Question

我试图让这个SQL查询正确但我不断获得一个空值的行。

所以我有两张桌子。

表A是chat表。

表B是wallpaper表。

chat可能或可能不会分配任何壁纸ID是外键。

我尝试使用以下查询将壁纸附加到聊天中：

const char *query = "SELECT wallpaper.* FROM wallpaper INNER JOIN chat ON wallpaper.id_wallpaper = chat.id_wallpaper AND (chat.id_chat = ? AND chat.id_wallpaper NOT NULL)";

在我的方法中：

-(Wallpaper *)getWallpaperForChatHeaderID:(NSString *)chatHeaderID
{
    sqlite3_stmt *statement;

    const char *query = "SELECT wallpaper.* FROM wallpaper INNER JOIN chat ON wallpaper.id_wallpaper = chat.id_wallpaper AND (chat.id_chat = ? AND chat.id_wallpaper NOT NULL)";

    Wallpaper *wallpaper = nil;

    if(sqlite3_prepare_v2(_database, query, -1, &statement, nil) == SQLITE_OK)
    {
        sqlite3_bind_int(statement, 1, [chatHeaderID intValue]);

        int statusCode = sqlite3_step(statement);

        if(statusCode == SQLITE_DONE)
        {
            long idWallpaper = sqlite3_column_int(statement, 0);

            char *dbFilename = (char *)sqlite3_column_text(statement, 1);

            ...

            wallpaper = [[Wallpaper alloc] initWithWallpaperID:idWallpaper ....];
        }
        else
        {
            NSLog(@"sql step status error code: %d", statusCode);
        }

        sqlite3_finalize(statement);
    }
    else
    {
        NSLog(@"Failed to get wallpaper for chat header with ID: %@", chatHeaderID);
    }

    return wallpaper;
}

但是这会让我返回一个wallpaper个实体，所有字段都为NULL。

聊天只能有1个壁纸。所以我只想获取一个wallpaper，它链接到chatHeaderID标识的聊天。

据我所知，SQL查询根据我传入的chat在chat表中找到了匹配的chatHeaderID条目，但似乎忽略了我的WHERE子句：

(chat.id_chat = ? AND chat.id_wallpaper NOT NULL)

我期待看到的是SQL查询此刻返回0个条目，因为我没有将任何壁纸条目与任何聊天条目链接。

有什么想法吗？：d

Answer 1

我的另一个问题是我自己解决了= /

不确定为什么上面的问题代码不起作用，但我发现了一项似乎可以产生预期结果的工作。

-(Wallpaper *)getWallpaperForChatHeaderID:(NSString *)chatHeaderID
{
    sqlite3_stmt *statement;

    const char *query = "SELECT wallpaper.* FROM wallpaper INNER JOIN chat ON wallpaper.id_wallpaper = chat.id_wallpaper WHERE chat.id_chat = ? AND chat.id_wallpaper NOT NULL";

    Wallpaper *wallpaper = nil;

    if(sqlite3_prepare_v2(_database, query, -1, &statement, nil) == SQLITE_OK)
    {
        sqlite3_bind_int(statement, 1, [chatHeaderID intValue]);

        while (sqlite3_step(statement) == SQLITE_ROW)
        {
            long idWallpaper = sqlite3_column_int(statement, 0);

            ...

            wallpaper = [[Wallpaper alloc] initWithWallpaperID:idWallpaper ...];
        }

        sqlite3_finalize(statement);
    }
    else
    {
        NSLog(@"Failed to get wallpaper for chat header with ID: %@", chatHeaderID);
    }

    return wallpaper;
}

这里的不同之处在于我使用：

while(... == SQLITE_ROW) 
{ 
    ... 
}

检查是否满足WHERE过滤条件的匹配行的条件：）

我想指出我已将上述SQL查询更改为使用：

SELECT wallpaper.* FROM wallpaper INNER JOIN chat ON wallpaper.id_wallpaper = chat.id_wallpaper WHERE chat.id_chat = ? AND chat.id_wallpaper NOT NULL

但我已经测试过：

SELECT wallpaper.* FROM wallpaper INNER JOIN chat ON wallpaper.id_wallpaper = chat.id_wallpaper AND (chat.id_chat = ? AND chat.id_wallpaper NOT NULL)

两者似乎都有效。

我会使用包含明确WHERE关键字的关键字，因为它看起来更具可读性。

SQLite INNER JOIN WHERE列不为空

1 个答案: