我尝试只插入两个整数变量到我的 sqlite 数据库。我创建了一个名为 ups.sqlite 的数据库,它有一个表 (upssTable),表有两列。但是当我打开 / Users / ds / Library / Application Support / iPhone Simulator / 5.0 / Applications / FCB4B455-4B7F-4C47-81B6-AC4121874596 /
- (IBAction)onClick:(id)sender {
NSFileManager *fileManager = [NSFileManager defaultManager];
NSError *error;
NSString *dbPath = [self getDBPath];
BOOL success = [fileManager fileExistsAtPath:dbPath];
if(!success) {
NSString *defaultDBPath = [[[NSBundle mainBundle] resourcePath] stringByAppendingPathComponent:@"ups.sqlite"];
success = [fileManager copyItemAtPath:defaultDBPath toPath:dbPath error:&error];
if (!success)
NSAssert1(0, @"Failed to create writable database file with message '%@'.", [error localizedDescription]);
}
NSLog(@"database path %@",dbPath);
if(!(sqlite3_open([dbPath UTF8String], &cruddb) == SQLITE_OK))
{
NSLog(@"An error has occured.");
}
if(sqlite3_open([dbPath UTF8String], &cruddb) ==SQLITE_OK){
NSString * str1 =@"1";
NSString * str2 =@"1";
const char *sql = "INSERT INTO upssTable (column1, column2) VALUES (?,?)";
NSInteger result = sqlite3_prepare_v2(cruddb,sql, -1, &stmt, NULL);
NSLog(@"upss %s\n", sqlite3_errmsg(cruddb));
if(result == SQLITE_OK)
{
sqlite3_bind_int(stmt, 1, [str1 integerValue]);
sqlite3_bind_int(stmt, 2, [str2 integerValue]);
}
else
{
NSAssert1(0, @"Error . '%s'", sqlite3_errmsg(cruddb));
}
sqlite3_reset(stmt);
sqlite3_finalize(stmt);
sqlite3_close(cruddb);
}
}
- (NSString *) getDBPath {
NSArray *paths = NSSearchPathForDirectoriesInDomains(NSDocumentDirectory , NSUserDomainMask, YES);
NSString *documentsDir = [paths objectAtIndex:0];
return [documentsDir stringByAppendingPathComponent:@"ups.sqlite"];
}
我怎样才能解决这个问题？请帮助我。谢谢你的回复。
6
你没有使用 sqlite3_step（）。试试这种方式...
sqlite3 *database;
dbPath=[self.databasePath UTF8String];
if(sqlite3_open(dbPath,&database)==SQLITE_OK)
{
const char *sqlstatement = "INSERT INTO upssTable (column1, column2) VALUES (?,?)";
sqlite3_stmt *compiledstatement;
if(sqlite3_prepare_v2(database,sqlstatement , -1, &compiledstatement, NULL)==SQLITE_OK)
{
NSString * str1 =@"1";
NSString * str2 =@"1";
sqlite3_bind_int(compiledstatement, 1, [str1 integerValue]);
sqlite3_bind_int(compiledstatement, 2, [str2 integerValue]);
if(sqlite3_step(compiledstatement)==SQLITE_DONE)
{
NSLog(@"done");
}
else NSLog(@"ERROR");
sqlite3_reset(compiledstatement);
}
else
{
NSAssert1(0, @"Error . '%s'", sqlite3_errmsg(cruddb));
}
sqlite3_close(database);
}
5
您正在创建 sqlite3 语句，但实际上并没有使用sqlite3_step()执行它。
你似乎也打开数据库两次？
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