Question

我正在一个项目上，需要设置以0、1或2开头的旧数据以简单地等于一个级别。

if (dr["Acct"].ToString().Substring(0,1) == "0"
    || dr["Acct"].ToString().Substring(0,1) == "1"
    || dr["Acct"].ToString().Substring(0,1) == "2")
{
    // then Level = Balance Sheet
}

我实质上是想说子字符串是0、1还是2，那么它应该等于称为资产负债表的水平。

Answer 1

如果您要检查多个字符串，我会将其存储在字符串数组中并检查其可读性

public class TeachersAdapter extends RecyclerView.Adapter<TeachersAdapter.TeacherAdapterViewHolder> {

private ProgressDialog LoadingBar;
FirebaseAuth teacherAuth;
DatabaseReference teacherRef;
FirebaseDatabase db;

public Context c;
public ArrayList<Modal_Teachers> arrayList;

public TeachersAdapter(Context c, ArrayList<Modal_Teachers> arrayList){
    this.c= c;
    this.arrayList = arrayList;

}
@Override
public int getItemCount() {
    return arrayList.size();
}

@Override
public long getItemId(int position) {
    return position;
}

@NonNull
@Override
public TeacherAdapterViewHolder onCreateViewHolder(@NonNull ViewGroup parent, int viewType) {
    View v = LayoutInflater.from(parent.getContext()).inflate(R.layout.list_teachers,parent, false);
    Button edit_t = v.findViewById(R.id.teacher_edit);
    Button delete_t = v.findViewById(R.id.teacher_edit);
    return new TeacherAdapterViewHolder(v);
}

@Override
public void onBindViewHolder(@NonNull final TeacherAdapterViewHolder holder, final int position) {
    final Modal_Teachers sTeachers = arrayList.get(position);
    holder.teacherName.setText(sTeachers.getName());
    holder.teacherDesignation.setText(sTeachers.getQualification());
    Picasso.get().load(sTeachers.getPhotourl()).into(holder.teacherImage);


public class TeacherAdapterViewHolder extends RecyclerView.ViewHolder{
    public CircleImageView teacherImage;
    public TextView teacherName,
            teacherDesignation;
    public Button edit_t, delete_t;

    public TeacherAdapterViewHolder(@NonNull View itemView) {

        super(itemView);
        teacherImage = itemView.findViewById(R.id.image_teacher);
        teacherName =  itemView.findViewById(R.id.teacher_nome);
        teacherDesignation = itemView.findViewById(R.id.teacher_designation);
        edit_t = itemView.findViewById(R.id.teacher_edit);
        delete_t = itemView.findViewById(R.id.teacher_delete);
    }
}
}

Answer 2

字符串操作非常昂贵，这可能是一种更具可读性的处理方式。

string acctVal = dr["Acct"]?.ToString();
char[] balanceSheetVals= {'0', '1', '2'};

if(!String.IsNullOrEmpty(acctVal))
{
   if(balanceSheetVals.contains(acctVal[0]))
   {
      Level = "Balance Sheet";
   }
   else
   {
      //defaults
   }
}
else
{
   //defaults
}

编辑：@Rotem的大多数道具都清理了很多

Answer 3

如果您要检查某个string是否以另一些string开头，我个人认为String.StartsWith比使用{{ 1}}。

String.Substring

编写使字符串等于其他内容的if语句的最佳方法是什么？

3 个答案: