CODEWITHSUNDEEP
javaandroidstring
John Jared
John Jared

Reputation: 800

Android using indexOf

I have this, tweet.title is a string that is equals to I love burgers.

        CharSequence i = tweet.title.indexOf(Integer.valueOf("I"));

        SpannableString WordtoSpan = new SpannableString(i);        
        WordtoSpan.setSpan(new ForegroundColorSpan(Color.RED), 0, 7, Spannable.SPAN_EXCLUSIVE_EXCLUSIVE);

        holder.txtTitle.setText(WordtoSpan);

Now I get this error from it Type mismatch: cannot convert from int to CharSequence

I wanna find the I in the string.

Upvotes: 2

Views: 25700

Answers (4)

Michael Yaworski
Michael Yaworski

Reputation: 13483

Code (tested and works)

CharSequence i = tweet.title.indexOf(Integer.valueOf("I"));

should be 

int index = tweet.title.indexOf("I"); // find int position of "I"

// set to be the String of where "I" is to plus 1 of that position
CharSequence i = tweet.title.substring(index, index + 1);

// Alternative to substring, you could use charAt, which returns a char
CharSequence i = Character.toString(tweet.title.charAt(index));

Explanation

indexOf(String) returns the int position of where that String is.
You gave Integer.valueOf("I") as the String to indexOf(String).

Integer.valueOf(String) converts a String to an Integer. Why would you give the indexOf(String) an Integer and why would you try to convert "I" to an Integer?

What you meant to do was this: CharSequence i = tweet.title.indexOf("I"); but that is also wrong because it will return an int (position in the String), hence the mismatch error.

You need to find the position of "I" in the tweet.title, so that's tweet.title.indexOf("I"). Then set the CharSequence to be tweet.title at that position up until that position +1 (so that you get just the one character I).

Upvotes: 3

Steve Benett
Steve Benett

Reputation: 12943

You can work with String because it implements the CharSequence interface. See the docs.

Because you want just one letter you can get the char at the index and make a String out of it:

    int i = tweet.title.indexOf("I");
    String letter = Character.toString(tweet.title.charAt(i));

    SpannableString wordtoSpan = new SpannableString(letter);        
    wordtoSpan.setSpan(
            new ForegroundColorSpan(Color.RED), 0, 7, Spannable.SPAN_EXCLUSIVE_EXCLUSIVE);

If you wanna get more chars of your String you can use the String.subString(int beginIndex, int endIndex) method.

F.e: you wanna get "burgers":

    int i = tweet.title.indexOf("b");
    String sub = title.subString(i, i + 6);
    SpannableString wordtoSpan = new SpannableString(sub);        
    wordtoSpan.setSpan(new ForegroundColorSpan(Color.RED), 0, 7, Spannable.SPAN_EXCLUSIVE_EXCLUSIVE);

Upvotes: 1

mttdbrd
mttdbrd

Reputation: 1831

String.indexOf returns an int not a CharSequence. Also, you can't get the index of the I using Integer.valueOf().

    int i = tweet.title.indexOf("I");

If you want to just get the "I" back do this:

    int i = tweet.title.indexOf("I");
    String s = tweet.title.substring(i, i + 1);

Edited: I fixed a couple of bugs in this code.

Upvotes: 4

Shayan Pourvatan
Shayan Pourvatan

Reputation: 11948

String.contains() which checks if the string contains a specified sequence of char values String.indexOf() which returns the index within the string of the first occurence of the specified character or substring (there are 4 variations of this method)

Upvotes: 3

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