无法从SQLite数据库中检索和查看结果

时间:2010-11-30 23:16:30

标签: iphone objective-c database sqlite

我正在尝试建立我作为iPhone开发人员的技能,目前我正在使用SQLite数据库。我在GroceryListAppDelegate.m类中创建了一个SQLite表:

- (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];

}

我在RootViewController.m类中进行了一次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];

}

我的executeSqlWithParameters()方法如下所示:

- (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;

}

当我构建并运行我的代码时,我没有得到任何结果,实际上,在我的SQL调用中,我应该在我的列表中得到苹果和橙子。但是,当我修改我的SQL代码时:

self.results = [appDelegate.database executeSql:@"SELECT * from GroceryItem"];

调用不同的方法:executeSql():

- (NSArray *) executeSql: (NSString *)sql {

return [self executeSql:sql withParameters: nil];

}

在这种情况下,我最终得到了结果:苹果和橘子。为什么是这样?我究竟做错了什么?为什么我从一个SQL调用获得结果,而不是从另一个SQL调用获得结果?

2 个答案:

答案 0 :(得分:0)

如果从模拟器或设备中提取.sqlite文件并使用sqlite3在命令行中打开它,查询是否有效?这将使您能够分别测试数据库创建代码和查询代码。

答案 1 :(得分:0)

非常感谢您的所有帮助,但我找到了解决方案。问题出在我的bindArguments()函数中:

- (void) bindArguments: (NSArray *) arguments toStatement: (sqlite3_stmt *) statement queryInfo: (NSDictionary *) queryInfo {

我在哪里获得了数字的绑定语句:

else if ([argument isKindOfClass:[NSNumber class]])
        {
            sqlite3_bind_double(statement, -1, [argument doubleValue]);
        }

我需要将-1更改为i。这个变量告诉SQL我将在语句中使用哪个变量。那么在这种情况下我们只有一个变量代表?它应该用5代替。