2015-02-09 124 views
1

我可以传递视图的名称作为函数的参数吗?例如:将视图的名称作为参数传递给函数

CREATE OR REPLACE FUNCTION example_test(test type_view) return void as $$ 
BEGIN 
    start_ts = CLOCK_TIMESTAMP(); 
    REFRESH MATERIALIZED VIEW test; 
    GET DIAGNOSTICS total_rows = ROW_COUNT; 
    INSERT INTO control_dw_monitoring (name, start_time, end_time, total) 
    VALUES ('view points that never contacted', start_ts, CLOCK_TIMESTAMP(), total_rows); 
END 
$$ language plpgsql; 
+0

你需要为这个动态SQL:http://www.postgresql.org/docs/current/static/plpgsql-statements.html#PLPGSQL -STATEMENTS-EXECUTING-DYN – 2015-02-09 12:27:10

+0

[那么你有答案吗?](http://meta.stackexchange.com/a/5235/169168) – 2015-07-24 03:21:25

回答

0

尝试以下方法:

CREATE FUNCTION refresh_view_by_name(view_name text) RETURNS VOID AS $$ 
BEGIN 
    EXECUTE 'REFRESH MATERIALIZED VIEW ' || view_name::regclass; 
END 
$$ LANGUAGE PLPGSQL; 
0

的视图的名称是一个标识符,而不是一个值。你不能参数化它,所以你需要动态SQL与EXECUTE - 你需要避免SQL注入。首先制定参数为regclass

CREATE OR REPLACE FUNCTION example_test(_v regclass) 
    RETURNS VOID AS 
$func$ 
DECLARE 
    start_ts timestamptz := clock_timestamp(); 
    total_rows int; 
BEGIN 
    EXECUTE 'REFRESH MATERIALIZED VIEW ' || _v; -- converted to text automatically 
    GET DIAGNOSTICS total_rows = ROW_COUNT; 

    INSERT INTO control_dw_monitoring (name, start_time, end_time, total) 
    VALUES (_v::text, start_ts, clock_timestamp(), total_rows); 
END 
$func$ LANGUAGE plpgsql; 

修正了其中的一些错误。
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