如何获得PostgreSQL数据库中的JSON值
在PostgreSQL数据库中有一列为JSON,要获得JSON中得数据可以用下面sql：
select orderno as OrderNo
,amount as Amount
,ordertime as OrderTime
, recordtype as RecordType
from jsonb_to_recordset(( –特定方法
select array_to_json(array_agg(data)) –转换成一个数组
from wallet_details
where id = @id
)::jsonb) as x(orderno text, amount numeric(16, 6), ordertime text, recordtype varchar(32));
如果你获得得数据是当前行，但是JSON中也要取出来几个值可以用下面的方式获得：
select pay_params::json->>’Key’ as Md5Key ,
pay_params::json->>’AppId’ as Appid ,
pay_params::json->>’MchId’ as Mchid ,
pay_params::json->>’SubMchId’ as Submchid ,
tenant_id as Tenant_Id
from spm_wallet_settings where id=’12’
补充：PostgreSql数据库sql语句取Json值
1：json字段实例：
{
“boxNum”: 0,
“orderNum”: 0,
“commentNum”: 0
}
A.取boxNum的值
1.1)select 字段名->‘boxNum’ from 表名;
1.2)select jsonb_extract_path_text字段名, ‘boxNum’) from 表名;
2：json字段实例：
{
“boxNum”: “0”,
“orderNum”: “0”,
“commentNum”: “0”
}
A.取boxNum的值,不带双引号。
2.1)select 字段名->>‘boxNum’ from 表名;
2.2)select jsonb_extract_path_text字段名, ‘boxNum’) from 表名;
3：json字段实例：
{
“unitPrices”: [{
“price”: 10.0,
“unitId”: “8”,
“unitName”: “500克”,
“unitAmount”: “0”,
“isPMDefault”: true,
“isHomeDefault”: true,
“originalPrice”: 10.0
}],
“productName”: “远洋 加拿大 螯龙虾 野生捕捞”,
“productType”: 1,
“skuPortRate”: {
“id”: “a6b83048⑶878⑷698⑻8c2⑵a9de288ac56”,
“cityId”: “2bf8c60c⑺89d⑷33a⑼1ae⑻e4ae3e587a4”,
“dynamicProperties”: [{
“name”: “死亡率”,
“propertiesId”: “f05bda8c-f27c⑷cc6-b97e-d4bd07272c81”,
“propertieValue”: {
“value”: “2.0”
}
}, {
“name”: “失水率”,
“propertiesId”: “ee9d95d7⑺e28⑷d54-b572⑷8ae64146c46”,
“propertieValue”: {
“value”: “3.0”
}
}]
},
“quotePriceAttribute”: {
“currencyName”: “人民币”
}
}
A.取quotePriceAttribute中的currencyName币制名称
select (字段名>>‘quotePriceAttribute’)::json->>‘currencyName’ from 表名；
B.取unitPrices中的price单价
select jsonb_array_elements((字段名->>‘unitPrices’)::jsonb)->>‘price’ from 表名；
C.取skuPortRate中的dynamicProperties的name为死亡率的propertieValue里面的value；
select bb->‘propertieValue’->>‘value’ as value from (
select jsonb_array_elements(((字段名->>‘skuPortRate’)::json->>‘dynamicProperties’)::jsonb) as bb from 表名) as dd where dd.bb @> ‘{“name”: “死亡率”}’;
4.json字段实例：
[{“name”: “捕捞方式”, “showType”: 4, “propertiesId”: “9a14e435⑼688⑷e9b-b254-0e8e7cee5a65”,
“propertieValue”: {“value”: “野生捕捞”, “enValue”: “Wild”}},
{“name”: “加工方式”, “showType”: 4, “propertiesId”: “7dc101df-d262⑷a75-bdca⑼ef3155b7507”,
“propertieValue”: {“value”: “单冻”, “enValue”: “Individual Quick Freezing”}},
{“name”: “原产地”, “showType”: 4, “propertiesId”: “dc2b506e⑹620⑷e83⑻ca1-a49fa5c5077a”,
“propertieValue”: {“value”: “爱尔兰”, “remark”: “”, “enValue”: “Ireland”}}]
–获得原产地
select
(SELECT ss->‘propertieValue’ as mm FROM
(SELECT jsonb_array_elements (dynamic_properties) AS ss FROM product
where id=a.id) as dd where dd.ss @> ‘{“name”: “原产地”}’)->>‘value’ as cuntry,
a.*
from product as a where a.id=‘633dd80f⑺250⑷65f⑻982⑺a7f01aaeeec’；
5：json例子:huren:[“aaa”,“bbb”,“ccc”…]
需求：取值aaa去““双引号”
select replace(cast(jsonb_array_elements(huren) as text), ‘”‘,”) from XXX limit 1
以上为个人经验，希望能给大家一个参考，也希望大家多多支持。如有毛病或未斟酌完全的地方，望不吝赐教。
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文章标题：如何获得PostgreSQL数据库中的JSON值
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