postgresql 双精度类型的输入语法无效
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Invalid input syntax for type double precision
提问by jovicbg
I've created functions and one of them is:
我创建了函数，其中之一是：
CREATE OR REPLACE FUNCTION core.cal_status(
er_v DOUBLE PRECISION,
cell_v DOUBLE PRECISION
) RETURNS DOUBLE PRECISION
LANGUAGE plpgsql
AS $$
DECLARE out DOUBLE PRECISION;
BEGIN
IF er_v < 15 THEN out := 'LOW_LOAD';
ELSEIF cell_v < 60 THEN out := 'LOW_LOAD';
ELSEIF er_v > 40 THEN out := 'CONGESTED';
ELSEIF cell_v > 80 THEN out := 'CONGESTED';
ELSEIF cell_v >60 and cell_v < 80 THEN out := 'HIGH_LOAD';
END IF;
RETURN out;
END;
$$;
When I call functions with this:
当我用这个调用函数时：
LEFT JOIN LATERAL core.cal_status(
er_v,
cell_v) status ON true
I got next error:
我得到下一个错误：
ERROR: invalid input syntax for type double precision: "LOW_LOAD" Where: PL/pgSQL function core.cal_status(double precision,double precision) line 4 at assignment
What is this about, I guess it's something with output type, but not sure.
这是关于什么的，我猜它与输出类型有关，但不确定。
回答by Kevin Miranda
The error happens because the variable outis of text type when you do out :='LOW_LOAD'and the function is expecting to return a DOUBLE PRECISION.
发生错误out是因为在您执行此操作时变量是文本类型，out :='LOW_LOAD'并且该函数期望返回 DOUBLE PRECISION。
Try
尝试
CREATE OR REPLACE FUNCTION core.cal_status(
er_v DOUBLE PRECISION,
cell_v DOUBLE PRECISION
) RETURNS TEXT
LANGUAGE plpgsql
AS $$
DECLARE out TEXT;
BEGIN
IF er_v < 15 THEN out := 'LOW_LOAD';
ELSEIF cell_v < 60 THEN out := 'LOW_LOAD';
ELSEIF er_v > 40 THEN out := 'CONGESTED';
ELSEIF cell_v > 80 THEN out := 'CONGESTED';
ELSEIF cell_v >60 and cell_v < 80 THEN out := 'HIGH_LOAD';
END IF;
RETURN out;
END;
$$;
