select sum(1) over() as total, sum(1) over(order by id) as idx from test_table;
假设test_table 结果为:
| total | idx |
| 10 | 1 |
| 10 | 2 |
| 10 | 3 |
| 10 | 4 |
| 10 | 5 |
| 10 | 6 |
| 10 | 7 |
| 10 | 8 |
| 10 | 9 |
| 10 | 10 |
select sum(1) over() as total, sum(1) over(order by id) as idx from test_table;
假设test_table 结果为:
| total | idx |
| 10 | 1 |
| 10 | 2 |
| 10 | 3 |
| 10 | 4 |
| 10 | 5 |
| 10 | 6 |
| 10 | 7 |
| 10 | 8 |
| 10 | 9 |
| 10 | 10 |

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