select title from phpcms_content
union all
select name from phpcms_c_news where name like '%123%'
结果是把title和name 放一张表中
用表的关联来实现，你可以参考下
select title from phpcms_content where title like '%123%' union all
select name from phpcms_c_news where name like '%123%'
select * from (
select title t_name from phpcms_content
union all
select name t_name from phpcms_c_news
)
where t_name like '%123%'
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