Question

我试图将这两个SQL查询连接在一起。我的数据位于https://policevideorequests.cartodb.com/tables/seattle_police_govqa_audit_trails，它有一个Postgresql SQL API。

SELECT 
    t1.customer_id, t1.c,
    t2.customer_name, t2.customer_email, t2.customer_email_domain 
FROM 
    (SELECT 
         a.customer_id, count(a.customer_id) as c 
     FROM 
         (SELECT customer_id, reference_no 
          FROM seattle_police_govqa_audit_trails 
          WHERE customer_id NOT IN (5, 0, -1) 
          GROUP BY customer_id, reference_no) a 
     GROUP BY 
         a.customer_id 
     ORDER BY 
         count(a.customer_id) DESC) t1 
INNER JOIN 
    (SELECT DISTINCT 
         customer_id,
         INITCAP(LOWER(SUBSTRING(new_value FROM 'Dear&nbsp;(.*?):</div>'))) as customer_name,
         LOWER(SUBSTRING(new_value FROM 'login:<b>(.*?)</b>')) as customer_email, 
         LOWER(SUBSTRING(new_value FROM 'login:<b>.*?@(.*?)</b>')) as customer_email_domain 
     FROM 
         seattle_police_govqa_audit_trails 
     WHERE 
         SUBSTRING(new_value FROM 'Dear&nbsp;(.*?):</div>') IS NOT NULL) t2 ON t1.customer_id = t2.customer_id 
ORDER BY 
    t1.c DESC

SELECT DISTINCT 
    t1.new_value as requester_type, t2.customer_id 
FROM 
    (SELECT 
         reference_no, new_value 
     FROM 
         seattle_police_govqa_audit_trails 
     WHERE 
         action_desc = 154) t1
INNER JOIN 
    (SELECT 
         reference_no, customer_id 
     FROM 
         seattle_police_govqa_audit_trails 
     WHERE 
         customer_id NOT IN (0, -1, 5)) t2 ON t1.reference_no = t2.reference_no

我尝试加入这两个人：

SELECT t1.customer_id,t3.requester_typer,t1.c,t2.customer_name,t2.customer_email,t2.customer_email_domain,t2.customer_email_domain_tld FROM (SELECT a.customer_id,count(a.customer_id) as c FROM (SELECT customer_id, reference_no FROM seattle_police_govqa_audit_trails WHERE customer_id NOT IN (5,0,-1) GROUP BY customer_id,reference_no) a GROUP BY a.customer_id ORDER BY count(a.customer_id) DESC) t1 
INNER JOIN (SELECT DISTINCT customer_id,INITCAP(LOWER(SUBSTRING(new_value FROM 'Dear&nbsp;(.*?):</div>'))) as customer_name,LOWER(SUBSTRING(new_value FROM 'login:<b>(.*?)</b>')) as customer_email, LOWER(SUBSTRING(new_value FROM 'login:<b>.*?@(.*?)</b>')) as customer_email_domain, LOWER(SUBSTRING(new_value FROM 'login:<b>.*?@.*?\.(.*?)</b>')) as customer_email_domain_tld FROM seattle_police_govqa_audit_trails WHERE SUBSTRING(new_value FROM 'Dear&nbsp;(.*?):</div>') IS NOT NULL) t2 
ON t1.customer_id = t2.customer_id ORDER BY t1.c DESC
INNER JOIN (SELECT DISTINCT t1.new_value as requester_type,t2.customer_id FROM (SELECT reference_no,new_value FROM seattle_police_govqa_audit_trails WHERE action_desc = 154) t1
INNER JOIN (SELECT reference_no,customer_id FROM seattle_police_govqa_audit_trails WHERE customer_id NOT IN (0,-1,5)) t2
ON t1.reference_no = t2.reference_no) as t3
ON t2.customer_id = t3.customer_id

我在INNER＆＃34;

附近收到错误＆＃34;语法错误

Answer 1

SQL查询中的问题是您试图将ORDER BY保留在中间位置。必须将ORDER BY子句一直移动到查询的后面，因为排序适用于整个查询，而不是其部分。

Answer 2

试试这个：

  SELECT t1.customer_id,t3.requester_typer,t1.c,t2.customer_name,t2.customer_email,t2.customer_email_domain,t2.customer_email_domain_tld FROM (SELECT a.customer_id,count(a.customer_id) as c FROM (SELECT customer_id, reference_no FROM seattle_police_govqa_audit_trails WHERE customer_id NOT IN (5,0,-1) GROUP BY customer_id,reference_no) a GROUP BY a.customer_id ORDER BY count(a.customer_id) DESC) t1 
    INNER JOIN (SELECT DISTINCT customer_id,INITCAP(LOWER(SUBSTRING(new_value FROM 'Dear&nbsp;(.*?):</div>'))) as customer_name,LOWER(SUBSTRING(new_value FROM 'login:<b>(.*?)</b>')) as customer_email, LOWER(SUBSTRING(new_value FROM 'login:<b>.*?@(.*?)</b>')) as customer_email_domain, LOWER(SUBSTRING(new_value FROM 'login:<b>.*?@.*?\.(.*?)</b>')) as customer_email_domain_tld FROM seattle_police_govqa_audit_trails WHERE SUBSTRING(new_value FROM 'Dear&nbsp;(.*?):</div>') IS NOT NULL) t2 
    ON t1.customer_id = t2.customer_id 
    INNER JOIN (SELECT DISTINCT t1.new_value as requester_type,t2.customer_id FROM (SELECT reference_no,new_value FROM seattle_police_govqa_audit_trails WHERE action_desc = 154) t1
    INNER JOIN (SELECT reference_no,customer_id FROM seattle_police_govqa_audit_trails WHERE customer_id NOT IN (0,-1,5)) t2
    ON t1.reference_no = t2.reference_no) as t3
    ON t2.customer_id = t3.customer_id
    ORDER BY t1.c DESC

加入两个有INNER JOINS的查询

2 个答案: