Question

我有两个数据库。

ch_coins.db是[NSBundle mainBundle]中tableview的只读数据。
文档中的User_data.sqlite

我尝试将两个数据库中的select组合并将数据填充到object。我的方法是这样的：

-(NSMutableArray*)returnSubCountries
{
NSString *path = [[[NSBundle mainBundle] resourcePath]
                  stringByAppendingPathComponent:databaseName];

NSArray *paths = NSSearchPathForDirectoriesInDomains (NSDocumentDirectory, NSUserDomainMask, YES);
NSString *documentsDir = [paths objectAtIndex:0];
NSString *userDB = [documentsDir stringByAppendingPathComponent:user_data];


NSMutableArray *subCountiresArr=[[NSMutableArray alloc]init];

if (sqlite3_open([path UTF8String], &database) == SQLITE_OK)
{

    /*const char *sqlSubCountries="SELECT subCountryID,subCountryName,subCountryComment,image,priority,hasRegions,navigationKey\
    FROM subCountries\
    ORDER BY priority ASC";*/
    const char *sqlSubCountries="SELECT subCountryID,subCountryName,subCountryComment,image,priority,hasRegions,navigationKey,usdb.quantity\
    attach database 'userDB' as usdb\
    INNER JOIN usdb on subCountries.subCountryID=usdb.refID\
    FROM subCountries\
    ORDER BY priority ASC";
    sqlite3_stmt *statement;
    int sqlResult = sqlite3_prepare_v2(database, sqlSubCountries, -1, &statement, NULL);
    if ( sqlResult== SQLITE_OK)
    {
        while (sqlite3_step(statement) == SQLITE_ROW)
        {
            SubCountry *sbCountryObj=[[SubCountry alloc]init];
            sbCountryObj.subCountryID=sqlite3_column_int(statement, 0);
            char *subCountryName=(char *)sqlite3_column_text(statement, 1);
            char *subCountryComment=(char *)sqlite3_column_text(statement, 2);
            char *image=(char *)sqlite3_column_text(statement, 3);
            sbCountryObj.priority=sqlite3_column_int(statement, 4);
            sbCountryObj.hasRegions=(sqlite3_column_int(statement, 5)==1);
            sbCountryObj.navigationKey=sqlite3_column_int(statement, 6);
            sbCountryObj.quantity=sqlite3_column_int(statement, 7);
            sbCountryObj.subCountryName=(subCountryName)?[NSString stringWithUTF8String:subCountryName]: @"";
            sbCountryObj.subCountryComment=(subCountryComment)?[NSString stringWithUTF8String:subCountryComment]: @"";
            sbCountryObj.image=(image)?[NSString stringWithUTF8String:image]: @"";
            [subCountiresArr addObject:sbCountryObj];
        }
        sqlite3_finalize(statement);
        sqlite3_close(database);
    }
}
else
{
    //[self dbConnectionError];
}
return subCountiresArr;
}

我一定是做错了。任何帮助深表感谢。感谢名单。

编辑：

-(NSMutableArray*)returnSubCountries
{
NSString *path = [[[NSBundle mainBundle] resourcePath]
                  stringByAppendingPathComponent:databaseName];

NSArray *paths = NSSearchPathForDirectoriesInDomains (NSDocumentDirectory, NSUserDomainMask, YES);
NSString *documentsDir = [paths objectAtIndex:0];
NSString *userDB = [documentsDir stringByAppendingPathComponent:user_data];


NSMutableArray *subCountiresArr=[[NSMutableArray alloc]init];

if (sqlite3_open([path UTF8String], &database) == SQLITE_OK)
{

    /*const char *sqlSubCountries="SELECT subCountryID,subCountryName,subCountryComment,image,priority,hasRegions,navigationKey\
    FROM subCountries\
    ORDER BY priority ASC";*/
    NSString *userDBName = [NSString stringWithFormat:@"attach database '%@' as usdb", userDB];
    const char *sqlAttachedDatabase = [userDBName UTF8String];

