iphone - iOS 中的 SQLite 插入错误

Question

我正在尝试使用 SQLite 执行插入语句，我确定我一定在做一些愚蠢的事情，但我不知道如何让它工作。

我收到的错误是:“错误准备语句”

这是我的代码:

NSString *json_string = [[NSString alloc] initWithData:responseData encoding:NSUTF8StringEncoding];
[responseData release];

// Create new SBJSON parser object
SBJsonParser *parser = [[SBJsonParser alloc] init];

// parse the JSON response into an object
NSDictionary *download = [parser objectWithString:json_string error:nil];
NSDictionary *buildings = [download objectForKey:@"buildings"];

// Open DB
sqlite3 *db = [RLSampleAppDelegate getNewDBConnection];

// Buildings
for (NSDictionary *building in buildings)
{
  NSDictionary *thisbuilding = [building objectForKey:@"building"];
  NSLog(@"%@", [thisbuilding objectForKey:@"buildingname"]);
  sqlite3_stmt *statement = nil;
  NSString* someString = [NSString stringWithFormat:@"INSERT INTO building (buildingid, buildingname) VALUES (\'%@\', \'%@\')", [thisbuilding objectForKey:@"buildingid"], [thisbuilding objectForKey:@"buildingname"]];

  NSLog(@"somestring: %@", someString);
  const char *sql = (const char *) someString;
  if(sqlite3_prepare_v2(db, sql, -1, &statement, NULL)!=SQLITE_OK)
   NSAssert1(0,@"Error preparing statement",sqlite3_errmsg(db));

  while(sqlite3_step(statement) == SQLITE_DONE){}

  sqlite3_finalize(statement);
 }

[json_string release];

Answer 1

您不能在 SQLite3 API 中使用 NSString 实例。您需要一个纯 C 字符串。

所以，尝试改变这个:

const char *sql = (const char *) someString;

进入:

const char *sql = [someString cStringUsingEncoding:[NSString defaultCStringEncoding]];