尝试在空对象引用上调用虚拟方法“java.lang.String”

我正在使用一个应用程序，但我有一个问题。当我按下后退按钮时，我的应用程序崩溃，并给出空指针异常错误。 我有一个列表视图，其中包含来自api的一些数据，我在api中使用Reformation2调用了这些数据。除了“后退”按钮外，一切正常。 以下是我所做的： 我的适配器

private Context context;
private Industry industries;

public IndustryAdapter(Context context, Industry industries) {
    this.industries = industries;
    this.context = context;
}

@Override
public int getCount() {
    return 1;
}

@Override
public Object getItem(int position) {
    return 1;
}

@Override
public long getItemId(int position) {
    return 0;
}

@Override
public View getView(int position, View convertView, ViewGroup parent) {
    if (convertView == null) {
        LayoutInflater inflater = ((Activity) context).getLayoutInflater();

        convertView = inflater.inflate(R.layout.row_item, parent, false);
    }

    TextView tvName = (TextView)convertView.findViewById(R.id.tvName);
    TextView tvDescription = (TextView)convertView.findViewById(R.id.tvDescription);

    String name = industries.getName();
    //industries.setName(name);
    String descr = industries.getDescription();
    //industries.setDescription(descr);

    tvName.setText(name);
    tvDescription.setText(descr);

    return convertView;
}

在我的活动中我有这个

CompaniesRetriver companiesRetriver = new CompaniesRetriver();
    this.recyclerView = (RecyclerView) findViewById(R.id.companies_list);
    SharedPreferences editor = getSharedPreferences(MY_PREFERENCE, MODE_PRIVATE);
    String token = editor.getString("token", "");
    final Context sfdsf = this;
    Callback<Companies> callback = new Callback<Companies>() {
        @Override
        public void onResponse(Call<Companies> call, Response<Companies> response) {
            progressBar.dismiss();
            if(response.isSuccessful()) {
                progressBar.dismiss();
                Companies companies = response.body();
                CompanyAdapter adapter = new CompanyAdapter(CompaniesListActivity.this, companies);
                recyclerView.setLayoutManager(new LinearLayoutManager(CompaniesListActivity.this));
                recyclerView.setAdapter(adapter);
                System.out.println(companies);
            } else {
                System.out.println(response.body());
            }
        }

        @Override
        public void onFailure(Call<Companies> call, Throwable t) {
            progressBar.dismiss();
            System.out.println(t.getLocalizedMessage());
        }
    };

    companiesRetriver.getCompanies(callback, token);

我的POJO课

@SerializedName("name")
    private String  name;
    @SerializedName("id")
    private int  id;
    @SerializedName("description")
    private String description;

    public Industry(String description, String name, int id) {
        this.description = description;
        this.name = name;
        this.id = id;
    }

    public String getDescription() {
        return description;
    }

    public void setDescription(String description) {
        this.description = description;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public int getId() {
        return id;
    }

    public void setId(int id) {
        this.id = id;
    }

---

感谢您的帮助

试着检查下一行

 String name = industries.getName();

这里的industries对象可能为null，这就是为什么会出现此错误。 如果您只需要显示一项，则不需要使用适配器，但如果有更多项，则可以按如下方式使用它：-

private Context context;
private ArrayList<Industry> industryList;

public IndustryAdapter(Context context, Industry industry) {
    industryList = new ArrayList();
    industryList.add(industry);
    this.context = context;
}
 public IndustryAdapter(Context context, ArrayList<Industry> industryList) {
    this.industryList=this.industryList;
    this.context = context;
}

@Override
public int getCount() {
    return industryList.size();
}

@Override
public Object getItem(int position) {
    return position;
}

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

@Override
public View getView(int position, View convertView, ViewGroup parent) {
    if (convertView == null) {
        LayoutInflater inflater = ((Activity) context).getLayoutInflater();

        convertView = inflater.inflate(R.layout.row_item, parent, false);
    }

    TextView tvName = (TextView)convertView.findViewById(R.id.tvName);
    TextView tvDescription = (TextView)convertView.findViewById(R.id.tvDescription);

    String name = industryList.get(position).getName();
    String descr = industryList.get(position).getDescription();


    tvName.setText(name);
    tvDescription.setText(descr);

    return convertView;
}

---

发布日志。在此字符串上，name=industries.getName；您面临的问题请发布您的pojo，以便我们了解您传递数据的问题和活动您提供了两个不同的问题。表示您提供了保存公司和行业适配器数据的活动。在该活动中，您使用的是CompanyAdapter，而不是IndustryAdapter。但是列表可以正常工作。我尝试更改为行业适配器，但不起作用。是的，行业为空。我有两个项目要显示，这很好，但当我按下后退按钮，我的应用程序crashes@EngineerEraldo您是否重写了onBackPressed方法？是的，我这样做了，但没有在onBackPressed方法中使用debug检查错误是否发生在转到onBackPressed方法之前或我们在其中时。

private Context context;
private ArrayList<Industry> industryList;

public IndustryAdapter(Context context, Industry industry) {
    industryList = new ArrayList();
    industryList.add(industry);
    this.context = context;
}
 public IndustryAdapter(Context context, ArrayList<Industry> industryList) {
    this.industryList=this.industryList;
    this.context = context;
}

@Override
public int getCount() {
    return industryList.size();
}

@Override
public Object getItem(int position) {
    return position;
}

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

@Override
public View getView(int position, View convertView, ViewGroup parent) {
    if (convertView == null) {
        LayoutInflater inflater = ((Activity) context).getLayoutInflater();

        convertView = inflater.inflate(R.layout.row_item, parent, false);
    }

    TextView tvName = (TextView)convertView.findViewById(R.id.tvName);
    TextView tvDescription = (TextView)convertView.findViewById(R.id.tvDescription);

    String name = industryList.get(position).getName();
    String descr = industryList.get(position).getDescription();


    tvName.setText(name);
    tvDescription.setText(descr);

    return convertView;
}