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我正在尝试从一个表中进行选择,其中字段名称如 P & L 但我收到了替换变量的提示。有没有办法在没有 Substitution 变量的情况下触发该查询并获得相同的结果?
我的选择声明:
select * from student_master_table where STU_MIDDLE_NAME like '%P & L%';
SqlDeveloper 提示:
请帮忙。
最简单的方法是将字符串分成两个字符串:
select * from student_master_table where STU_MIDDLE_NAME like '%P &'||' L%';
你必须逃避'L'
设置定义关闭;