我正在尝试构建我作为iPhone开发人员的技能,目前我正在使用SQLite数据库。我创建了一个SQLite表在我GroceryListAppDelegate.m类如下:无法检索和查看SQLite数据库的结果
- (void)applicationDidFinishLaunching:(UIApplication *)application {
self.database = [[[ISDatabase alloc] initWithFileName:@"TestDB.sqlite"] autorelease];
if(![[database tableNames] containsObject:@"GroceryItem"]) {
[database executeSql:@"create table GroceryItem(primaryKey integer primary key autoincrement, name text NOT NULL, number integer NOT NULL)"];
[database executeSql:@"insert into GroceryItem (name, number) values ('apples', 5)"];
[database executeSql:@"insert into GroceryItem (name, number) values ('oranges', 3)"];
}
[window addSubview:navigationController.view];
[window makeKeyAndVisible];
}
我对我做RootViewController.m类SQL调用:
- (void)viewDidLoad {
[super viewDidLoad];
GroceryList1AppDelegate *appDelegate = (GroceryList1AppDelegate *)[[UIApplication sharedApplication] delegate];
self.results = [appDelegate.database executeSqlWithParameters:@"SELECT * from GroceryItem where number < ?", [NSNumber numberWithInt:6], nil];
}
我executeSqlWithParameters()方法是这样的:
- (NSArray *) executeSql:(NSString *)sql withParameters: (NSArray *) parameters {
NSMutableDictionary *queryInfo = [NSMutableDictionary dictionary];
[queryInfo setObject:sql forKey:@"sql"];
if (parameters == nil) {
parameters = [NSArray array];
}
//we now add the parameters to queryInfo
[queryInfo setObject:parameters forKey:@"parameters"];
NSMutableArray *rows = [NSMutableArray array];
//log the parameters
if (logging) {
NSLog(@"SQL: %@ \n parameters: %@", sql, parameters);
}
sqlite3_stmt *statement = nil;
if (sqlite3_prepare_v2(database, [sql UTF8String], -1, &statement, NULL) == SQLITE_OK) {
[self bindArguments: parameters toStatement: statement queryInfo: queryInfo];
BOOL needsToFetchColumnTypesAndNames = YES;
NSArray *columnTypes = nil;
NSArray *columnNames = nil;
while (sqlite3_step(statement) == SQLITE_ROW) {
if (needsToFetchColumnTypesAndNames) {
columnTypes = [self columnTypesForStatement:statement];
columnNames = [self columnNamesForStatement:statement];
needsToFetchColumnTypesAndNames = NO;
}
id row = [[NSMutableDictionary alloc] init];
[self copyValuesFromStatement:statement toRow:row queryInfo:queryInfo columnTypes:columnTypes columnNames:columnNames];
[rows addObject:row];
[row release];
}
}
else {
sqlite3_finalize(statement);
[self raiseSqliteException:[[NSString stringWithFormat:@"failed to execute statement: '%@', parameters: '%@' with message: ", sql, parameters] stringByAppendingString:@"%S"]];
}
sqlite3_finalize(statement);
return rows;
}
当我构建和运行我的代码,我没有得到任何结果,而事实上,给我的SQL调用,我应该得到苹果,橘子和我的列表中。然而,当我修改我的SQL代码如下:
self.results = [appDelegate.database executeSql:@"SELECT * from GroceryItem"];
它调用不同的方法:的ExecuteSQL():
- (NSArray *) executeSql: (NSString *)sql {
return [self executeSql:sql withParameters: nil];
}
在这种情况下,我最终得到的结果:苹果,橙子。为什么是这样?我究竟做错了什么?为什么我从一个SQL调用中获取结果,而不是从另一个调用中获得结果?
非常感谢您的回复。但是,我将如何从模拟器中提取.sqlite文件,并使用sqlite3在命令行上打开它?再次感谢你的帮助。 – syedfa 2010-12-02 03:34:42