Question

WITH company_size(comp_id, employee_count) AS (
SELECT comp_id, count(*)
FROM position NATURAL JOIN works
GROUP BY comp_id),

total_salaries(total_salaries) AS( 
    SELECT sum(pay_rate)
    from position)

SELECT comp_id, employee_count, total_salaries
FROM company_size NATURAL JOIN total_salaries
WHERE employee_count = (SELECT max(employee_count) 
                        FROM company_size);

所以这是我正在处理的查询。我得到了工作，以便在雇员人数方面获得最大雇主，但现在我必须找到从该特定公司支付的工资和工资总额。我有总薪资，但这只是给了我所有公司的总和。如何才能从那个公司获得总工资？

Answer 1

我没有您的表格可以使用，但我希望class LoginView(View): def get(self, request): return render(request, "login.html", {}) def post(self, request): login_form = LoginForm(request.POST) if login_form.is_valid(): user_name = request.POST.get("username", "") pass_word = request.POST.get("password", "") user = authenticate(username=user_name, password=pass_word) if user is not None: if user.is_active: login(request, user) return HttpResponseRedirect(request.META.get('HTTP_REFERER')) else: return render(request, "login.html", {"msg": "用户未激活！"}) else: return render(request, "login.html", {"msg": "用户名或密码错误！"}) else: return render(request, "login.html", {"login_form": login_form})在NATURAL JOIN和company_size之间工作，他们都需要有一个列才能加入上。实际上，total_salaries返回一行包含单行。也许如果它在total_salaries上分组，你会得到更好的结果。

或者为了让它更加轻松，只需将comp_id移出sum(pay_rate)并进入total_salaries查询即可。然后，您可以取消company_size查询并加入total_salaries：

company_size

Answer 2

我认为这是你正在寻找的东西。您获得了所有公司的总和，因为您只是在total_salaries查询中选择了值，而您并未将其加入任何内容。添加id字段和group by将聚合数据，然后你只需要将它连接到某个东西。希望有所帮助。

WITH company_size(comp_id, employee_count) AS (
SELECT comp_id, count(*)
FROM position NATURAL JOIN works
GROUP BY comp_id),

total_salaries(total_salaries, comp_id) AS( 
    SELECT sum(pay_rate), comp_id
    FROM position
    GROUP BY comp_id)

SELECT comp_id, employee_count, total_salaries
FROM company_size cs INNER JOIN total_salaries ts ON cs.comp_id = ts.comp_id
WHERE employee_count = (SELECT max(employee_count) 
                        FROM company_size);

如何让此查询正常工作？

2 个答案: