Question

我想进行查询并使用用户在edittext中输入的值，因为condition是where子句。在数据库helper.java类中：查询如下：

Databasehelper.java：

new Computed(1, 2)()()

在ExaminatFragment.java中：

public Cursor patientHisto ( ) {
        SQLiteDatabase db = this.getWritableDatabase();
        Cursor res = db.rawQuery (" select dateap,diagnostic,prestreatment from patients_table, appointments_table, examinations_table where patients_table.id = appointments_table.idp and appointments_table.idap = examinations_table.id_ap and ID = ?  order by name, familyname asc"  , new String[] {ExaminatFragment.value} );
        return res;
    }

运行应用程序并单击按钮时，会出现以下错误：

@Override
    public View onCreateView(LayoutInflater inflater, ViewGroup parent, Bundle savedInstanceState){
    View v = inflater.inflate(R.layout.fragment_examinat, parent, false);
    myDB = new DatabaseHelper(getActivity());

    id_p = (EditText)v.findViewById(R.id.id_p_text);

    String value = id_p.getText().toString();

    getPatHisto( );
    return v;
    }   


public void getPatHisto( ) {



    showhp.setOnClickListener(

            new View.OnClickListener() {

            @Override
            public void onClick(View v) {
              Cursor res = myDB.patientHisto();
            if (res.getCount() == 0){

            showMessage("Patient history", "No patient history found");
            return;

                    }


    StringBuffer buffer = new StringBuffer();
    while (res.moveToNext()){
                            buffer.append("Dateap:"+res.getString(0)+"\n");
                            buffer.append("Diagnostic:"+res.getString(1)+"\n");
                            buffer.append("Prestreatment:"+res.getString(2)+"\n\n");




                        }
    showMessage("Patient's history", buffer.toString());
                    }


                }
                );
                }


public void showMessage(String title, String Message){

        AlertDialog.Builder builder = new AlertDialog.Builder(getActivity());
        builder.setCancelable(true);
        builder.setTitle(title);
        builder.setMessage(Message);
        builder.show();

    }

请帮助..提前谢谢

Answer 1

您指定为本地变量 value，而不是分配给类成员：

 String value = id_p.getText().toString();

应该是

 value = id_p.getText().toString();

否则ExaminatFragment.value将在SQL语句

中为null

您可能希望更改数据库方法以接受参数：

public Cursor patientHisto ( String value ) {
    SQLiteDatabase db = this.getWritableDatabase();
    Cursor res = db.rawQuery (" select dateap,diagnostic,prestreatment from patients_table, appointments_table, examinations_table where patients_table.id = appointments_table.idp and appointments_table.idap = examinations_table.id_ap and ID = ?  order by name, familyname asc"  , new String[] {value} );
    return res;
}

并且不依赖于 static 字段来传递该值。您对该方法的调用将更改为

 Cursor res = myDB.patientHisto(value);

java.lang.IllegalArgumentException：索引1处的绑定值为null

1 个答案: