需要创建一个Json_Object,其中可以包含多个嵌套的Json_objects,Json_arrays和Json_arrayaggs。
我用一些虚拟数据创建了这个表来演示这个问题:
create table test_tbl(
test_col1 varchar2(20),
test_col2 varchar2(20),
test_col3 varchar2(20),
test_col4 varchar2(20),
test_col5 varchar2(20),
test_col6 varchar2(20)
);
insert into test_tbl values('val0', 'val1', 'val2', 'val7', 'val11', 'val12');
insert into test_tbl values('val0', 'val3', 'val4', 'val7','val11', 'val12');
insert into test_tbl values('val0', 'val5', 'val6', 'val7','val13', 'val14');
insert into test_tbl values('val0', 'val5', 'val6', 'val7','val11', 'val12');
insert into test_tbl values('val0', 'val5', 'val6', 'val8','val11','val12');
insert into test_tbl values('val1', 'val9', 'val10', 'val7','val11', 'val12');
insert into test_tbl values('val1', 'val9', 'val10', 'val7','val13', 'val14');
当使用以下查询来创建Json_object时:
SELECT JSON_OBJECT (
'output' VALUE JSON_ARRAYAGG(
JSON_OBJECT(
'common' VALUE test_col1,
'list' VALUE JSON_ARRAYAGG(
JSON_OBJECT(
'key1' VALUE test_col2,
'key2' VALUE test_col3
)
),
'anotherlist' VALUE JSON_ARRAYAGG(
JSON_OBJECT(
'key1' VALUE test_col5,
'key2' VALUE test_col6
)
)
)
)
)
FROM (
SELECT DISTINCT
test_col1, test_col2, test_col3, test_col5, test_col6
FROM test_tbl
WHERE test_col4 = 'val7'
)
GROUP BY
test_col1
这会导致json在聚合数组中使用重复的键,值对-
{
"output": [
{
"common": "val0",
"list": [
{
"key1": "val5",
"key2": "val6"
},
{
"key1": "val3",
"key2": "val4"
},
{
"key1": "val1",
"key2": "val2"
},
{
"key1": "val5",
"key2": "val6"
}
],
"anotherlist": [
{
"key1": "val13",
"key2": "val14"
},
{
"key1": "val11",
"key2": "val12"
},
{
"key1": "val11",
"key2": "val12"
},
{
"key1": "val11",
"key2": "val12"
}
]
},
{
"common": "val1",
"list": [
{
"key1": "val9",
"key2": "val10"
},
{
"key1": "val9",
"key2": "val10"
}
],
"anotherlist": [
{
"key1": "val11",
"key2": "val12"
},
{
"key1": "val13",
"key2": "val14"
}
]
}
]
}
而我期望的Json是:
{
"output": [
{
"common": "val0",
"list": [
{
"key1": "val5",
"key2": "val6"
},
{
"key1": "val3",
"key2": "val4"
},
{
"key1": "val1",
"key2": "val2"
}
],
"anotherlist": [
{
"key1": "val13",
"key2": "val14"
},
{
"key1": "val11",
"key2": "val12"
}
]
},
{
"common": "val1",
"list": [
{
"key1": "val9",
"key2": "val10"
}
],
"anotherlist": [
{
"key1": "val11",
"key2": "val12"
},
{
"key1": "val13",
"key2": "val14"
}
]
}
]
}
提前感谢您对如何获得上述预期JSON的任何建议。
1条答案
按热度按时间wn9m85ua1#
对第一对列使用一个
DISTINCT
子查询,然后对另一对列使用第二个DISTINCT
子查询,并对公共test_col1
使用JOIN
:输出: