谁能告诉我为什么我无法在 sqlite 中插入图像。有时我得到如下错误：1）（sqlite3_step（stm）== SQLITE_DONE），将值设为21。2）存储在数组中而不是数据库中。我是目标 c 的新手，所以如果你能发回代码，那就太好了，因为我几乎搜索了每个网站。谢谢！
-(IBAction)submitDetails:(id)sender
{
sqlite3_stmt *stm;
const char *dbpath = [databasePath UTF8String];
if (sqlite3_open(dbpath, &connectDB) == SQLITE_OK)
{
UIImage *contactImage = imageView.image;
NSData *imageData = UIImageJPEGRepresentation(contactImage, 100);
NSUserDefaults *defaults = [NSUserDefaults standardUserDefaults];
[defaults setObject:imageData forKey:@"image"];
[defaults synchronize];
NSLog(@"Data saved");
NSString *insertSQL = [NSString stringWithFormat: @"INSERT INTO STUDENTS (name,salary, gid,photo) VALUES (\"%@\", \"%@\", \"%@\",?)", name.text,salary.text,gid.text,imageData] ;
const char *insert_stmt = [insertSQL UTF8String];
sqlite3_prepare_v2(connectDB, insert_stmt, -1, &stm, NULL);
NSArray *paths =NSSearchPathForDirectoriesInDomains(NSDocumentDirectory, NSUserDomainMask, YES);
NSString *documentsDirec = [paths objectAtIndex:0];
NSString *imgePath = [documentsDirec stringByAppendingPathComponent:@"note.sqlite"];
if(sqlite3_open([imgePath UTF8String], &database) == SQLITE_OK){
const char *sql = "insert into images (images) values (?)";
if(sqlite3_prepare_v2(database, sql, -1, &addStmt, NULL) == SQLITE_OK){
UIImage *edtedImae = [info objectForKey:UIImagePickerControllerOriginalImage];
NSData *dataForImage = UIImagePNGRepresentation(edtedImae);
sqlite3_bind_blob(addStmt, 1, [dataForImage bytes], [dataForImage length], SQLITE_TRANSIENT);
NSLog(@"1st ..... %i",sqlite3_step(stm));
NSLog(@" 2nd.... %i",SQLITE_DONE);
if (sqlite3_step(stm) == SQLITE_DONE)
{
NSString* aName=name.text;
NSString* aSalary=salary.text;
NSString* aGid=gid.text;
UIImage* aPhoto=[UIImage imageWithData:imageData];
Person *person = [[Person alloc] initWithName:aName salary:aSalary gid:aGid photo:aPhoto];
[rootObject.list addObject:person];
[person release];
status.text = @"Contact added";
NSLog(@"contact added %@ %@",name.text,gid.text);
NSLog(@"the count in person list is %i",[rootObject.list count]);
name.text = @"";
gid.text = @"";
salary.text = @"";
}
else
{
status.text = @"Failed to add contact";
}
sqlite3_finalize(stm);
sqlite3_close(connectDB);
}
}