iOS Developer Question:

Dealing with sqlite database in iOS:
☛ 1. Create database : sqlite3 AnimalDatabase.sql
☛ 2.Create table and insert data in to table :
CREATE TABLE animals ( id INTEGER PRIMARY KEY, name VARCHAR(50), description TEXT, image VARCHAR(255) );

INSERT INTO animals (name, description, image) VALUES ('Elephant', 'The elephant is a very large animal that lives in Africa and Asia', 'http://dblog.com.au/wp-content/elephant.jpg');

☛ 3. Create new app --> Add SQLite framework and database file to project
☛ 4. Read the database and close it once work done with database :
// Setup the database object
sqlite3 *database;

// Init the animals Array
animals = [[NSMutableArray alloc] init];

// Open the database from the users filessytem
if(sqlite3_open([databasePath UTF8String], &database) == SQLITE_OK) {
// Setup the SQL Statement and compile it for faster access
const char *sqlStatement = "select * from animals";
sqlite3_stmt *compiledStatement;
if(sqlite3_prepare_v2(database, sqlStatement, -1, &compiledStatement, NULL) == SQLITE_OK) {
// Loop through the results and add them to the feeds array
while(sqlite3_step(compiledStatement) == SQLITE_ROW) {
// Read the data from the result row
NSString *aName = [NSString stringWithUTF8String:(char *)sqlite3_column_text(compiledStatement, 1)];
NSString *aDescription = [NSString stringWithUTF8String:(char *)sqlite3_column_text(compiledStatement, 2)];
NSString *aImageUrl = [NSString stringWithUTF8String:(char *)sqlite3_column_text(compiledStatement, 3)];

// Create a new animal object with the data from the database
Animal *animal = [[Animal alloc] initWithName:aName description:aDescription url:aImageUrl];

// Add the animal object to the animals Array
[animals addObject:animal];

[animal release];
}
}
// Release the compiled statement from memory
sqlite3_finalize(compiledStatement);

}
sqlite3_close(database);