MySQL中用于构造多个JOIN的正确语法是什么

Question

我是SQL的新手,并尝试一起构建6个表的连接.这是我的示例代码:

SELECT cr.COMMUNICATIONS_ID, cr.CONSUMER_ID, cr.ACTION_LOG_ID, nc.CONSUMER_ID, cal.CONSUMER_ID, cal.TIPS_AMOUNT, cal.LAST_MOD_TIME, cirm.CONSUMER_RATING, ces.EXPERT_SCORE, cim.CONSUMER_INTEREST_EXPERT_ID, scs.SIMILARITY
FROM COMMUNICATION_RELVANCE AS cr
JOIN network_communications AS nc
ON cr.COMMUNICATIONS_ID=nc.COMMUNICATIONS_ID
JOIN consumer_action_log AS cal
ON cr.ACTION_LOG_ID=cal.ACTION_LOG_ID
JOIN communication_interest_mapping AS cim
ON cr.COMMUNICATIONS_ID=cim.COMMUNICATIONS_ID
JOIN consumer_interest_rating_mapping AS cirm
ON cr.CONSUMER_ID=cirm.CONSUMER_ID
    AND cim.CONSUMER_INTEREST_EXPERT_ID=cirm.CONSUMER_INTEREST_ID
JOIN consumer_expert_score AS ces
ON nc.CONSUMER_ID=ces.CONSUMER_ID
    AND cim.CONSUMER_INTEREST_EXPERT_ID=CONSUMER_EXPERT_ID
JOIN survey_customer_similarity AS scs
ON nc.CONSUMER_ID=scs.CONSUMER_ID
    AND cal.CONSUMER_ID=scs.CONSUMER_ID;

在开始调试之前,我想确保我的代码中没有任何批发问题我的语法.

Answer 1

这看起来很好.尝试运行它以查看是否有任何错误.

几点:

• 您没有WHERE子句,因此它将返回所有行.这是故意的吗？
• 你错过了一张桌子的名字:COMMUNICATION_RELVANCE.你可能应该在编写太多代码之前解决这个问题.我知道这听起来很迂腐,但这是因为在初始开发过程中修复它可能很容易.如果你留下太长时间,人们会编写依赖于拼写错误的代码,最终可能会因更改名称和更新所有现有代码而变得太困难/昂贵.然后你会被它坚持下去,它会惹恼所有必须在未来几年维护你的代码的人.