将 mySQL 查询转换为 KNEX

Convert mySQL query to KNEX

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();

更准确地说是这一行:

SELECT id,reviewed,log_reference,CONVERT(notification USING utf8),create_time,update_time,store,user_id

我有这个 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:转换(使用 utf8 的通知)。

查询无效,因为转换在括号中。怎么用knex写呢?

通常如何在 KNEX 语法中包含 SQL 函数调用?

您可以使用 raw 在 Knex 查询中包含 SQL 函数调用,就像您在 where:

中所做的那样
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()'))

这是@Veve 的答案的固定版本,其中包含正确的标识符引用和 knex.raw 语法:

knex('store_failure_log')
        .select('id', 'reviewed', 'log_reference', knex.raw('CONVERT(?? USING utf8)', ['notification']), 'create_time', 'update_time', 'store', 'user_id')
        .where('reviewed', 0)
        .where(knex.raw('?? BETWEEN NOW() - INTERVAL 18 HOUR AND NOW()', ['create_time']))

https://runkit.com/embed/lh2i1qif7obx