【发布时间】:2011-09-24 11:55:42
【问题描述】:
我正在将图像转换为数据并将图像保存在数据库中,如下所示 NSData *imageData = UIImagePNGRepresentation(imgView.image);
[dataArray addObject:imageData];
并将数据检索到图像中,如下所示
NSData *imdata = [[tableArray objectAtIndex:indexPath.row] objectForKey:@"Image"];
用于保存的 SQlite 代码:
if (sqlite3_open([databasePath UTF8String], &database)== SQLITE_OK)
{
NSString *statement;
sqlite3_stmt *compiledstatement;
int Id;
NSString *Name;
NSData *imgData;
Id = [[recordArray objectAtIndex:0] intValue];
Name = [recordArray objectAtIndex:1];
imgData = [recordArray objectAtIndex:2];
statement = [[NSString alloc]initWithFormat:@"insert into ImageTable values ('%d','%@','%@')",Id,Name,imgData] ;
const char *sqlstatement = [statement UTF8String];
if (sqlite3_prepare_v2(database, sqlstatement, -1, &compiledstatement, NULL)== SQLITE_OK) {
if (SQLITE_DONE!=sqlite3_step(compiledstatement) ) {
NSAssert1(0,@"Error when inserting %s",sqlite3_errmsg(database));
}
else {
NSLog(@"Data inserted Successfully");
}
//[recordDict release];
}
else {
NSLog(@"Failed with error");
NSAssert1(0,@"Error when creation %s",sqlite3_errmsg(database));
}
}
//sqlite3_finalize(compiledstatement);
获取的Sqlite代码:
- (NSMutableArray *) getRecord
{
[self checkAndCreateDatabase];
sqlArray=[[NSMutableArray alloc]init ];
if (sqlite3_open([databasePath UTF8String], &database) == SQLITE_OK)
{
const char *sql = "select * from ImageTable";
sqlite3_stmt *selectstmt;
if(sqlite3_prepare_v2(database, sql, -1, &selectstmt, NULL) == SQLITE_OK)
{
while(sqlite3_step(selectstmt) == SQLITE_ROW)
{
//[sqlDict retain];
sqlDict =[[NSMutableDictionary alloc]init ];
NSString *Id = [NSString stringWithUTF8String:(char *)sqlite3_column_text(selectstmt, 0)];
//[barcodeArray addObject:prdbcode];
[sqlDict setObject:Id forKey:@"Id"];
//[prdbcode release];
NSString *name = [NSString stringWithUTF8String:(char *)sqlite3_column_text(selectstmt, 1)];
//[nameArray addObject:prdname];
[sqlDict setObject:name forKey:@"Name"];
//[prdname release];
NSData *image = [NSString stringWithUTF8String:(char *)sqlite3_column_text(selectstmt, 2)];
//NSData *data = [[NSData alloc] initWithBytes:sqlite3_column_blob(selectstmt, 1) length:sqlite3_column_bytes(selectstmt, 1)];
//[descArray addObject:prdDesc];
[sqlDict setObject:image forKey:@"Image"];
// [prdDesc release];
[sqlArray addObject:sqlDict];
}
sqlite3_finalize(selectstmt);
}
sqlite3_close(database);
//
}
return sqlArray;
}
[imView setImage:[UIImage imageWithData:imdata]];
[cell addSubview:imView];
Id 和 Name 已检索但图像未显示。 该应用程序终止而没有任何错误消息.. 请告诉我解决方案
【问题讨论】:
-
这与 SQLite 有什么关系?
-
类似问题:stackoverflow.com/questions/10192945/…希望你能解决你的问题。
标签: iphone database xcode uiimage