我正在将一系列查询转换为Knex语法。我对此查询有疑问:
SELECT id,reviewed,log_reference,CONVERT(notification USING utf8),create_time,update_time,store,user_id
FROM store_failure_log
WHERE reviewed = 0
AND create_time BETWEEN NOW() - INTERVAL 18 HOUR AND NOW();
更确切地说是这行:
SELECT id,reviewed,log_reference,CONVERT(notification USING utf8),create_time,update_time,store,user_id
我有此Knex:
knex('store_failure_log')
.select('id', 'reviewed', 'log_reference', 'CONVERT(notification USING utf8)', 'create_time', 'update_time', 'store', 'user_id').convert('notification USING utf8')
.where('reviewed', 0)
.where(knex.raw('create_time BETWEEN NOW() - INTERVAL 18 HOUR AND NOW()'))
产生此SQL查询:
select `id`, `reviewed`, `log_reference`, `CONVERT(notification USING utf8)`, `create_time`, `update_time`, `store`, `user_id` from `store_failure_log` where `reviewed` = 0 and create_time BETWEEN NOW() - INTERVAL 18 HOUR AND NOW()
问题出在:Convert(使用utf8的通知)。
查询无效,因为转换位于括号内。如何使用knex编写它?
通常,我如何在KNEX语法中包括SQL函数调用?
您可以使用raw在您的Knex查询中包括SQL函数调用,就像您已经在where
中完成的操作一样:
knex('store_failure_log')
.select(knex.raw('id', 'reviewed', 'log_reference', 'CONVERT(notification USING utf8)', 'create_time', 'update_time', 'store', 'user_id'))
.where('reviewed', 0)
.where(knex.raw('create_time BETWEEN NOW() - INTERVAL 18 HOUR AND NOW()'))