Question

给定一个如下所示的对象：

class Products {
    ArrayList<Integer> numbers;

    public Products(ArrayList<Integer> numbers){}
        this.numbers = numbers;
    }
}

让我说我像这样实例化它：

ArrayList<Integer> numbers1 = new ArrayList();
numbers1.add(23432);
numbers1.add(5555); 
numbers1.add(323423);
Products products1 = new Products(numbers1);
ArrayList<Integer> numbers2 = new ArrayList();
numbers2.add(234222);
numbers2.add(55665); 
numbers2.add(3278663);
Products products2 = new Products(numbers2);
ArrayList<Integer> numbers3 = new ArrayList();
numbers3.add(266622);
numbers3.add(52225); 
numbers3.add(7763);
Products products3 = new Products(numbers3);

使用 RxJava ，如何将products1，product2和product3转换为整数流？

所以给定像products1,products2,products3这样的流我需要 RxJava 来构建这个流：

23432,5555,323423,234222,55665,3278663,266622,52225,7763

Answer 1

您可以将<form action="create.php" method="POST"> <label for="student_id"> <input type="hidden" name="student_id" value="<?= $_GET['student_id'] ?>"> </label><br> <label for="id">Student's name <input type="text" value="<?= $note->getStudentFullName($_GET['student_id']); ?>" disabled> </label><br> <label for="title">Title <input type="text" name="title"> </label><br> <label for="description">Description<br> <textarea style="width:500px; height: 200px;" name="description"></textarea> </label><br> <label for="visibility">Visibility <select name="visibility[]"> <?php foreach ($visibilities as $visibility) { ?> <option value="<?= $visibility['visibility_id'] ?>"><?= $visibility['name'] ?></option> <?php } ?> </select> </label><br> <button type="submit" name="submit">Submit</button> </form>与concatMap()

一起使用

Observable.from(Iterable)

rxjava - 如何从包含整数列表的observable流构建整数流

1 个答案: