sql例子

select * from plat_material_resource
where stl_url LIKE '/data1/upload%'
--截取字符串
UPDATE plat_material_resource
SET stl_url = RIGHT (
	stl_url,
	LENGTH(stl_url) - LENGTH('/data1/upload')
)
WHERE
	stl_url LIKE '/data1/upload%'

　　

转载于:https://www.cnblogs.com/ephuizi/p/5194589.html