Question

我使用Parse.com，我有2个表：食谱和成分，现在在1个食谱中我有很多成分在解析中我使用成分表中的父连接表，我正在尝试构建Recige的ArrayList（每个配方包含成分的ArrayList）

这样的事情：

 ParseQuery<ParseObject> query = ParseQuery.getQuery("Recipe");

        query.orderByDescending("createdAt");
        query.include("Ingredient");

        query.findInBackground(new FindCallback<ParseObject>() {
            public void done(List<ParseObject> RecipeList, ParseException e) {

//What should i write here ???
------------------------------

}

非常感谢

配料表： ObjectID，NAME，UNIT，parent

Answer 1

由于成分只属于食谱清单，我认为收件人可以参考属于他们列表的成分。

如果您的参考是指向配方的所有成分的指针，那么这应该足够了

    ParseQuery<ParseObject> query = ParseQuery.getQuery("Recipies");
    query.include("ingredients"); //Name of your ingredients table
    query.findInBackground(new FindCallback<ParseObject>() {

        @Override
        public void done(List<ParseObject> objects, ParseException e) {
            if (e == null) {
                for(int index = 0; index < objects.size(); i++){
                    objects.get(i); //This is a recipe
                    objects.get(i).get("name of the column that hold the pointer"); //This is your ingredient
                }
            } else {
                Log.e("", e.getMessage());
            }
            functionCallback.done(objects, e);
        }

    });

有关'include'查询的更多信息，请访问here（在关系查询下）

如果您将引用存储在数组中，您可以尝试执行类似这样的操作

    ParseQuery<ParseObject> ingredientQuery = ParseQuery.getQuery("ingredients");

    ParseQuery<ParseObject> query = ParseQuery.getQuery("Recipies");
    query.findInBackground(new FindCallback<ParseObject>() {

        @Override
        public void done(List<ParseObject> objects, ParseException e) {
            if (e == null) {
                for(int index = 0; index < objects.size(); i++){
                    objects.get(i); //This is a recipe
                    String[] pointers = objects.get(i).get("name of the column that hold the pointer"); 
                    //This is your ingredient reference (if you did 'recipe.put("ingredients", ingredient);', 
                    //the reference that would be saved is the id of the ingredient.
                    ingredientsQuery.whereContainedIn("objectId", pointers);
                    ingredientsQuery.findInBackground(new FindCallback<ParseObject>() {

                    @Override
                    public void done(List<ParseObject> ingredients, ParseException e) {
                    if (e == null) {
                        for(int index = 0; index < ingredients.size(); i++){
                            ingredients.get(i); //This is a ingredient
                        }
                    } else { 
                        Log.e("", e.getMessage());
                    }
               }
            } else { 
                Log.e("", e.getMessage());
            }
            functionCallback.done(objects, e);
        }

    });

认为应该这样做！

parse.com android关系查询

1 个答案: