在Fragment函数中getActivity（）返回null(getActivity() returns null in Fragment function)

 首先，我们来做一些测试数据： 
 
create table client (client_id integer not null primary key auto_increment,
                     name varchar(64));
create table portfolio (portfolio_id integer not null primary key auto_increment,
                        client_id integer references client.id,
                        cash decimal(10,2),
                        stocks decimal(10,2));
insert into client (name) values ('John Doe'), ('Jane Doe');
insert into portfolio (client_id, cash, stocks) values (1, 11.11, 22.22),
                                                       (1, 10.11, 23.22),
                                                       (2, 30.30, 40.40),
                                                       (2, 40.40, 50.50);
 
 如果您不需要投资组合ID，那将很容易： 
 
select client_id, name, max(cash + stocks)
from client join portfolio using (client_id)
group by client_id

+-----------+----------+--------------------+
| client_id | name     | max(cash + stocks) |
+-----------+----------+--------------------+
|         1 | John Doe |              33.33 | 
|         2 | Jane Doe |              90.90 | 
+-----------+----------+--------------------+
 
 由于您需要投资组合ID，事情变得更加复杂。 让我们一起做 首先，我们将编写一个返回每个客户端的最大投资组合值的子查询： 
 
select client_id, max(cash + stocks) as maxtotal
from portfolio
group by client_id

+-----------+----------+
| client_id | maxtotal |
+-----------+----------+
|         1 |    33.33 | 
|         2 |    90.90 | 
+-----------+----------+
 
 然后，我们将查询投资组合表，但是使用前一个子查询的连接，以便仅保留其总值为客户端最大值的投资组合： 
 
 select portfolio_id, cash + stocks from portfolio 
 join (select client_id, max(cash + stocks) as maxtotal 
       from portfolio
       group by client_id) as maxima
 using (client_id)
 where cash + stocks = maxtotal

+--------------+---------------+
| portfolio_id | cash + stocks |
+--------------+---------------+
|            5 |         33.33 | 
|            6 |         33.33 | 
|            8 |         90.90 | 
+--------------+---------------+
 
 最后，我们可以加入客户端表（如你所做），以便包括每个客户端的名称： 
 
select client_id, name, portfolio_id, cash + stocks
from client
join portfolio using (client_id)
join (select client_id, max(cash + stocks) as maxtotal
      from portfolio 
      group by client_id) as maxima
using (client_id)
where cash + stocks = maxtotal

+-----------+----------+--------------+---------------+
| client_id | name     | portfolio_id | cash + stocks |
+-----------+----------+--------------+---------------+
|         1 | John Doe |            5 |         33.33 | 
|         1 | John Doe |            6 |         33.33 | 
|         2 | Jane Doe |            8 |         90.90 | 
+-----------+----------+--------------+---------------+
 
 请注意，这将为John Doe返回两行，因为他具有两个完全相同总值的投资组合。 为了避免这种情况，并选择一个任意的顶级投资组合，在GROUP BY子句上标记： 
 
select client_id, name, portfolio_id, cash + stocks
from client
join portfolio using (client_id)
join (select client_id, max(cash + stocks) as maxtotal
      from portfolio 
      group by client_id) as maxima
using (client_id)
where cash + stocks = maxtotal
group by client_id, cash + stocks

+-----------+----------+--------------+---------------+
| client_id | name     | portfolio_id | cash + stocks |
+-----------+----------+--------------+---------------+
|         1 | John Doe |            5 |         33.33 | 
|         2 | Jane Doe |            8 |         90.90 | 
+-----------+----------+--------------+---------------+

First, let's make some test data:
 
create table client (client_id integer not null primary key auto_increment,
                     name varchar(64));
create table portfolio (portfolio_id integer not null primary key auto_increment,
                        client_id integer references client.id,
                        cash decimal(10,2),
                        stocks decimal(10,2));
insert into client (name) values ('John Doe'), ('Jane Doe');
insert into portfolio (client_id, cash, stocks) values (1, 11.11, 22.22),
                                                       (1, 10.11, 23.22),
                                                       (2, 30.30, 40.40),
                                                       (2, 40.40, 50.50);
 
If you didn't need the portfolio ID, it would be easy:
 
select client_id, name, max(cash + stocks)
from client join portfolio using (client_id)
group by client_id

+-----------+----------+--------------------+
| client_id | name     | max(cash + stocks) |
+-----------+----------+--------------------+
|         1 | John Doe |              33.33 | 
|         2 | Jane Doe |              90.90 | 
+-----------+----------+--------------------+
 
Since you need the portfolio ID, things get more complicated. Let's do it in steps. First, we'll write a subquery that returns the maximal portfolio value for each client:
 
select client_id, max(cash + stocks) as maxtotal
from portfolio
group by client_id

+-----------+----------+
| client_id | maxtotal |
+-----------+----------+
|         1 |    33.33 | 
|         2 |    90.90 | 
+-----------+----------+
 
Then we'll query the portfolio table, but use a join to the previous subquery in order to keep only those portfolios the total value of which is the maximal for the client:
 
 select portfolio_id, cash + stocks from portfolio 
 join (select client_id, max(cash + stocks) as maxtotal 
       from portfolio
       group by client_id) as maxima
 using (client_id)
 where cash + stocks = maxtotal

+--------------+---------------+
| portfolio_id | cash + stocks |
+--------------+---------------+
|            5 |         33.33 | 
|            6 |         33.33 | 
|            8 |         90.90 | 
+--------------+---------------+
 
Finally, we can join to the client table (as you did) in order to include the name of each client:
 
select client_id, name, portfolio_id, cash + stocks
from client
join portfolio using (client_id)
join (select client_id, max(cash + stocks) as maxtotal
      from portfolio 
      group by client_id) as maxima
using (client_id)
where cash + stocks = maxtotal

+-----------+----------+--------------+---------------+
| client_id | name     | portfolio_id | cash + stocks |
+-----------+----------+--------------+---------------+
|         1 | John Doe |            5 |         33.33 | 
|         1 | John Doe |            6 |         33.33 | 
|         2 | Jane Doe |            8 |         90.90 | 
+-----------+----------+--------------+---------------+
 
Note that this returns two rows for John Doe because he has two portfolios with the exact same total value. To avoid this and pick an arbitrary top portfolio, tag on a GROUP BY clause:
 
select client_id, name, portfolio_id, cash + stocks
from client
join portfolio using (client_id)
join (select client_id, max(cash + stocks) as maxtotal
      from portfolio 
      group by client_id) as maxima
using (client_id)
where cash + stocks = maxtotal
group by client_id, cash + stocks

+-----------+----------+--------------+---------------+
| client_id | name     | portfolio_id | cash + stocks |
+-----------+----------+--------------+---------------+
|         1 | John Doe |            5 |         33.33 | 
|         2 | Jane Doe |            8 |         90.90 | 
+-----------+----------+--------------+---------------+