从iPhone上的SQLite3获取数据时出现问题

我一直试图从一个表返回数据，之前已经访问了两个表，但在本例中，它进入while语句，但没有分配任何值，因为所有内容都设置为null

代码是：

NSMutableArray *all_species = [[NSMutableArray alloc] init];
sqlite3 *db_species;
int dbrc_species;
Linnaeus_LiteAppDelegate *appDelegate = (Linnaeus_LiteAppDelegate*) [UIApplication sharedApplication].delegate;
const char* dbFilePathUTF8 = [appDelegate.dbFilePath UTF8String];
dbrc_species = sqlite3_open (dbFilePathUTF8, &db_species);
if (dbrc_species) {
    return all_species;
}
sqlite3_stmt *dbps_species;
const char *queryStatement = "SELECT species_id, species_name, species_latin, species_genus FROM \
                                linnaeus_species;";
if (sqlite3_prepare_v2 (db_species, queryStatement, -1, &dbps_species, NULL) == SQLITE_OK) {
    sqlite3_bind_int(dbps_species, 1, [the_species_id intValue]);
    while (sqlite3_step(dbps_species) == SQLITE_ROW) {
        Species *species = [[Species alloc] init];
        NSLog(@"%@", sqlite3_column_int(dbps_species, 0));
        [species setSpecies_id:[[NSNumber alloc] initWithInt:sqlite3_column_int(dbps_species, 0)]];
        char *new_name = (char *) sqlite3_column_text(dbps_species, 1);
        [species setSpecies_name:nil];
        if (new_name != NULL) {
            [species setSpecies_name:[NSString stringWithUTF8String:(char *) sqlite3_column_text(dbps_species, 1)]];
        }
        char *new_latin = (char *) sqlite3_column_text(dbps_species, 2);
        [species setSpecies_latin:nil];
        if (new_latin != NULL) {
            [species setSpecies_latin:[NSString stringWithUTF8String:(char *) sqlite3_column_text(dbps_species, 2)]];
        }
        [species setSpecies_genus:[NSNumber numberWithInt:sqlite3_column_int(dbps_species, 3)]];

        [species setEdited:0];
        [all_species addObject:species];
        [species release];
    }
    sqlite3_finalize(dbps_species);
}
else {
    sqlite3_close(db_species);
}

我还尝试过使用NSLog（@“Data:%@”，sqlite3_column_text（dbps_species，1））；它会导致EXC_BAD_访问错误，这表明它可能与内存有关，但我不明白为什么

NSLog(@"Data: %@", sqlite3_column_text(dbps_species, 1));

由于

sqlite3\u column\u text

的结果是C字符串（

char*

），而不是

NSString*

，因此将导致

EXC\u BAD\u访问

。要打印C字符串，您需要

%s

格式说明符：

NSLog(@"Data: %s", sqlite3_column_text(dbps_species, 1));

---

另外，不要浪费时间调用

sqlite3\u column\u text两次，例如

    char *new_name = (char *) sqlite3_column_text(dbps_species, 1);
    [species setSpecies_name:nil];
    if (new_name != NULL) {
        [species setSpecies_name:[NSString stringWithUTF8String:new_name]];
    }

您还可以尝试使用FMDB类。这使得使用sqlite变得更加容易

谢谢您的回复：）这就解释了为什么我当时无法调试它！我认为%@可以允许任何内容类型：（现在我已经能够进行调试，以表明我能够解决问题-谢谢！：D