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.
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.
Thanks Shawn Downs and biesior.
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")
To show other options I have used an enum of possible values of age.
In your model there will be 2 fields
Model code
Class User{
public String name;
public int age;
}
Controller code
.
.
.
Form<User> userForm = Form.form(User.class);
User user = new User();
user.name = "Albert";
user.age = 19;
userForm.fill(user);
.
.
.
Util code
package utils;
import java.util.HashMap;
import java.util.LinkedHashMap;
public class DropdownUtils {
public static HashMap<String, String> getAgeList(int ageLimit){
LinkedHashMap<String, String> ageList = new LinkedHashMap<>();
for(Integer i=0; i<=ageLimit; i++){
ageList.put(i.toString(), i.toString());
}
return ageList;
}
}
View code
.
.
.
<form>
@helper.inputText(userForm("name"))
@helper.select(userForm("age"),
helper.options(utils.DropdownUtils.getAgeList(25)))
</form>
.
.
.
来源:https://stackoverflow.com/questions/29389201/how-to-prefill-a-dropdown-using-scala-template-and-play-framework