无法使用mysql 3联接查询获得完整数据

Question

我在域表中有1500条记录，但是通过使用这个查询只能得到1215条记录。如何修改此查询以提供所需的结果和更好的性能

SELECT 
  d.id, d.domain_name, d.action, d.comment, d.agent_email, 
  d.assigned_date, d.added_date, dd.registered_on, dd.expiry_date,
  dd.updated_date, dd.acquire_price, dd.acquire_date, dd.email,
  dd.effective_price, dd.registrar, dd.status, dd.servers, 
  count(l.lead_domain), d.domainer_email, d.current_status, 
  d.undeveloped, d.sedo, d.afternic, d.flippa, d.uniregistry, 
  d.go_daddy, d.domr, d.minimum_offer, d.buy_it_now_price, 
  d.way_to_find_leads, dd.tlds_taken 
FROM domains d 
right join domains_data dd on d.domain_name=dd.domain_name 
left outer join lead_domains l on d.domain_name=l.domain_name 
  and d.domainer_email=l.domainer_email 
group by d.domain_name 
having d.action='all' and d.domainer_email='abc@gmail.com' 
order by d.added_date desc;

Answer 1

@Harshal根据我的理解，我不认为您提供的查询会因为按规则分组不被适当遵守而起作用。

SELECT  d.id,
        d.domain_name,
        d.action,
        d.comment,
        d.agent_email,
        d.assigned_date,
        d.added_date,
        dd.registered_on,
        dd.expiry_date,
        dd.updated_date,
        dd.acquire_price,
        dd.acquire_date,
        dd.email,
        dd.effective_price,
        dd.registrar,
        dd.status,
        dd.servers,
        (SELECT COUNT(lead_domain) FROM lead_domains WHERE domain_name = d.domain_name AND domainer_email = d.domainer_email) AS lead_domain,
        d.domainer_email,
        d.current_status,
        d.undeveloped,
        d.sedo,
        d.afternic,
        d.flippa,
        d.uniregistry,
        d.go_daddy,
        d.domr,
        d.minimum_offer,
        d.buy_it_now_price,
        d.way_to_find_leads,
        dd.tlds_taken
FROM domains d
LEFT JOIN domains_data dd 
       ON d.domain_name = dd.domain_name
WHERE d.action = 'all'
      AND d.domainer_email = 'abc@gmail.com'
ORDER BY
    d.added_date DESC;

由于你没有提供任何预期的结果，我只是猜测我的解决方案会奏效。

Answer 2

您的结果取决于domains_data表。

由于您正确地将domains加入到domains_data中，所以domains_data中的条目将被接受。如果您希望考虑domains的条目，请使用左联接，如下所示。

SELECT d.id, d.domain_name, d.action, d.comment, d.agent_email, d.assigned_date, d.added_date, dd.registered_on, dd.expiry_date, dd.updated_date, dd.acquire_price, dd.acquire_date, dd.email, dd.effective_price, dd.registrar, dd.status, dd.servers, COUNT(l.lead_domain), d.domainer_email, d.current_status, d.undeveloped, d.sedo, d.afternic, d.flippa, d.uniregistry, d.go_daddy, d.domr, d.minimum_offer, d.buy_it_now_price, d.way_to_find_leads, dd.tlds_taken
FROM domains d
LEFT JOIN domains_data dd ON d.domain_name=dd.domain_name
LEFT OUTER JOIN lead_domains l ON d.domain_name=l.domain_name AND d.domainer_email=l.domainer_email
GROUP BY d.domain_name
HAVING d.action='all' AND d.domainer_email='abc@gmail.com'
ORDER BY d.added_date DESC;

如果仍然无法获得所需的结果，请检查having条件。

Answer 3

这是我的建议，在加入其他表格之前先过滤。

SELECT d.id
    ,d.domain_name
    ,d.action
    ,d.comment
    ,d.agent_email
    ,d.assigned_date
    ,d.added_date
    ,dd.registered_on
    ,dd.expiry_date
    ,dd.updated_date
    ,dd.acquire_price
    ,dd.acquire_date
    ,dd.email
    ,dd.effective_price
    ,dd.registrar
    ,dd.status
    ,dd.servers
    ,t1.cnt
    ,d.domainer_email
    ,d.current_status
    ,d.undeveloped
    ,d.sedo
    ,d.afternic
    ,d.flippa
    ,d.uniregistry
    ,d.go_daddy
    ,d.domr
    ,d.minimum_offer
    ,d.buy_it_now_price
    ,d.way_to_find_leads
    ,dd.tlds_taken 
FROM domains d 
inner join
    (select domain_name, count(1) as cnt 
        from domain
        where
            action='all' and domainer_email='abc@gmail.com'
        group by domain_name
     ) as t1 on t1.domain_name = d.domain_name
right join domains_data dd on d.domain_name=dd.domain_name 
left outer join lead_domains l on d.domain_name=l.domain_name and d.domainer_email=l.domainer_email 
order by d.added_date desc;