我们在我们的项目中使用了以下cypher查询(只是相关的部分,有一堆匹配,以及在此之前的位置):
"RETURN entry.id AS id,\n" +
...
" CASE WHEN ljCase1 IS NOT NULL THEN {id: ljCase1.id, guide: false} \n" +
" ELSE CASE WHEN ljCase2 IS NOT NULL THEN {id: ljCase2.id, guide: true}\n" +
" ELSE CASE WHEN ljCase3 IS NOT NULL THEN {id: ljCase3.id, guide: true} END\n" +
" END\n" +
" END AS learningJourney\n" +
...它工作正常,但我觉得这并不像它应该做的那样干净。有没有办法简化这部分查询?
发布于 2021-09-21 10:03:30
也许是这样,使用COALESCE来避免嵌套情况。
WITH COALESCE(ljCase1,ljCase2,ljCase3) AS myCase
WITH CASE myCase
WHEN x THEN ..
WHEN y THEN ..
ELSE ..
ENDhttps://stackoverflow.com/questions/69266827
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