Postgresql函数动态表quote_ident错误

Question

我创建了一个动态函数。我动态地获取表名的一部分。该功能已成功创建。但是当我执行该功能。我收到一个错误。我怎么解决这个问题？我打电话跟

select * from dwgcould.getlatlngcenter(2000653); 


CREATE OR REPLACE FUNCTION dwgcould.getlatlngcenter(IN pro_id integer, 
    OUT lat_center double precision, OUT lng_center double precision) 
AS $$ 

BEGIN 
     EXECUTE 'SELECT st_x(st_centroid(st_transform(geom,4326))) as lng_center ,st_y(st_centroid(st_transform(geom,4326))) as lat_center 
     FROM dwgcould.adpes_v1_' || quote_ident(pro_id) || '_line limit 1'; 
END; 
$$ LANGUAGE plpgsql; 

功能的错误代码是

ERROR: function quote_ident(integer) does not exist LINE 2: FROM dwgcould.adpes_v1_' || quote_ident(pro_id) || '_line... ^ HINT: No function matches the given name and argument types. You might need to add explicit type casts. QUERY: SELECT 'SELECT st_x(st_centroid(st_transform(geom,4326))) as lng_center ,st_y(st_centroid(st_transform(geom,4326))) as lat_center FROM dwgcould.adpes_v1_' || quote_ident(pro_id) || '_line limit 1' CONTEXT: PL/pgSQL function dwgcould.getlatlngcenter(integer) line 4 at EXECUTE statement ********** Error ********** ERROR: function quote_ident(integer) does not exist SQL state: 42883 Hint: No function matches the given name and argument types. You might need to add explicit type casts. Context: PL/pgSQL function dwgcould.getlatlngcenter(integer) line 4 at EXECUTE statement

此外，我如何检查表是否存在？

Answer 1

更好地使用的格式，例如：

CREATE OR REPLACE FUNCTION dwgcould.getlatlngcenter(IN pro_id integer, 
    OUT lat_center double precision, OUT lng_center double precision) 
AS $$ 

BEGIN 
    if (select count(1) from pg_tables where tablename = format('adpes_v1_%s_line',pro_id)) < 1 then 
    raise info '%','NO SUCH TABLE!'; 
    return; 
    end if; 
    EXECUTE format('SELECT * FROM dwgcould.adpes_v1_%s_line limit 1',pro_id) into lat_center,lng_center; 
    return; 
END; 
$$ LANGUAGE plpgsql; 

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