这是代码:
ListAdapter adapter = new ListAdapter(this, jenres, pics);
(alt + enter)
没有帮助,它说“ListAdapter中的ListAdapter()
无法应用于:
预期的实际
参数参数
!!
答案 0 :(得分:0)
因为你没有在ListAdapter类中拥有所需的匹配构造函数。将构造函数创建为:
public ListAdapter(Context context, yourJenresType jenres, yourPicsType pics) {
// do stuff
}
答案 1 :(得分:0)
在您的主要课程中,签名: -
ListAdapter adapter = new ListAdapter(Activityname.this, jenres, pics);
在ListAdpater中: -
public class ListAdapter extends YourAdapter{
Context context;
VariableType jenres;
VariableType pics;
//Constructor:
public ListAdapter(Context context, VariableType jenres, VariableType pics){
this.context= context;
this. jenres=jenres;
this.pics=pics;
}
//Adapter methods below
}