我正在处理Mysql查询以检索JSON对象
Select JSON_OBJECT('id', students.id, 'name', students.name) as test from students;
输出
"test": {
"id": null,
"name": null
}预期产出
test: null
如何检查测试对象是否有空值,然后返回空值?
发布于 2021-04-20 10:26:05
如何检查测试对象是否有空值,然后返回空值
形式上(不了解逻辑):
SELECT CASE WHEN students.id IS NULL AND students.name IS NULL
THEN CAST( 'null' AS JSON )
ELSE JSON_OBJECT('id', students.id, 'name', students.name)
END AS test
FROM students;https://stackoverflow.com/questions/67176674
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