php - 使用简单的类似查询获取数据问题

Question

这是我的查询

SELECT DISTINCT *
 FROM job_detail, apply 
WHERE job_detail.job_id = apply.job_id 
AND apply.end_user_id =1 
AND job_detail.jobtitle =  'barber'
AND (job_detail.Zip LIKE  '%900%' 
OR job_detail.city LIKE  '%900%' 
OR job_detail.country LIKE '%900%')

它返回给我 id 37 的记录...但我获取的数据必须为空

Answer 1

使用显式连接语法并看看会得到什么

select a.*
From job_detail j
Join apply a
    On j.job_id = a.job_id 
    AND a.end_user_id =1 
    AND j.jobtitle =  'barber'
    AND (j.Zip LIKE  '%900%' 
    OR j.city LIKE  '%900%' 
    OR j.country LIKE '%900%')