How to prefill a dropdown using scala template and

Question

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.

Answer 1

In your model there will be 2 fields

Model code

Class User{
   public String name;
   public int age;
}

Controller code

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.
Form<User> userForm = Form.form(User.class);
User user = new User();
user.name = "Albert";
user.age = 19;

userForm.fill(user);
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.
.

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

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<form>
@helper.inputText(userForm("name"))
@helper.select(userForm("age"),
                helper.options(utils.DropdownUtils.getAgeList(25)))
</form>
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.
.

Answer 2

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.