从查询id中得到的输出重复它内部的数据是多个
select json_arrayagg(json_object("network", basics_profiles.network, "username", basics_profiles.username, "url", basics_profiles.url)) as profiles from basics_profiles;好的,在这里,当我运行这段代码时,在工作部分,即使只有2个数据。它会加倍。如果只有单一的数据。那就没有问题了。
输出时,有更多的tan 1数据:
[
{
"url": "twitter.com/naruto",
"network": "Twitter",
"username": "uzumakinaruto"
},
{
"url": "twitter.com/naruto",
"network": "Twitter",
"username": "uzumakinaruto"
},
{
"url": "instagram.com/naruto",
"network": "Instagram",
"username": "uzumakinaruto"
},
{
"url": "instagram.com/naruto",
"network": "Instagram",
"username": "uzumakinaruto"
}
]我想要这样的东西
[
{
"url": "twitter.com/naruto",
"network": "Twitter",
"username": "uzumakinaruto"
},
{
"url": "instagram.com/naruto",
"network": "Instagram",
"username": "uzumakinaruto"
}
]你可以试试这张桌子:
create table basics_profiles(network text, url text, username text);
insert into basics_profiles values("twitter","jhaajdka.com","naruto");
insert into basics_profiles values("instagram","jhasdasdasdsdd.com","sasuke"); 发布于 2022-02-07 04:50:32
谢谢你帮我。我有问题了。使用about代码,如果我有多行数据,它将显示单个行中的所有数据,对我来说,问题是当连接两个表时,数据会被一次又一次地重复。
select
basics_information.*,
(select json_arrayagg(json_object("name", interests.name, "keywords", interests.keywords))
from interests
where basics_information.id = interests.resumeId) as interests
from basics_information
left join interests on basics_information.id = interests.resumeId
group by basics_information.id;发布于 2022-02-05 09:38:59
https://stackoverflow.com/questions/70996664
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