Counting none. Characters and words in strings in PL/SQL

Given a string of arbitrary length, the task is to count the number of characters and words in the string using PL/SQL.

PL/SQL is a combination of SQL and procedures. Features of programming language. It was developed by Oracle Corporation in the early 1990s to enhance SQL functionality. PL/SQL is one of three key programming languages ​​in embedded systems Oracle Database, as well as SQL itself and Java.

In PL/SQL block we have DECLARE block for declaring variables used in Programming, we have BEGIN block where we can write the logic for the given problem,

For example

Input − string str = “Tutorials Point”
Output− count of characters is: 15
      Count of words are: 2

Explanation-: In the given string, we have total 2 words, So the number of words is 2, and in these words we have 14 characters, plus 1 character to represent a space in the given string.

Input − string str = “Honesty is the best policy”
Output − count of characters is: 26
      Count of words are: 5

Explanation - In the given string we have total 5 words so the number of words is 5 and out of these words we have 24 characters plus 4 characters for four spaces in the given string.

The method used in the following program is as follows

• Enter a string of any length and store it in a variable, such as str

• Calculate the length of a string using the length() function, which returns an integer value based on the number of letters in the string (including spaces).

< li>

Traverse the loop from i to 0 until the length of the string str

• Use the function substr(), which will return the string The number of substrings is the number of words in the string

• and, each loop iteration increases the number of characters until the length of the string is reached.

  < /li>

• Print the number of characters and words in the string.

Example

DECLARE
      str VARCHAR2(40) := 'Tutorials Point';
      nchars NUMBER(4) := 0;
      nwords NUMBER(4) := 1;
      s CHAR;
BEGIN
   FOR i IN 1..Length(str) LOOP
      s := Substr(str, i, 1);
      nchars:= nchars+ 1;
      IF s = ' ' THEN
      nwords := nwords + 1;
      END IF;
END LOOP;
dbms_output.Put_line('count of characters is:'
   ||nchars);

dbms_output.Put_line('Count of words are: '
   ||nwords);
END;

Output

If we run the above code, it will generate the following output -

count of characters is: 15
Count of words are: 2

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