i. But you do have a y.

if (y == len) // not 'i'

[SwiftStriker00]

String cStr = "c";
		char c = cStr.toCharArray()[0];

This post has been edited by SwiftStriker00: 20 January 2010 - 09:53 AM

[erik.price]

String c = st.Substring(y,y+1);

substring is lowercased

Also, since c is a String, you cannot convert it to a char just by doing this:

String c = st.Substring(y,y+1);
char ch = c;

You could, as SwiftStriker suggested, do this:

String a = "a";
char s = a.toCharArray()[0];

[Locke]

Nobody suggests charAt()? Interesting.

[NeoTifa]

They already did in his/her other 6 threads s/he made about the same topic.

[Locke]

I know I already did in one of them.

But since we're talking about accessing individual characters, I'm surprised that didn't come up here. It would involve another variable to continuously go through the string, but meh.

[SwiftStriker00]

charAt() is too obvious of an answer

[g00se]

boolean valid = st.matches("[A-Za-z]+");

[Locke]

That's the easy one, but not for beginners.

[pbl]

The String class has a method to return an array of char[] from a String

char[] digit = string.toCharArray();

[macosxnerd101]

Also, please post your question in the body of your post, not in the title where it gets truncated.