Java (Android) - can't detect first character of string?

Question

Not sure if I've missed something really obvious. I know for sure that my String is as follows:

1This is a test message

I'm trying to detect whether the first character is '1', so here's some of my code:

    //This outputs '1'
    Toast noCmd = Toast.makeText(Play.this, decodedMessage.substring(0,1), Toast.LENGTH_SHORT);
    noCmd.show();

        if (decodedMessage.charAt(0) == 1) {

            noCmd = Toast.makeText(Play.this, "This should show up", Toast.LENGTH_SHORT);
            noCmd.show();

            noCmd = Toast.makeText(Play.this, finalMessage + " from " + sender,     Toast.LENGTH_SHORT);
            noCmd.show();
        }

    if (decodedMessage.substring(0,1) == "1") {

        noCmd = Toast.makeText(Play.this, "This should show up", Toast.LENGTH_SHORT);
        noCmd.show();

        noCmd = Toast.makeText(Play.this, finalMessage + " from " + sender, Toast.LENGTH_SHORT);
        noCmd.show();
    }

As you can see, I'm trying two methods to get the toasts inside the if statement to show up. Weirdly, when the code is run, only the top (unconditional) toast displays.

Any ideas?

chiastic-security · Accepted Answer · 2014-10-05T06:31:38.230

For the first one, the char is '1'. What's currently happening in your code is that because you're comparing a char with an integer, the char is being converted to an int using its character code. For a 1, that comes out as 49, which is not equal to the integer 1. You need to compare the char you're retrieving from the String with the char representing a digit "1", and that means you need to write it as '1'.

For the second one, you need to use .equals() to test for String equality, rather than ==. If you take two String objects s and t that have the same content, then you still will find that s==t will come out as false, unless they happen to be pointing at the same bit of memory (i.e., they're the same instance). To check whether they have the same content, you check

s.equals(t)

rather than

s==t

So, in summary, make the first one

if (decodedMessage.charAt(0) == '1') {
    //toast stuff
}

and the second one

if ("1".equals(decodedMessage.substring(0,1))) {
    //toast stuff
}

The reason, by the way, for not writing

if (decodedMessage.substring(0,1).equals("1")) {
    //toast stuff
}

instead is that if the String on which you call .equals() is null then you'll end up with a NullPointerException, which usually you want to avoid. Actually in this case it would be fine, because the substring() call won't return null, but in the general case if you want to test whether s and "something" are equal then you use

"something".equals(s)

rather than

s.equals("something")

just in case s is null.

Thank you! Accepting this as the answer as it was the first, and most comprehensive (for other readers). I knew there was something wrong... I wasn't using .equals()... — , Oct 04 '14 at 19:55
Nice explanation but Java uses Unicode, not ASCII. Java's `char` datatype stores one UTF-16 code-unit, one or two of which encode a Unicode codepoint. One base codepoint and zero or more combining codepoints form a grapheme, which is what people think of when using the term "character." The digit 1 is encoded as one `char` so can be written as a char literal '1'. — Tom Blodget, Oct 05 '14 at 02:36

score 1 · Answer 2 · edited May 23 '17 at 12:11

1

1 is an integer with value 1. If you want the ASCII 1, use '1' in single quotes which has the integer value of 49.

For comparing strings, use equals() and not ==. See How do I compare strings in Java?

edited May 23 '17 at 12:11

Community

1
1

answered Oct 04 '14 at 19:44

laalto

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score 0 · Answer 3 · answered Oct 04 '14 at 19:44

0

To compare strings you need to use equals method:

if("1".equals(decodedMessage.charAt(0))){

}

answered Oct 04 '14 at 19:44

eleven

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Java (Android) - can't detect first character of string?

3 Answers3