我有以下型号:
class Model(models.Model):
creator = models.ForeignKey(User,related_name='com_creator',on_delete=models.SET_NULL, blank=True, null=True)
username = models.CharField(max_length=60,default="")
created = models.DateTimeField(auto_now_add=True)
body = models.TextField(max_length=10000,default=" ")
subtype = models.CharField(_("SubType"),max_length=100)
typ = models.CharField(_("Type"),max_length=50)
plus = models.ManyToManyField(User,related_name='com_plus', verbose_name=_('Plus'), blank=True)
is_anonymous = models.BooleanField(_('Stay Anonymous'), blank=True, default=False)
typ和subtype中的值是代码,如:" 6_0_p",(因为原始值非常长,所以我使用代码和dict转换为人类可读的形式)。
问题:如何在django admin中拦截这些值并将其转换为人类可读的形式?
这是我到目前为止所尝试的:
class ModelAdmin(admin.ModelAdmin):
model = Model
extra = 1
exclude = ("username","creator","plus" )
readonly_fields = ('subtype','typ','is_anonymous','created')
fields = ('body','subtype','typ')
def typ(self, obj):
self.typ = "ppp"
obj.typ = "ppp"
return "pppp"
我尝试返回对象,self,其他一些值。我还试图在不使用可调用的情况下设置值,只是声明" typ =' xxx'"。没有。我可能不明白这一切是如何运作的......
任何想法都将受到赞赏。
答案 0 :(得分:2)
我认为您的代码不起作用,因为方法名称与字段相同。
你必须更改字段的名称,比如字段元组中的_typ,然后添加一个名为_typ的方法,它返回任何东西或某些obj attr。
即:
class ModelAdmin(admin.ModelAdmin):
model = Model
extra = 1
exclude = ("username","creator","plus" )
readonly_fields = ('subtype','typ','is_anonymous','created')
fields = ('body','subtype','_typ')
def _typ(self, obj):
return "pppp"
答案 1 :(得分:1)
您需要创建一个与您的方法名称对应的public class Test
{
public static boolean find()
{
int x = 10;
int y = 20;
int z = 30;
change(x,y,z); // call helper method to change this method's variables.
System.out.println(x); // should be 20
System.out.println(y); // should be 22
System.out.println(z); // should be 15
}
//helper method to be called from find() method
private static void change(int changeX ,int changeY,int changeZ)
{
//change find() variables.
x = changeX * 2;
y = changeY + 2;
z = changeZ /2;
}
}
,然后从该方法返回任何内容。
readonly_field