Question

我目前正在从事一项任务，并且要求通过从JSON读取值来更新大约2000至4000条记录。我优化了JSON部分，但目前，我正在逐一更新每个记录。有人可以建议最好的方法来仅通过一次查询而不是将其运行2000至4000次来更新所有2000至4000条记录吗？这是我的示例代码

    APEX_JSON.PARSE(V_OUTPUT_DATA); 
plan_count := apex_json.get_count('plan'); 
IF plan_count > 0 THEN 
    FOR I in 1..plan_count LOOP 
        activities_count := APEX_JSON.get_count(p_path => 'plan['||i||'].activities'); 
        IF activities_count > 0 THEN 
            FOR j in 2..(activities_count-1) LOOP 
                V_TASK_ID := APEX_JSON.get_varchar2(p_path => 'plan['||i||'].activities['||j||'].task_id'); 
                V_SEQ := APEX_JSON.get_number(p_path => 'plan['||i||'].activities['||j||'].sequence');
                UPDATE TABLE_NAME 
                SET ROUTE_SEQUENCE = V_SEQ, UPDATED_BY = 'SYSTEM',UPDATED_ON = SYSTIMESTAMP 
                WHERE TASK_ID = V_TASK_ID; 
            END LOOP; 
            COMMIT; 
        END IF; 
    END LOOP;       
END IF;

我应该使用2D数组并将其用于批量更新，还是可以遵循某些其他方法？

添加了示例JSON

{
"plan": [{
    "vehicle_id": "vehicle_1",
    "activities": [{
        "sequence": 0,
        "timestamp": "2017-11-10T09:48:19Z",
        "location_id": "depot"
    },
    {
        "sequence": 1,
        "timestamp": "2017-11-10T09:50:07Z",
        "task_id": "465427",
        "location_id": "465427",
        "travel_distance": 1099,
        "travel_duration": "00:01:48"
    },
    {
        "sequence": 2,
        "timestamp": "2017-11-10T09:50:10Z",
        "task_id": "443951",
        "location_id": "443951",
        "travel_distance": 26,
        "travel_duration": "00:00:03"
    },
    {
        "sequence": 3,
        "timestamp": "2017-11-10T09:50:25Z",
        "task_id": "165760",
        "location_id": "165760",
        "travel_distance": 152,
        "travel_duration": "00:00:15"
    },
    {
        "sequence": 4,
        "timestamp": "2017-11-10T09:51:34Z",
        "task_id": "459187",
        "location_id": "459187",
        "travel_distance": 705,
        "travel_duration": "00:01:09"
    }]
  }]
}

Answer 1

假设您的桌子是这样的：

  create table table_name 
  ( 
     id number(12) primary key,
     route_sequence number(12),
     updated_by varchar2(30),
     updated_on timestamp(9)  
  )

并且json对象就是这样的：

  {
       "activities": 
        [
            {"task_id": 1, "sequence" : 10},
            {"task_id": 2, "sequence" : 20},
            {"task_id": 3, "sequence" : 30},
            {"task_id": 4, "sequence" : 40},
            {"task_id": 5, "sequence" : 50},
        ]
  }

您可以使用“ JSON_TABLE” sql运算符（oracle 12的新功能，请参见https://docs.oracle.com/database/121/SQLRF/functions092.htm#SQLRF56973）在SQL中直接查询json数据，然后可以在数据库中使用此类查询来利用它“合并”语句：

这条SQL语句可以满足您的需求：

 merge into table_name t
 using
 (
      select * 
      from JSON_TABLE(
              '{
                   "activities": 
                   [
                       {"task_id": 1, "sequence" : 10},
                       {"task_id": 2, "sequence" : 20},
                       {"task_id": 3, "sequence" : 30},
                       {"task_id": 4, "sequence" : 40},
                       {"task_id": 5, "sequence" : 50},
                   ]
              }', 
              '$."activities"[*]'

              COLUMNS(    
                       V_TASK_ID  NUMBER   PATH '$.task_id',
                       V_SEQ    NUMBER   PATH '$.sequence'
                  )
              )
 ) json_data
 on (json_data.v_task_id = t.id)
 when matched then 
    update set 
      ROUTE_SEQUENCE = V_SEQ,
      UPDATED_BY = 'SYSTEM',
      UPDATED_ON = SYSTIMESTAMP

编辑：现在您已经发布了实际的json示例：

要使我的示例处理您的数据，您只需替换

 '$."activities"[*]'

与此行：

 '$."plan"[0]."activities"[*]'

如果“计划”数组项包含多个元素，事情可能会变得更加复杂，但是仍然可以做到。

编辑2：如何处理嵌套对象（即：“计划”包含多个对象时该怎么做

让我们说要处理的json字符串就是这个

    '{
        "plan": 
         [
              {
                  "vehicle_id": "vehicle_1",
                  "activities": 
                   [
                       {
                           "sequence": 1,
                           "task_id": "465427"
                       },
                       {
                           "sequence": 2,
                           "task_id": "443951"
                       }
                   ]    
              }
              ,
              {
                  "vehicle_id": "vehicle_2",
                  "activities": 
                   [
                       {
                           "sequence": 3,
                           "task_id": "165760"
                       },
                       {
                           "sequence": 4,
                           "task_id": "459187"
                       }
                   ]
               }
          ]
    }'

（我在示例中不会重复：我只会在代码中写

如果您对读取vehicle_id字段不感兴趣，并且希望获得所有活动详细信息的平面视图（无论哪个“计划”对象包含它们，您都可以从此更改根对象选择器字符串

 '$."plan"[0]."activities"[*]'

对此：

 '$."plan"[*]."activities"[*]'

因此，此查询：

 select * 
 from JSON_TABLE
   (
       <json_string_here>,
       '$."plan"[*]."activities"[*]'
        COLUMNS(    
                V_TASK_ID  NUMBER   PATH '$.task_id',
                V_SEQ    NUMBER   PATH '$.sequence'
        )
   )

将遍历所有计划对象的所有“活动”对象，但仅返回“ task_id”和“ sequence”列。

如果您还想在所有行上重复对应的车辆ID列，则必须使用此表达式逐步提高根选择器的水平

'$."plan"[*]'

在“ columnns”子句中，您可以使用“嵌套路径”语法来表示您希望同时扩展子对象的列：

   select * 
   from JSON_TABLE
   (
        <json_string_here>,
        '$."plan"[*]'
        COLUMNS
        ( 
            VEHICLE  varchar2(20) PATH '$."vehicle_id"',
            NESTED PATH '$."activities"[*]'
               COLUMNS
               (
                     V_TASK_ID  NUMBER   PATH '$.task_id',
                     V_SEQ    NUMBER   PATH '$.sequence'
                )
        )
   )

Oracle 12.1.0.2中的批量更新？

1 个答案: