我的应用程序中有tableview，因为当我单击该行时，视图打开，包含两个图像查看器，其中两个图像从数据库中形成两个不同的表，一个显示图像的图像查看器取决于单击的行，其中第二个为所有行显示相同的图像。这是我的代码，
来自数据库的第一张图片，
-(void)Readthesqlitefileforname:(NSString *)brandname
{
sqlite3 *database;//database object
NSString *docpath=[self doccumentspath];
const char *ch=[docpath UTF8String];//string to constant char UTF8string main part to connect DB
if (sqlite3_open(ch, &database)==SQLITE_OK)
{
NSString *strstmt=[NSString stringWithFormat:@"select name,nik,dob,study,phone,mail,fsong,dp from scrap where name = '%@'",brandname];
const char *chstmt=[strstmt UTF8String];
sqlite3_stmt *sqlstmt;//to execute the above statement
if (sqlite3_prepare_v2(database, chstmt, -1, &sqlstmt, NULL)==SQLITE_OK)
{
while (sqlite3_step(sqlstmt)==SQLITE_ROW) {
const char *ch=(char *)sqlite3_column_text(sqlstmt, 0);
bnnam=[NSString stringWithFormat:@"%s",ch];
const char *ch1=(char *)sqlite3_column_text(sqlstmt, 1);
urladdr=[NSString stringWithFormat:@"%s",ch1];
const char *ch2=(char *)sqlite3_column_text(sqlstmt, 2);
ydob=[NSString stringWithFormat:@"%s",ch2];
const char *ch3=(char *)sqlite3_column_text(sqlstmt, 3);
ystud=[NSString stringWithFormat:@"%s",ch3];
const char *ch4=(char *)sqlite3_column_text(sqlstmt, 4);
yphon=[NSString stringWithFormat:@"%s",ch4];
const char *ch5=(char *)sqlite3_column_text(sqlstmt, 5);
ymail=[NSString stringWithFormat:@"%s",ch5];
const char *ch6=(char *)sqlite3_column_text(sqlstmt,6);
yface=[NSString stringWithFormat:@"%s",ch6]; //NSLog(@"%@",urladdr);
//const char *ch5=(char *)sqlite3_column_text(sqlstmt, 4);
NSUInteger legnth=sqlite3_column_bytes(sqlstmt, 7);
if (legnth>0)
{
NSData *dt=[NSData dataWithBytes:sqlite3_column_blob(sqlstmt, 7) length:legnth];
clsimg=[UIImage imageWithData:dt];//converting data to image
NSLog(@"image 1 %@",clsimg);
}
else
{
clsimg=nil;
}
}
}
sqlite3_finalize(sqlstmt);
}
sqlite3_close(database);
}
来自数据库的第二张图片，
-(void)Readsqlitefile
{
sqlite3 *database;//database object
NSString *docpath=[self doccumentspath];//get sqlite path
const char *ch=[docpath UTF8String];
if (sqlite3_open(ch, &database)==SQLITE_OK)
{
const char *chstmt="select * from scrapsign";
sqlite3_stmt *sqlstmt;
if (sqlite3_prepare_v2(database, chstmt, -1, &sqlstmt, NULL)==SQLITE_OK)
{
while (sqlite3_step(sqlstmt)==SQLITE_ROW)
{
NSUInteger legnt=sqlite3_column_bytes(sqlstmt, 0);
if (legnt>0) {
NSData *dt=[NSData dataWithBytes:sqlite3_column_blob(sqlstmt, 0) length:legnt];
mysign=[UIImage imageWithData:dt];
}
else
{
mysign=nil;
}
}
}
sqlite3_finalize(sqlstmt);
}
sqlite3_close(database);
}
行点击的图像设置
- (void)tableView:(UITableView *)tableView didSelectRowAtIndexPath:(NSIndexPath *)indexPath
{
scrapview *detailViewController = [[scrapview alloc] initWithNibName:@"scrapview" bundle:nil];
[self Readthesqlitefileforname:[ar objectAtIndex:indexPath.row]];
[self.navigationController pushViewController:detailViewController animated:YES];
if (clsimg !=nil)
{
detailViewController.imgv.image=clsimg;
}
[self Readsqlitefile];
if (mysign !=nil)
{
detailViewController.imgv2.image=mysign;//second image viewer
NSLog(@"%@",mysign);
}
detailViewController.txt1.text=urladdr;
detailViewController.txt2.text=ydob;
detailViewController.txt3.text=ystud;
detailViewController.txt4.text=yphon;
detailViewController.txt5.text=ymail;
detailViewController.txt.text=yface;
detailViewController.txt7.text=bnnam;
}
单击该行时，视图应在两个图像查看器上显示不同的图像，但只有一个图像视图仅在单击时更改，第二个图像视图为所有行单击显示相同的图像。请帮我解决问题。