Django 管理员 : populate the field based on previous field value

Question

我在django admin中有一个模型如下

ChoiceA= (
      ("on-false","on-false"),
       ("on-true","on-true"),
     )

ChoiceB =  (
        ("always","always"),
        ("never","never"),
       )
   id = models.CharField(verbose_name="Field",max_length=32)
   type = models.CharField(verbose_name="Expression",max_length=32)
   action = models.CharField(max_length=32, choices=x)

现在基于用户输入的类型，即如果用户输入 type = "a"，则操作的选择应设置为 ChoiceA，如果用户输入 type = "b"，则操作的选择应设置为 ChoiceB。我怎样才能在 Django Admin 中实现这一点？

编辑:

action_change.js

jQuery(document).ready(function(){
$("#id_type").change( function(event) {
$.ajax({
        "type"     : "POST",
        "url"      : "/action_choices/",
        "dataType" : "json",
        "cache"    : false,
        "error"   :  alert("hello"),  
        "success"  : function(json) {
            $('#id_action >option').remove();
            for(var j = 0; j < json.length; j++){
                $('#id_action').append($('<option></option>').val(json[j][0]).html(json[j][1]));
            }
        }

});
});
});

Answer 1

您可以使用 Ajax 和 jQuery 实现它:

模型.py:

type   = models.CharField(verbose_name="Expression",max_length=32)
action = models.CharField(max_length=32, choices = (('', ''), ))

管理员.py:

class MyModelAdmin(admin.ModelAdmin):
    list_display = ('type', )

    class Media:
        js = ['/static/js/action_change.js']

admin.site.register(MyModel, MyModelAdmin)

urls.py:

url(r'^action_choices/', 'myproject.myapp.views.action_choices'),

views.py:

def action_choices(request): 
    action_list = []
    ChoiceA = ("on-false", "on-true")
    ChoiceB = ("always", "never")

    action_type = request.GET.get('action_type')
    if str(action_type).lower() == 'a':
        choices = ChoiceA
    elif str(action_type).lower() == 'b':
        choices = ChoiceB
    else:
        choices = ()

    [action_list.append((each,each)) for each in choices]
    json = simplejson.dumps(action_list)
    return HttpResponse(json, mimetype='application/javascript')

在静态文件夹中创建包含以下内容的文件 action_change.js，并在 ModelAdmin 的 class Media 中定义正确的路径。

action_change.js

(function($){   
    $(function(){
        $(document).ready(function() {
            $('#id_type').bind('keyup', type_change);           
            $('#id_action >option').show();
        });
});  
})(django.jQuery);

// based on the type, action will be loaded

var $ = django.jQuery.noConflict();

function type_change()
{
    var action_type = $('#id_type').val();
    $.ajax({
            "type"     : "GET",
            "url"      : "/action_choices/?action_type="+action_type,
            "dataType" : "json",
            "cache"    : false,
            "success"  : function(json) {
                $('#id_action >option').remove();
                for(var j = 0; j < json.length; j++){
                    $('#id_action').append($('<option></option>').val(json[j][0]).html(json[j][1]));
                }
            }           
    })(jQuery);
}

这应该适用于您提出的方案。我在下面给出我的建议:

模型.py

type   = models.CharField(verbose_name="Expression",max_length=32, choices = (('a', 'a'), ('b', 'b'), ))
action = models.CharField(max_length=32, choices = (('', ''), ))

action_change.js(第 5 行)

$('#id_type').bind('change', type_change);