如何使用Scala模板和Play框架预填充下拉菜单 [英] How to prefill a dropdown using scala template and play framework
问题描述
我正在为我的项目使用scala模板和Play 2.0框架.假设我有一个用户表单,其中包含诸如名称(文本字段),年龄(下拉列表)之类的字段.创建用户时,我将姓名填写为dave,并将年龄选择为25.
I am using scala template and Play 2.0 framework for my project. Let's say I have a user form with fields like name (textfield), age (dropdown). While creating the user I filled name as dave and selected age as 25.
现在在我的编辑屏幕上,我希望我的值被预填充,我知道如何使用文本字段(即,将值设置为userForm('name')),但是下拉菜单呢?怎么做.
Now on my edit screen, I want my values to be prefilled, i know how to do it with textfield (i.e. set value as userForm('name')) but what about the dropdown? how to do it.
推荐答案
感谢Shawn Downs和biesior.
Thanks Shawn Downs and biesior.
好吧,我们可以使用@select
scala帮助器类来显示预填充的结果.
喜欢.
Well, we can use @select
scala helper class to show the pre-filled result.
like.
@select(userForm("age"),models.Age.values().toList.map(v => (v.getDisplayName(), v.getDisplayName())),'id->"age")
为显示其他选项,我使用了可能的age枚举.
To show other options I have used an enum of possible values of age.
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