    const char *sqlSubCountries="SELECT subCountryID,subCountryName,subCountryComment,image,priority,hasRegions,navigationKey,usdb.quantity\
    sqlAttachedDatabase userDB as usdb\
    INNER JOIN usdb on subCountries.subCountryID=usdb.refID\
    FROM subCountries\
    ORDER BY priority ASC";
    sqlite3_stmt *statement;
    int sqlResult = sqlite3_prepare_v2(database, sqlSubCountries, -1, &statement, NULL);
    if ( sqlResult== SQLITE_OK)
    {
        while (sqlite3_step(statement) == SQLITE_ROW)
        {
            SubCountry *sbCountryObj=[[SubCountry alloc]init];
            sbCountryObj.subCountryID=sqlite3_column_int(statement, 0);
            char *subCountryName=(char *)sqlite3_column_text(statement, 1);
            char *subCountryComment=(char *)sqlite3_column_text(statement, 2);
            char *image=(char *)sqlite3_column_text(statement, 3);
            sbCountryObj.priority=sqlite3_column_int(statement, 4);
            sbCountryObj.hasRegions=(sqlite3_column_int(statement, 5)==1);
            sbCountryObj.navigationKey=sqlite3_column_int(statement, 6);
            sbCountryObj.quantity=sqlite3_column_int(statement, 7);
            sbCountryObj.subCountryName=(subCountryName)?[NSString stringWithUTF8String:subCountryName]: @"";
            sbCountryObj.subCountryComment=(subCountryComment)?[NSString stringWithUTF8String:subCountryComment]: @"";
            sbCountryObj.image=(image)?[NSString stringWithUTF8String:image]: @"";
            [subCountiresArr addObject:sbCountryObj];
        }
        sqlite3_finalize(statement);
        sqlite3_close(database);
    }
    else
    {
        NSLog(@"%s: prepare failed: %s", __FUNCTION__, sqlite3_errmsg(database));
    }
}
else
{
    //[self dbConnectionError];
}
return subCountiresArr;
}

Answer 1

您应该执行以下操作：

运行SQL“attach database”full db path'as usdb“，其中”full db path“是数据库文件的完整路径
准备并执行SELECT语句（从中删除“attach database ...”）
在不需要时分离数据库“分离数据库'完整数据库路径'”

Answer 2

您在SQL中将数据库路径指定为userDB，但这不是数据库文件的名称。也就是说，其中包含完整路径名的变量的名称，但您可能实际上必须使用该文件名构建SQL。例如：

NSString *sql = [NSString stringWithFormat:@"attach database '%@' as usdb", userDB];
const char *sqlAttachedDatabase = [sql UTF8String];

执行。然后，作为单独的语句，您可以执行使用usdb别名的SELECT SQL。当你完成后，分离数据库。

此外，我注意到您正在检查结果是否为SQLITE_OK（这是好的）。但是，如果不是，那么您没有显示任何诊断信息。例如，您的SQL错误，但您没有显示有意义的错误消息。因此，如果sqlite3_prepare_v2未能返回SQLITE_OK，您应该：

NSLog(@"%s: prepare failed: %s", __FUNCTION__, sqlite3_errmsg(database));

如果你在遇到错误时这样做，你就能弄明白什么是错的。没有它，你就会失明。

一个工作示例，我的“作者”数据库位于捆绑包中，而我的“book”数据库位于Documents中：

int rc;

NSString *docsPath = NSSearchPathForDirectoriesInDomains(NSDocumentDirectory, NSUserDomainMask, YES)[0];
NSString *bookFilename = [docsPath stringByAppendingPathComponent:@"book.sqlite"];

NSString *authorFilename = [[NSBundle mainBundle] pathForResource:@"author" ofType:@"sqlite"];

sqlite3 *database;

if ((rc = sqlite3_open([authorFilename UTF8String], &database)) != SQLITE_OK)
{
    NSLog(@"%s: open failed: %s (%d)", __FUNCTION__, sqlite3_errmsg(database), rc);
    return;
}

NSString *sql = [NSString stringWithFormat:@"attach database '%@' as userdb;", bookFilename];

if ((rc = sqlite3_exec(database, [sql UTF8String], NULL, NULL, NULL)) != SQLITE_OK)
    NSLog(@"%s: attach failed: %s (%d)", __FUNCTION__, sqlite3_errmsg(database), rc);

sqlite3_stmt *statement;

sql = @"select book.*, author.* from userdb.book inner join author on author_id = book_author_id;";

if ((rc = sqlite3_prepare_v2(database, [sql UTF8String], -1, &statement, NULL)) != SQLITE_OK)
    NSLog(@"%s: prepare failed: %s (%d)", __FUNCTION__, sqlite3_errmsg(database), rc);

while ((rc = sqlite3_step(statement)) == SQLITE_ROW)
{
    // do whatever you want row by row
    NSLog(@"Row");
}

if (rc != SQLITE_DONE)
    NSLog(@"%s: step failed: %s (%d)", __FUNCTION__, sqlite3_errmsg(database), rc);

sqlite3_finalize(statement);

sql = @"detach database userdb;";

if ((rc = sqlite3_exec(database, [sql UTF8String], NULL, NULL, NULL)) != SQLITE_OK)
    NSLog(@"%s: detach failed: %s (%d)", __FUNCTION__, sqlite3_errmsg(database), rc);

sqlite3_close(database);

iOS 6：尝试从不同路径的2个数据库中进行选择

2 个答案: