tarantool对并不能获取所有元组

secondary:select（{1，'RU'，2}）

box.space.test.index.secondary

---

局部r=''表示k，box.space.test.index.secondary中的元组：pairs（{1，'RU'，2}）do r=r..'，find（'..tuple[1]..'）if（tuple[7]仅返回唯一的（通过索引中的字段）元组？…或者tarantool将内部指针移动到列表中的下一个元组，当我们：按主键删除元组时…？我自己找到答案。tarantool在循环期间重建索引，所以当我成对迭代时，我跳过一些记录。正确的方法是通过选择获取元组的部分，迭代它删除每个元组，获取新部分…而数据存在。

---
- - [47, 6, 1, 'RU', 2, 11, 6]
  - [44, 3, 1, 'RU', 2, 101, 6]
  - [42, 1, 1, 'RU', 2, 189, 6]
  - [34, 3, 1, 'RU', 2, 260, 5]
  - [49, 8, 1, 'RU', 2, 290, 6]
  - [41, 0, 1, 'RU', 2, 303, 6]
  - [37, 6, 1, 'RU', 2, 494, 5]
  - [35, 4, 1, 'RU', 2, 633, 5]
  - [45, 4, 1, 'RU', 2, 694, 6]
  - [43, 2, 1, 'RU', 2, 780, 6]
  - [46, 5, 1, 'RU', 2, 833, 6]
  - [40, 9, 1, 'RU', 2, 870, 5]
  - [48, 7, 1, 'RU', 2, 927, 6]
  - [50, 9, 1, 'RU', 2, 930, 6]
...

---
- unique: false
  parts:
  - type: NUM
    fieldno: 3
  - type: STR
    fieldno: 4
  - type: NUM
    fieldno: 5
  - type: NUM
    fieldno: 6
  id: 2
  space_id: 512
  name: secondary
  type: TREE
...

---
- ', found (47), found (44), found (42), found (34), delete(34), found (41), found(37), delete(37), found (45), found (43), found (46), found (40), delete(40), found(50)'
...

---
- - [47, 6, 1, 'RU', 2, 11, 6]
  - [44, 3, 1, 'RU', 2, 101, 6]
  - [42, 1, 1, 'RU', 2, 189, 6]
  - [49, 8, 1, 'RU', 2, 290, 6]
  - [41, 0, 1, 'RU', 2, 303, 6]
  - [35, 4, 1, 'RU', 2, 633, 5]
  - [45, 4, 1, 'RU', 2, 694, 6]
  - [43, 2, 1, 'RU', 2, 780, 6]
  - [46, 5, 1, 'RU', 2, 833, 6]
  - [48, 7, 1, 'RU', 2, 927, 6]
  - [50, 9, 1, 'RU', 2, 930, 6]
...

---
- ', found (47), found (44), found (42), found (49), found (41), found (35), delete(35), found (43), found (46), found (48), found (50)'
...

---
- - [47, 6, 1, 'RU', 2, 11, 6]
  - [44, 3, 1, 'RU', 2, 101, 6]
  - [42, 1, 1, 'RU', 2, 189, 6]
  - [49, 8, 1, 'RU', 2, 290, 6]
  - [41, 0, 1, 'RU', 2, 303, 6]
  - [45, 4, 1, 'RU', 2, 694, 6]
  - [43, 2, 1, 'RU', 2, 780, 6]
  - [46, 5, 1, 'RU', 2, 833, 6]
  - [48, 7, 1, 'RU', 2, 927, 6]
  - [50, 9, 1, 'RU', 2, 930, 6]
...