Question

我有3张桌子，第一张是申请人

Ap_id       Username          Email            password
 1          Elen_zey     Alan@gmail.com        123456
 2          Morgan_jam    Morgan@gmail.com     443556
 3          Alex_Hales     Alex@gmail.com      534545

申请人个人信息

Ap_id       FirstName     LastName        Contact       CNIC
 1           Alan          Zey          214343555   4663-14334-1
 2           Morgan        Jam          984209482   2663-14334-4
 3           Alex          Hales        343453453   7663-14334-2

申请人专业信息

Ap_id    duration    organization     Designation   Total Experience
 1         10          10 Pearls       Intern          20 Months
 1         10          10 Pearls       DBA             20 Months
 2         36          Radium Inc       SQA             36 Months
 3         46          Sper Inc        Intern          46 Months

我想检索申请人的电子邮件，联系电话，姓名和姓氏，其中总体验相等且超过10个月。我试过这个查询

SELECT GROUP_CONCAT(DISTINCT CONCAT(aper.FirstName," ",aper.LastName)) AS 
Name, GROUP_CONCAT(DISTINCT ap.Email), GROUP_CONCAT(DISTINCT aper.Contact), 
GROUP_CONCAT(DISTINCT apro.totalExperience) AS Experience FROM 
applicant_personal_info aper INNER JOIN applicant ap ON aper.ApplicantID = 
ap.ApplicantID INNER JOIN applicant_professional_info apro ON 
aper.ApplicantID = apro.ApplicantID WHERE apro.totalExperience >=10

我得到这样的结果

Name : 'Alan Zey, Morgan Jam, Alex Hales',
Email : 'Alan@gmail.com, Morgan@gmail.com,Alex@gmail.com',
Contact : '214343555,984209482,343453453',
Experience : '20,36,46'

但我想这样结果

Name : Alan Zey,
Email : Alan@gmail.com,  
Contact : 214343555,
Experience : 20

Name : Morgan Jam,
Email : Mirgan@gmail.com,  
Contact : 984209482   ,
Experience : 36

Name : Alex Hales,
Email : Alex@gmail.com,  
Contact : 343453453,
Experience : 46

由于申请人专业信息中有多条记录，因此可以使用任何记录，因为在总体经验栏中，所有持续时间列的总数与申请人ID相同。

我怎样才能实现这个目标？

Answer 1

摆脱所有GROUP_CONCAT()次调用，他们将所有记录合并为一行，并以逗号分隔值。

使用GROUP BY aper.ApplicantID为每位申请人获取一行，并使用MAX(apro.totalExperience)在最终结果中获得所有专业信息记录的最高体验。

SELECT DISTINCT CONCAT(aper.FirstName," ",aper.LastName) AS Name, ap.Email, aper.Contact, MAX(apro.totalExperience) AS Experience
FROM applicant_personal_info aper 
INNER JOIN applicant ap ON aper.ApplicantID = ap.ApplicantID 
INNER JOIN applicant_professional_info apro ON aper.ApplicantID = apro.ApplicantID 
WHERE apro.totalExperience >= 10
GROUP BY aper.ApplicantID

代表同一个id检索数据时遇到错误

1 个答案: