嗨我的問題在加載UITableView使用SQLite。我有這個代碼!問題加載UITableView
數據庫的結構:
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我有這樣的功能:
-(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;
我的問題是:
我有時間問題,從數據庫中值NSInteger的發揮帶來一個整數值,他們總是帶來零和數據庫有是列的值。
當我嘗試上傳一個使用Subtitle風格的UITableView並播放從數據庫中獲得的myvariable的值時,它不會添加任何內容,也不會添加任何格式,從而帶來正常的效果。
我想做到以下幾點:
根據上述我的數據庫結構,我填寫了詳細的UITableView,與圖像,標題和標題(如果可能的話根據圖片標題或星價值)是所有需要完成的工作,並通過點擊通話詳細信息完成在線。
感謝