下面哪个是错误的助手函数view调用？
A: return view("",["name"=>$myname,"id"=>$myid]);
B: return view("",["name"=>myname]);
C: return view("user/login");
D: return view("");
A: return view("",["name"=>$myname,"id"=>$myid]);
B: return view("",["name"=>myname]);
C: return view("user/login");
D: return view("");
举一反三
- 一个控制操作show定义在下面的控制器文件中：application/index/controller/Index.php一个视图模板文件stu.html保存在下面的路径中：application/index/view/index如果要在该操作中渲染视图模板文件，哪个是正确的渲染语句。 A: return view(""); B: return view("show"); C: return view("stu"); D: return view("index");
- 下面定义视图的SQL语句，哪一个是错误的？（ ） A: create view VIEW1 as select id,name,sex from users B: create view VIEW2 as select count(*) as studentTotal,classname from student C: create view VIEW3 as select name,price from D: create view VIEW4 as select logdate,price,1.1*parice as ‘涨价’ from shop
- 下面关于“CREATE VIEW v_goods AS SELECT id, name FROM goods”描述错误的是
- 在tb_name表中创建一个名为name_view的视图,并设置视图的属性为name、pwduser,执行语句是什么?( ) A: CREATE VIEW name_view(name,pwd,user)AS SELECT name.pwd,user FROM tb_mame; B: SHOW VIEW name_view(name,pwd,user) AS SELECT name,pwd,user FROM tb_mame: C: DROP VIEW name_view(name,pwd,user) AS SELECT name,pwd,user FROM tb_mame: D: SELECT * FROM name_view(name,pwd,user) AS SELECT name,pwd,user FROM tb_ name:
- 下面关于修改视图的SQL语句，那个是正确的（ ）。 A: alter view VIEW01 as select * from users order by age desc B: alter view VIEW02 as select name,avg(score) as avgScore from studentScore C: alter view VIEW03 as select top 10 id,name,password from admin order by id D: alter view VIEW04 as select * from stuent,class where student.classid=class.classid order by class.classid