1. String Manipulation

2. Java String class  The String class represents character strings “Tammy Bailey”  All strings (arrays of characters) in Java programs are implemented as instances of this class  Declaration String name;  Assignment name = “Tammy Bailey”;  Declare and initialize String name = “Tammy Bailey”;

3. Concatenation operator  Strings can be concatenated (“added together”) using the concatenation operator + String first = “Tammy”; String last = “Bailey”; String name = first + “ ” + last; String greeting = “Hello ” + first;  The string variable name contains the string “Tammy Bailey”  The string variable greeting contains the string “Hello Tammy”

4. String indexing and length  Characters in strings are indexed the same as arrays –first index is 0 –last index is one less than the number of characters  The length of a string is the number of characters in the string –letters, numbers, spaces, punctuation, etc. –the length method returns the length of a string  Example String name = “Tammy Bailey”; int len = name.length(); –The value of len is 12

5. String characters  char charAt(int index) –returns the character at the specified index  String replace(char old,char new) –returns new string resulting from replacing all occurrences of the old character with new  String toLowerCase() –returns new string with all characters in lowercase  Examples String name = “Tammy Bailey”; char c = name.charAt(8); String s1 = name.replace(‘y’,‘i’); String s2 = name.toLowerCase();

6. String indices  int indexOf(String s) –returns the index of the first occurrence of the specified substring  int indexOf(String s, int index) –returns the index of the first occurrence of the specified substring, starting at the specified index  Examples String name = “Tammy Bailey”; int idx1 = name.indexOf(“y”); int idx2 = name.indexOf(“mm”); int idx3 = name.indexOf(“Z”); int idx4 = name.indexOf(“y”,5); idx1=4 idx2=2 idx3=-1 idx4=11

7. Substrings  String substring(int startIdx) –returns a new string that is a substring of the string –the substring begins with the character at startIdx and extends to the end of the string  String substring(int startIdx, int endIdx) –returns a new string that is a substring of the string –the substring begins at startIdx and extends to the character at index endIdx-1 –the length of the substring is endIdx-startIdx

8. Substring examples String name = “Tammy Bailey”; String sub1 = name.substring(8); String sub2 = name.substring(0,3); String sub3 = name.substring(name.indexOf(‘B’)); String sub4 = name.substring(0,name.indexOf(“ ”)); sub1 = “iley” sub2 = “Tam” sub3 = “Bailey” sub4 = “Tammy”

9. Strings and loops  What does the string name contain after exiting the while-loop? String name = “Tammy Lynn Bailey”; int index = 0; while(index != -1) { name = name.substring(index); index = name.indexOf(“ ”); }

10. Another example  What is the value of count after exiting the while-loop? String s = “How many words are there?”; int count = 0; int index = 0; while(index != -1) { count++; index = s.indexOf(“ ”, index+1); }

11. Parsing strings  String email = “tammy@cs.duke.edu”; String login = email.substring(0,email.indexOf(“@”)); String host = email.substring(email.indexOf(“@”)+1);  String name = “Bailey, Tammy”; String first = name.substring(name.indexOf(“,”)+1); String last = name.substring(0,name.indexOf(“,”));

12. Pig Latin  If a word begins with a consonant, move all of the characters up to the first vowel to the end of the word and add “ay” –bad becomes adbay  If a word begins with a vowel, write the word and add “way” –add becomes addway  Y is not a vowel. –yes becomes esyay  If there are no vowels in the word, just write the word –why becomes why

13. Pig Latin Converter  How to design a program to convert words to Pig Latin?  Need to find first occurrence of a vowel –word begins with vowel, or –word begins with consonant and we move all letters up to first vowel, or –there are no vowels in the word  Idea: loop through characters in word until vowel is found or we go through all characters (no vowels)  String methods to use – length, indexOf, charAt

14. Subroutine FindFirstVowel int FindFirstVowel(String word) { char c; int index = -1; word = word.toLowerCase(); while(index < word.length()) { c = word.charAt(index); if(c==‘a’||c==‘e’||c==‘i’||c==‘o’||c==‘u’) return index; index++; } return –1; } returns index of first vowel found in word or –1 if word contains no vowels

15. Conversion cases  Use the integer returned by FindFirstVowel to determine how to convert the word to Pig Latin  Three cases – FindFirstVowel returns 0 word begins with vowel – FindFirstVowel returns –1 no vowels in world – FindFirstVowel returns positive integer word begins with a consonant know index of first occurrence of vowel

16. Subroutine Convert String Convert(String s) { int i = FindFirstVowel(s); if( i == 0 ) { /* word begins with vowel */ } else if( i > 0 ) { /* word begins with consonant but has a vowel */ } else { /* word has no vowels */ }

17. Text Areas  A TextArea is similar to a TextField –Differences: has both rows and columns, has scrollbars, can append text  Declare a TextArea object TextArea tArea;  Initialize tArea with 5 rows and 60 columns tArea = new TextArea(5,60);  Set text in TextArea tArea.setText(“Hello”);  Append text to existing text in tArea tArea.append(“ there”); – tArea now contains the text “Hello there”

18. Newline character  The newline character ‘\n’ terminates a line of text –any text following the newline character begins on the next line –used in display/formatting of output text  Example –a TextArea object displays multiple lines of text –can append the newline character to the end of the output string –subsequent output will display on a new line tArea.setText(“Hello.”); tArea.append(“\n”); tArea.append(“How are you?”);