Question

我的数据库中有两个表。我想检查表1中的条件。如果它成为真，那么我必须在Table2中生成两个记录。我正在使用OracleSQLDeveloper，我想为它编写一个PL / SQL脚本。我能做到这一点的最佳方式是什么？

Answer 1

在学习完教程并尝试学习PL / SQL的基础之后，它应该看起来像这样：

max.cpp: In instantiation of 'const int max0<32>::value':
max.cpp:17:51:   recursively required from 'const int max1<16>::value'
max.cpp:17:51:   required from 'const int max1<15>::value'
max.cpp:28:28:   required from here
max.cpp:4:52: warning: integer overflow in expression [-Woverflow]
     static const int value = max0<bits - 1>::value * 2 + 1;
                              ~~~~~~~~~~~~~~~~~~~~~~^~~
max.cpp:4:22: error: overflow in constant expression [-fpermissive]
     static const int value = max0<bits - 1>::value * 2 + 1;
                      ^~~~~
max.cpp:4:22: error: overflow in constant expression [-fpermissive]
max.cpp:4:22: error: overflow in constant expression [-fpermissive]
max.cpp: In instantiation of 'const int max0<914>::value':
max.cpp:17:51:   recursively required from 'const int max1<16>::value'
max.cpp:17:51:   required from 'const int max1<15>::value'
max.cpp:28:28:   required from here
max.cpp:4:52: fatal error: template instantiation depth exceeds maximum of 900 (use -ftemplate-depth= to increase the maximum)
     static const int value = max0<bits - 1>::value * 2 + 1;
                              ~~~~~~~~~~~~~~~~~~~~~~^~~
compilation terminated.

希望这有帮助

Answer 2

按照J. Chomel的建议，在PL / SQL中进行，可读性很好。但是你也可以在纯SQL中实现它，如果出于某种原因这是必要的：

'layouts.menu'

显然，类似的东西不具有可读性。

在PL / SQL脚本中创建新记录

2 个答案: