检查字符串是否包含特定单词

Question

So how do you check if a string has a particular word in it?

So this is my code:

a.setOnClickListener(new View.OnClickListener() {

        @Override
        public void onClick(View arg0) {
            // TODO Auto-generated method stub
            if(d.contains("Hey")){
                c.setText("OUTPUT: SUCCESS!"); 
            }else{
                c.setText("OUTPUT: FAIL!");  
            }
        }
    });

I'm getting an error.

Answer 1

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.

Answer 2

.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 .

Answer 3

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.*

Answer 4

Using contains

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

if (sentence.toLowerCase().contains(search.toLowerCase())) {

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

} else {

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

}

Answer 5

The other answer (to date) appear to check for substrings rather than words. Major difference.

With the help of this article , I have created this simple method:

static boolean containsWord(String mainString, String word) {

    Pattern pattern = Pattern.compile("\\b" + word + "\\b", Pattern.CASE_INSENSITIVE); // "\\b" represents any word boundary.
    Matcher matcher = pattern.matcher(mainString);
    return matcher.find();
}

Answer 6

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

Compare the line of string in given string

if ((sentence.toLowerCase().trim()).equals(search.toLowerCase().trim())) {
System.out.println("not found");
}
else {  
    System.out.println("I found the keyword"); 
} 

Answer 7

It's been correctly pointed out above that finding a given word in a sentence is not the same as finding the charsequence, and can be done as follows if you don't want to mess around with regular expressions.

boolean checkWordExistence(String word, String sentence) {
    if (sentence.contains(word)) {
        int start = sentence.indexOf(word);
        int end = start + word.length();

        boolean valid_left = ((start == 0) || (sentence.charAt(start - 1) == ' '));
        boolean valid_right = ((end == sentence.length()) || (sentence.charAt(end) == ' '));

        return valid_left && valid_right;
    }
    return false;
}

Output :

checkWordExistence("the", "the earth is our planet"); true
checkWordExistence("ear", "the earth is our planet"); false
checkWordExistence("earth", "the earth is our planet"); true

PS Make sure you have filtered out any commas or full stops beforehand.

Answer 8

Solution-1: - If you want to search for a combination of characters or an independent word from a sentence.

String sentence = "In the Name of Allah, the Most Beneficent, the Most Merciful."
if (sentence.matches(".*Beneficent.*")) {return true;}
else{return false;}

Solution-2: - There is another possibility you want to search for an independent word from a sentence then Solution-1 will also return true if you searched a word exists in any other word. For example, If you will search cent from a sentence containing this word ** Beneficent** then Solution-1 will return true. For this remember to add space in your regular expression.

String sentence = "In the Name of Allah, the Most Beneficent, the Most Merciful."
if (sentence.matches(".* cent .*")) {return true;}
else{return false;}

Now in Solution-2 it wll return false because no independent cent word exist.

Additional : You can add or remove space on either side in 2nd solution according to your requirements.

Answer 9

if (someString.indexOf("Hey")>=0) 
     doSomething();

Answer 10

Maybe this post is old, but I came across it and used the "wrong" usage. The best way to find a keyword is using .contains , example:

if ( d.contains("hello")) {
            System.out.println("I found the keyword");
}