1. Methods in the String class Manipulating text in Java
Strings
Methods in the String class
Manipulating text in Java

2. The String Class
Like all standard java classes, there is a javadoc for the String class that can be helpful to you.
The methods our class is responsible for are:
length()
indexOf( String )
indexOf( String, int )
substring( int, int )
substring( int )

3. Strings A string is a series of characters, also known as an array.
Arrays are 0-based meaning we start counting at 0 rather than one
Ex.
Character M i c r o s f t
Index 1 2 3 4 5 6 7 8

4. Length
Strings can tell you how many characters they have by using the length method:
Ex.
String name1 = new String( “Sally” );
String name2 = new String( “John” );
String name3 = new String( “Bob” );
String name4 = new String( “Mary Sue” );
System.out.println( name1.length() );//prints 5
System.out.println( name2.length() );//prints 4
System.out.println( name3.length() );//prints 3
System.out.println( name4.length() );//prints 8
White space counts when asking for length!

5. Substrings
Strings have a method called substring that can give you pieces of the string
substring takes the starting index and one past the stopping index
Ex.
Character M i c r o s f t
Index 1 2 3 4 5 6 7 8
“Microsoft”.substring( 0, 5 );
returns “Micro”
“Microsoft”.substring( 5, 9 );
returns “Soft”
“Microsoft”.substring( 3, 7 );
returns “roso”

6. Substring substring has another form:
if you give substring the starting index only, it goes to the end of the String
“Microsoft”.substring( 3 ); //returns “rosoft”
“Microsoft”.substring( 5 ); //returns “soft”
“Microsoft”.substring( 7 ); //returns “ft”

7. indexOf
The indexOf method returns the index where a substring is found
Character M i c r o s f t
Index 1 2 3 4 5 6 7 8
“Microsoft”.indexOf( “Micro” );
returns 0
“Microsoft”.indexOf( “soft” );
returns 5
“Microsoft”.indexOf( “icro” );
returns 1

8. indexOf Character M i c r o s f t Index 1 2 3 4 5 6 7 8
When a character is not found indexOf returns -1, an invalid index, to let the caller know
When a character or substring can be found more than once, indexOf returns the first occurrence
Character M i c r o s f t
Index 1 2 3 4 5 6 7 8
“Microsoft”.indexOf( “m” );
returns -1
“Microsoft”.indexOf( “apple” );
Returns -1
“Microsoft”.indexOf( “o” );
returns 4

9. indexOf indexOf has another parameter
You can tell it where to start looking
Character M i c r o s f t
Index 1 2 3 4 5 6 7 8
“Microsoft”.indexOf( “o”, 5 );
returns 6;//skips past the first ‘o’ by starting at index 5
“Microsoft”.indexOf( “s”, 3 );
returns 5;// still finds the first s
“Microsoft”.indexOf( “M”, 5 );
returns -1;// no ‘M’ after index 5

10. The Last Index
Finding the last index in a String means you’ll need to use the length() method
Remember, Strings are 0-based!
The last index is length() – 1
Ex.
String name = new String( “Billy” );
String lastChar = new String();
int lastIndex = name.length() – 1;
lastChar = name.substring( lastIndex );
System.out.println( lastChar );//prints ‘y’

11. Other String Methods
I encourage you to refer to the JavaDoc online for Strings, there are many methods that you may find helpful as we create new projects.
We’ve discussed these for comparison with keys in Greenfoot:
equals
compareTo
Some suggestions that can be helpful:
replace
trim
valueOf
toUpperCase
toLowerCase
matches
You’ll also need to learn about a very useful programming concept called regular expressions in order to use this method

12. Example

13. Example
Write a method to check if a password meets the following criteria
Contains a number
Contains an uppercase letter
Is at least 9 letters long

14. Example
public boolean passwordOK( String pwd ) { boolean hasUppercase = false; for( int n = 0; n < pwd.length() - 1; n++ ) String ch = pwd.substring(n, n+1); if(ch.compareTo(“A”) >= 0 && ch.compareTo(“Z”) <= 0) hasUppercase = true; } boolean hasDigit = false; if(ch.compareTo(“0”) >= 0 && ch.compareTo(“9”) <= 0) hasDigit = true; return pwd.length() >= 9 && hasDigit && hasUppercase;

15. Example
Write a method to count the number of times the word “Java” appears in a string

16. Example
public int count(String words) { String target = “Java”; int foundAt = words.indexOf( target ); int count = 0; while( foundAt >= 0 ) count++; foundAt = words.indexOf( target, foundAt + target.length()); } return count;

17. Example
Let’s say * is a wildcard and we want to know if a String contains a particular substring
public int countPattern( String source, String target )
countPattern( “Hello World”, “orl” )  1
countPattern( “He**o World”, “Help”)  1
countPattern( “H**lo World”, “He” )  2
(once for He, and **)

18. Example
public int countPattern(String source, String target) { int count = 0; for(int n = 0; n < source.length() - 1; n++) int matches = 0; for(int k = 0; k < target.length() – 1 && n + k < source.length() - 1; k++) String sourceChar = source.substring(n + k, n + k +1); String targetChar = target.substring(k, k+1); if(sourceChar.compareTo(“*”) == 0)//wildcard matches++; else if( sourceChar.compareTo( targetChar ) == 0 ) else k = target.length();//stop the loop  can also use break but not as elegant } if(matches == target.length()) count++; return count;