postgresql-函数上出现空值错误

我正在尝试解决此

错误：当尝试选择我的函数时，空值不能格式化为SQL标识符

：

select * from store_keys();
ERROR:  null values cannot be formatted as an SQL identifier
CONTEXT:  SQL statement "SELECT string_agg(
    format('SELECT %1$I, count(email_store_key), email_store_key, form_created_datetime
            FROM %1$I where email_store_key=0 GROUP BY 3, 4', tbl_name),
    ' UNION ')            FROM information_schema.tables
  WHERE table_schema = 'abc_dev_sch_1234'
    AND table_name LIKE 'fact_%'"

它使用的基本查询不会产生

空值

。那么这是从哪里来的

select count(email_store_key), email_store_key, form_created_datetime
FROM <table_name> where email_store_key=0 GROUP BY email_store_key, form_created_datetime;

---

我想是简单的打字错误。您的

format

调用使用的是输出变量

tbl\u name

，而不是列

table\u name

。顺便说一句，我认为您的第一个格式说明符需要是

%1$L

，而不是

%1$I

@NickBarnes在此上下文中，%I和%L之间唯一显著的区别是前者在必要时是双引号；两者都产生一个字符串值。该错误是由于%I不接受从未初始化的“tbl_name”输出参数传入的空值。确实是简单的打字。@Patrick:

%L

生成一个带引号的值，如

quote\u literal（）

DROP FUNCTION store_keys();
CREATE OR REPLACE FUNCTION store_keys()
RETURNS TABLE (tbl_name varchar, count_keys bigint, email_store_key integer, form_created_datetime timestamp)
AS $$
DECLARE
  qry text;
BEGIN
  SELECT string_agg(
    format('SELECT %1$I, count(email_store_key), email_store_key, form_created_datetime
            FROM %1$I where email_store_key=0 GROUP BY 3, 4', tbl_name),
    ' UNION ') INTO qry
  FROM information_schema.tables
  WHERE table_schema = 'abc_dev_sch_1234'
    AND table_name LIKE 'fact_%';

  RETURN QUERY EXECUTE qry;
END
$$ LANGUAGE plpgsql;