To check if string contains particular word

Not as complicated as they say, check this you will not regret.

String sentence = "Check this answer and you can find the keyword with this code";
String search  = "keyword";

if ( sentence.toLowerCase().indexOf(search.toLowerCase()) != -1 ) {

   System.out.println("I found the keyword");

} else {

   System.out.println("not found");

}

You can change the toLowerCase() if you want.


You can use regular expressions:

if (d.matches(".*Hey.*")) {
    c.setText("OUTPUT: SUCCESS!");
} else {
    c.setText("OUTPUT: FAIL!");  
}

.* -> 0 or more of any characters

Hey -> The string you want

If you will be checking this often, it is better to compile the regular expression in a Pattern object and reuse the Pattern instance to do the checking.

private static final Pattern HEYPATTERN = Pattern.compile(".*Hey.*");
[...]
if (HEYPATTERN.matcher(d).matches()) {
    c.setText("OUTPUT: SUCCESS!");
} else {
    c.setText("OUTPUT: FAIL!");  
}

Just note this will also match "Heyburg" for example since you didn't specify you're searching for "Hey" as an independent word. If you only want to match Hey as a word, you need to change the regex to .*\\bHey\\b.*


.contains() is perfectly valid and a good way to check.

(http://docs.oracle.com/javase/1.5.0/docs/api/java/lang/String.html#contains(java.lang.CharSequence))

Since you didn't post the error, I guess d is either null or you are getting the "Cannot refer to a non-final variable inside an inner class defined in a different method" error.

To make sure it's not null, first check for null in the if statement. If it's the other error, make sure d is declared as final or is a member variable of your class. Ditto for c.

Tags:

Java

Android