我如何將物件從一個類傳遞到另一個類？

我有 3 個課程MainActivity：homePage和createPage; 和一個串列List<Recipe> recipeList = new ArrayList<>()。MainActivity

用戶輸入homePagefrom MainActivity。從homePage，用戶可以輸入createPage和創建新配方。這個新配方旨在傳回給MainActivity.

我在網上搜索了一下，發現有包裹。但是當我嘗試時，我得到了 NullPointerException。

createPage串列傳遞位置的代碼

ArrayList<Recipe> rList = new ArrayList<>();
Recipe r = new Recipe(...);
rList.add(r)
Intent i = new Intent();
Bundle b = new Bundle();
b.putParcelableArrayList("recipe", (ArrayList<? extends Parcelable>) rList);
i.putExtras(b);
i.setClass(createPage.this, homePage.class);
startActivity(i);

homePage接收串列的位置的代碼。

有什么問題getIntent()嗎？因為從MainActivityto移動時homePage，它不會收到捆綁包。這是導致錯誤嗎？

Intent intent = getIntent();
Bundle b = this.getIntent().getExtras();
if (b != null) {
    Recipe r = b.getParcelable("recipe");
    recipeList.add(r);
}

Recipe類代碼

public class Recipe implements Parcelable {

private String name;
private String description;
private String ingredients;
private int duration;
private String steps;
private int thumbnail;

protected Recipe(Parcel in) {
    name = in.readString();
    description = in.readString();
    ingredients = in.readString();
    duration = in.readInt();
    steps = in.readString();
    thumbnail = in.readInt();
}

public static final Creator<Recipe> CREATOR = new Creator<Recipe>() {
    @Override
    public Recipe createFromParcel(Parcel in) {
        return new Recipe(in);
    }

    @Override
    public Recipe[] newArray(int size) {
        return new Recipe[size];
    }
};

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

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

public String getIngredients() {
    return ingredients;
}
public void setIngredients(String ingredients) {
    this.ingredients = ingredients;
}

public int getDuration() {
    return duration;
}
public void setDuration(int duration) {
    this.duration = duration;
}

public String getSteps() { return steps; }
public void setSteps(String steps) { this.steps = steps; }

public int getThumbnail() { return thumbnail; }

public Recipe() {}

public Recipe(String name, int duration, String ingredients, String description, String steps, int thumbnail) {
    this.name = name;
    this.description = description;
    this.ingredients = ingredients;
    this.duration = duration;
    this.steps = steps;
    this.thumbnail = thumbnail;
}

@Override
public int describeContents() {
    return 0;
}

@Override
public void writeToParcel(Parcel parcel, int i) {
    parcel.writeString(name);
    parcel.writeString(description);
    parcel.writeString(ingredients);
    parcel.writeInt(duration);
    parcel.writeString(steps);
    parcel.writeInt(thumbnail);
}

}

uj5u.com熱心網友回復：

您正在鍵下寫入Parcelable整個陣列"recipe"

b.putParcelableArrayList("recipe", (ArrayList<? extends Parcelable>) rList);

但另一方面，您不是在尋找串列，而是Recipe在同一鍵下尋找單個專案

Recipe r = b.getParcelable("recipe");

你應該使用getParcelableArrayList，或者如果你只有一個Recipe用于傳遞給另一個Activity只是使用putParcelable（不列出）

uj5u.com熱心網友回復：

或者，您可以使用可序列化，這將不那么復雜。

供參考：https ://stackoverflow.com/a/2736612/9502601

盡管 Parcellables 更快，但如果您想要一個不太復雜的解決方案，那么您可以選擇它。

用于 Serializable 和 Parcelable 之間的比較。 https://stackoverflow.com/a/23647471/9502601

uj5u.com熱心網友回復：

您可以為此使用這個 gson Lib

implementation 'com.google.code.gson:gson:2.8.9'

有意圖地發送資料

Recipe r = new Recipe(...);
String recipeString = new Gson().toJson(r);
intent.putExtra("recipe",recipeString);

// For ArrayList
ArrayList<Recipe> recipeList = new ArrayList<>();
String recipeString = new Gson().toJson(recipeList);
intent.putExtra("recipeList",recipeString);

從 Intent 接收資料

Recipe r = new Gson().fromJson(intent.getStringExtra("recipe"), Recipe.class);

// For Array List
Type listType = new TypeToken<ArrayList<Recipe>>(){}.getType();
ArrayList<Recipe> recipeList = new Gson().fromJson(intent.getStringExtra("recipeList"),listType);

標籤：爪哇 安卓 安卓工作室 可打包的

上一篇：AndroidStudio出現錯誤IOException：為androidsdk31創建檔案時不允許操作

下一篇：Selenium，將瀏覽器從Chrome更改為Edge