我如何纠正我的错误？

Question

SELECT 
      hospital_general_information.provider_id,
      hosptial_name hospital_associated_infection,
      provider_id,     
      readmission_complications_deaths.provider_id
FROM
      hosptial_general_information
JOIN  hospital_associated_infection 
      ON hospital_general_information.provider_id = hosptial_associated_infection.provider_id
JOIN  hospital_associated_infection 
      ON hospital_associated_infection.provider_id = readmission_complications_deaths.provider_id

Answer 1

首先..。您需要使您的查询Readable..Readable是帮助您找到像Typo或Missing这样的Error的特性。

select 
   hospital_general_information.provider_id, 
   hosptial_name hospital_associated_infection,
   provider_id, 
   readmission_complications_deaths.provider_id 
from 
   hosptial_general_information 
   join hospital_associated_infection 
     on hospital_general_information.provider_id=hosptial_associated_infection.provider_id 
   join hospital_associated_infection 
     on hospital_associated_infection.provider_id=readmission_complications_deaths.provider_id

这是你的第一个查询。我发现了一些错误

hosptial_name hospital_associated_infection

您在那里有空间，Typo什么时候应该是hospital，成为hosptial，调用该列是错误的。语法是table.column

provider_id,

您调用provider_id，我看到您的join使用这个column，所以它也会产生错误，Ambiguous Error，因为provider_id超过1，您应该确定谁应该是provider_id。

hospital_associated_infection.provider_id=readmission_complications_deaths.provider_id

这是你的第二个加入条件..。你只分配了两个表hospital_associated_infection和hospital_general_information，那么readmission_complications_deaths是谁？你必须分配这个表。

你有那么多Typo在Hospital..。

所以如果这是你的查询应该是..。

select 
   hospital_general_information.provider_id, 
   hospital_associated_infection.hospital_name,
   hospital_associated_infection.provider_id, 
   readmission_complications_deaths.provider_id 
from 
   hospital_general_information 
   join hospital_associated_infection 
     on hospital_general_information.provider_id=hospital_associated_infection.provider_id 
   join readmission_complications_deaths
     on hospital_associated_infection.provider_id=readmission_complications_deaths.provider_id

我建议你用aliasing做你的桌子。因为你的表名太长了。我推荐使用字母(如hgi )表示hospital_general_information

Answer 2

1066个“名字太冗长”吗？

抱歉的。认真地说，当两次连接到同一个表时，请给每个副本一个别名：

select  gi.provider_id,
        hosptial_name AS hospital_associated_infection,
        provider_id,
        rcd.provider_id
    from  hosptial_general_information AS gi
    join  readmission_complications_deaths AS rcd
       ON ???
    join  hospital_associated_infection AS ai_1
       ON ai_1.provider_id = gi.provider_id
    join  hospital_associated_infection AS ai_2 
       ON ai_2.provider_id = rcd.provider_id 

或者是错误?需要readmission_complications_deaths而不是第二次引用hospital_associated_infection。

哦，有没有缺个逗号？