|| ' was paid on' || :new.paydate);测试它:set status=2car_ticket表:cid int, --- the car that violated parkingregulations
lid int, --- lot where violation took
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Table1和Table2id tid2 200Table2100 A我想从Table1拿出任何Table2中不存在的tid的id。请帮助我如何写一个这样的查询,这样的方式,它将减少时间。select id from Table1 where tid not in (select tid from Table2)
select a.id from Table1 a inner joinTable2 b o
in a view, one field has the following computed value, which takes a lot of time to return results fromthe database:
tid); print count_term_get_node_count($term); ?我如何缓存整个视图的结果,我想要每天晚上用cron作业运行count_term_get_node_count,并计算词汇表中存在的每个术语的结果?