让我们说我们有一个公式,指的是另一张纸,如
=HYPERLINK("http://www.w3c.org";AnotherSheet!A1)
是否有一个简单的方法getFormula()但已经执行了引用?
因此函数的结果应该是=HYPERLINK("http://www.w3c.org";"W3C")
AnotherSheet!A1="W3C"
。
答案 0 :(得分:0)
"一种简单的方法"是主观的。以下是如何使用脚本执行此操作的示例:
SELECT
date(survey_results.created_at),
json_build_object(
'high', ROUND(COUNT(*) FILTER (WHERE (scores#>>'{medic,categories,food_insecurity}' in('high'))) * 1.0 / COUNT(*) FILTER (WHERE (scores#>>'{medic,categories,food_insecurity}' in('high','medium','low'))) * 100,2),
'medium', ROUND(COUNT(*) FILTER (WHERE (scores#>>'{medic,categories,food_insecurity}' in('medium'))) * 1.0 / COUNT(*) FILTER (WHERE (scores#>>'{medic,categories,food_insecurity}' in('high','medium','low'))) * 100,2),
'low', ROUND(COUNT(*) FILTER (WHERE (scores#>>'{medic,categories,food_insecurity}' in('low'))) * 1.0 / COUNT(*) FILTER (WHERE (scores#>>'{medic,categories,food_insecurity}' in('high','medium','low'))) * 100,2)
) as food_insecurity,
json_build_object(
'high', ROUND(COUNT(*) FILTER (WHERE (scores#>>'{medic,categories,motivation}' in('high'))) * 1.0 / COUNT(*) FILTER (WHERE (scores#>>'{medic,categories,motivation}' in('high','medium','low'))) * 100,2),
'medium', ROUND(COUNT(*) FILTER (WHERE (scores#>>'{medic,categories,motivation}' in('medium'))) * 1.0 / COUNT(*) FILTER (WHERE (scores#>>'{medic,categories,motivation}' in('high','medium','low'))) * 100,2),
'low', ROUND(COUNT(*) FILTER (WHERE (scores#>>'{medic,categories,motivation}' in('low'))) * 1.0 / COUNT(*) FILTER (WHERE (scores#>>'{medic,categories,motivation}' in('high','medium','low'))) * 100,2)
) as motivation
FROM survey_results
GROUP BY date(survey_results.created_at);
相关
答案 1 :(得分:-1)
是的,您只需使用getValue()函数而不是getFormula()。
斯特凡