嗨我的问题在加载UITableView使用SQLite。我有这个代码!问题加载UITableView
数据库的结构:
modelID integer PK autoincrement
model varchar
avaliable int
image1 blob
image2 blob
image3 blob
.H
#import <UIKit/UIKit.h>
@interface DBModel : NSObject {
NSNumber *modelID;
NSString *model;
NSNumber *avaliable;
}
@property (nonatomic, retain) NSNumbe *modelID;
@property (nonatomic, retain) NSString *model;
@property (nonatomic, retain) NSNumbe *avaliable;
-(id)initWithName:(NSString *)n modelID:(NSInteger *)mid avaliable:(NSNumber *)aval;
@end
.M
#import "DBModel.h"
@implementation DBModel
@synthesize modelID, model, avaliable;
-(id)initWithName:(NSString *)n modelID:(NSInteger *)mid avaliable:(NSInteger *)aval {
self.name = n;
self.modelID= mid;
self.avaliable= aval ;
return self;
}
@end
的AppDelegate我有这样的功能:
-(void) readModeDatabase {
sqlite3 *database;
aryModel = [[NSMutableArray alloc] init];
if(sqlite3_open([databasePath UTF8String], &database) == SQLITE_OK) {
const char *sqlStatement = "select * from models";
sqlite3_stmt *compiledStatement;
if(sqlite3_prepare_v2(database, sqlStatement, -1, &compiledStatement, NULL) == SQLITE_OK) {
while(sqlite3_step(compiledStatement) == SQLITE_ROW) {
NSInteger *aModelID = sqlite3_column_int(compiledStatement, 1);
NSString *aModel = [NSString stringWithUTF8String:(char *)sqlite3_column_text(compiledStatement, 2)]; NSInteger *aAvaliable = sqlite3_column_int(compiledStatement, 3);
//Here I need load on image in my DBModel.
// Create a new animal object with the data from the database
DBModel *model = [[DBModel alloc] initWithName:aMode modelID:aModelID avaliable:aAvaliable];
[aryModel addObject:model];
[model release];
}
}
sqlite3_finalize(compiledStatement);
}
sqlite3_close(database);
}
事件的cellForRowAtIndexPath
// Set up the cell
DBModelAppDelegate *appDelegate = (DBModelAppDelegate *)[[UIApplication sharedApplication] delegate];
DBModel *models = (DBModel *)[appDelegate.aryModel objectAtIndex:indexPath.row];
cell.textLabel.text = models.model;
cell.detailTextLabel.text = [models.avaliable stringValue];
return cell;
我的问题是:
我有时间问题,从数据库中值NSInteger的发挥带来一个整数值,他们总是带来零和数据库有是列的值。
当我尝试上传一个使用Subtitle风格的UITableView并播放从数据库中获得的myvariable的值时,它不会添加任何内容,也不会添加任何格式,从而带来正常的效果。
我想做到以下几点:
根据上述我的数据库结构,我填写了详细的UITableView,与图像,标题和标题(如果可能的话根据图片标题或星价值)是所有需要完成的工作,并通过点击通话详细信息完成在线。
感谢