我想在Postgresql 9.4中访问一个嵌套的jsonb字段。但是我很难根据嵌套的jsonb值检索记录。有没有人有过这个成功?如何在Postgresql 9.4中访问嵌套的JSON?
首先,我创建的表:
CREATE TABLE Meal ( id INT , recipe JSONB );
其次,我插入以下记录到dbo.Meal: (我通过jsonlint.com跑了JSON和它回来了有效)
INSERT INTO Meal (id, recipe) VALUES ( 1, '{ "meal": [{ "calories" : 900, "serves" : [{"min": 2, "max": 4}], "fruit" : [{"id": 1, "qty": 2}, {"id": 4, "qty": 3}], "veggie" : [{"id": 4, "qty": 1}, {"id": 2, "qty": 10}] }] }');
第三,我想下面的查询在此基础上的正在其卡路里(无)来检索该记录:
这些返回0记录:
SELECT * FROM Meal ...
WHERE recipe::json#>>'{meal, calories}' = '900';
WHERE recipe::json->>'{meal, calories}' = '900';
WHERE recipe::json->>'meal[calories]' = '900';
WHERE recipe::json->>'{meal[calories]}' = '900';
WHERE recipe::json#>>'{meal[calories]}' = '900';
WHERE recipe::json#>>'{meal.calories}' = '900';
WHERE recipe::json->>'{meal.calories}' = '900';
WHERE recipe::jsonb#>>'{meal, calories}' = '900';
WHERE recipe::jsonb->>'{meal, calories}' = '900';
WHERE recipe::jsonb#>>'{meal[calories]}' = '900';
WHERE recipe::jsonb->>'{meal[calories]}' = '900';
WHERE recipe::jsonb->>'meal[calories]' = '900';
WHERE recipe::jsonb#>'{meal, calories}' = '900';
WHERE recipe::jsonb->'{meal, calories}' = '900';
WHERE recipe::jsonb->'meal[calories]' = '900';
WHERE recipe::jsonb#>'{meal[calories]}' = '900';
WHERE recipe::jsonb->'{meal[calories]}' = '900';
WHERE recipe::jsonb#>>'{meal.calories}' = '900';
WHERE recipe::jsonb#>'{meal.calories}' = '900';
WHERE recipe::jsonb->>'{meal.calories}' = '900';
WHERE recipe::jsonb->'{meal.calories}' = '900';
这些结果以失败(不正确的语法):
SELECT * FROM Meal ...
WHERE recipe::json#>'{meal, calories}' = '900';
WHERE recipe::json->'{meal, calories}' = '900';
WHERE recipe::json#>>'meal[calories]' = '900';
WHERE recipe::json#>'meal[calories]' = '900';
WHERE recipe::json->'meal[calories]' = '900';
WHERE recipe::json#>'{meal[calories]}' = '900';
WHERE recipe::json->'{meal[calories]}' = '900';
WHERE recipe::json#>'{meal.calories}' = '900';
WHERE recipe::json->'{meal.calories}' = '900';
WHERE recipe::jsonb#>>'meal[calories]' = '900';
WHERE recipe::jsonb#>'meal[calories]' = '900';
如果您有任何建议,我将不胜感激听他们。
我可能已经想通了。根据OP中的表,以下2个查询正确返回记录: (1)SELECT * FROM Meal WHERE recipe @>'{“meal':[{”calories“:900}]}':: jsonb ; (2)SELECT * FROM Meal WHERE recipe @>'{“meal':[{”calories“:900}]}':: jsonb AND recipe @>'{'meal':[{”fruit“:[{{ “ID”:4}]}]}':: jsonb; 并且下面的查询不会返回记录(这是因为meal.fruit.id = 40不存在是正确的): (3)SELECT * FROM Meal WHERE recipe @>'{“meal”:[{“calories “:900}]}':: jsonb AND recipe @>'{”meal“:[{”fruit“:[{”id“:40}]}]}':: jsonb; 希望这有助于! – bjones1831
在测试版发布时间不到72小时的情况下提出一个新功能问题,如果有的话,不会在SO上产生许多明智的答案。在这种情况下的问题最好在postgresql的邮件列表中问一下, - 黑客想起来了。 – Patrick