保存的UIImage作爲二進制SQLite和與imageWithData

取回我節省了一些圖片到SQLite數據庫在我的應用程序。我用圖形SQLite客戶端訪問數據庫，它正確保存了我的圖像的二進制文件。問題在於從數據庫中檢索它。這是什麼樣子：保存的UIImage作爲二進制SQLite和與imageWithData

保存

-(void)insertObject:(Object *)object 
{ 
    const char* sqliteQuery = "INSERT INTO OBJECTS (OBJECTID, OBJECTIMAGE, OBJECTNAME) VALUES (?,?,?)"; 

sqlite3_stmt* statement; 

if(sqlite3_prepare_v2(databaseHandle, sqliteQuery, -1, &statement, NULL) == SQLITE_OK) 

{ 

    sqlite3_bind_text(statement, 1, [object.id UTF8String], -1, SQLITE_TRANSIENT); 

    sqlite3_bind_blob(statement, 2, [object.imgData bytes], [object.imgData length], SQLITE_STATIC); 

    sqlite3_bind_text(statement, 3, [object.name UTF8String], -1, SQLITE_TRANSIENT); 

    sqlite3_step(statement); 

} 

else NSLog(@"SaveBody: Failed from sqlite3_prepare_v2. Error is: %s", sqlite3_errmsg(databaseHandle)); 

sqlite3_finalize(statement); 
}

檢索

-(Object *)retrieveObjectWithID:(NSString*)id 
{ 

NSString* sqliteQuery = [NSString stringWithFormat:@"SELECT OBJECTIMAGE,OBJECTNAME FROM OBJECTS WHERE OBJECTID = '%@'", id];  
NSLog(@"sqlitequery: %@",sqliteQuery); 
sqlite3_stmt* statement; 

Object *object = [[Object alloc]init]; 

object.id = id; 

if(sqlite3_prepare_v2(databaseHandle, [sqliteQuery UTF8String], -1, &statement, NULL) == SQLITE_OK) 

{ 

    if(sqlite3_step(statement) == SQLITE_ROW) 

    { 

     NSUInteger length = sqlite3_column_bytes(statement, 1); 

     object.imgData = [NSData dataWithBytes:sqlite3_column_blob(statement, 1) length:length]; 
     object.name = [NSString stringWithFormat:@"%s",sqlite3_column_text(statement, 2)]; 

    } 

} 


// Finalize and close database. 

sqlite3_finalize(statement); 


return object; 
}

這是對象

@interface Object : NSObject 


@property(nonatomic,strong) NSString *id; 
@property(nonatomic,strong) NSString *name; 
@property(nonatomic,strong) NSData *imgData; 

@end

來源

2012-07-03 Lucas Pereira

是什麼store.storeImgData？ –

什麼是你的數據檢索遇到的問題？ –

它檢索我一個微小的十六進制，這並不讓我建立從 –

請檢查下面的代碼，

NSData *data = [[NSData alloc] initWithBytes:sqlite3_column_blob(statement, 0) length:sqlite3_column_bytes(statement, 0)]; 

object.imgData = data;

代替，

NSUInteger length = sqlite3_column_bytes(statement, 1); 

     object.imgData = [NSData dataWithBytes:sqlite3_column_blob(statement, 1) length:length];

來源

2012-07-03 18:50:09

它給了我同樣的事情。相反，從一個巨大的十六進制的'UIImagePNGRepresentation'它返回我這個'<50524f44 55544f53>' –

@LucasPereira：我已經找到了問題，要指定圖像名稱爲圖像數據。我更新了上面的代碼。尤爾數據是OBJECTIMAGE行，所以你定義使用0而不是1 sqlite3_column_blob（聲明，0） –

是它的工作？或者你有相同的問題？\ –

保存的UIImage作爲二進制SQLite和與imageWithData

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