我用这段代码来验证
DELETE
这句话,但我相信你知道更好的方法:
CREATE OR REPLACE FUNCTION my_schema.sp_delete_row_table(table_name character varying
, id_column character varying
, id_value integer)
RETURNS integer AS
$BODY$
DECLARE
BEFORE_ROWS integer;
AFTER_ROWS integer;
BEGIN
EXECUTE 'SELECT count(*) FROM ' || TABLE_NAME INTO BEFORE_ROWS;
EXECUTE 'DELETE FROM ' || TABLE_NAME || ' WHERE ' || ID_COLUMN || ' = ' || (ID_VALUE)::varchar;
EXECUTE 'SELECT count(*) FROM ' || TABLE_NAME INTO AFTER_ROWS;
IF BEFORE_ROWS - AFTER_ROWS = 1 THEN
RETURN 1;
ELSE
RETURN 2;
END IF;
EXCEPTION WHEN OTHERS THEN
RETURN 0;
END;
$BODY$
LANGUAGE plpgsql VOLATILE;
如何改进这段代码?我需要它在 Postgres 8.4、9.1 和 9.2 中工作。
您不能将
FOUND
与 EXECUTE
一起使用。 手册:
特别注意
改变了EXECUTE
的输出, 但不会改变GET DIAGNOSTICS
。FOUND
还有更多问题,最重要的是允许 SQL 注入。我建议:
CREATE OR REPLACE FUNCTION my_schema.sp_delete_row_table(table_name regclass
, id_column text
, id_value int
, OUT del_ct int)
LANGUAGE plpgsql AS
$func$
BEGIN
EXECUTE format('DELETE FROM %s WHERE %I = $1', table_name, id_column);
USING id_value; -- assuming integer columns
GET DIAGNOSTICS del_ct = ROW_COUNT; -- directly assign OUT parameter
EXCEPTION WHEN OTHERS THEN
del_ct := 0;
END
$func$;
format()
或与 quote_ident()
! 进行字符串连接
相关:
查看名为
found
和 row_count
的变量:
http://www.postgresql.org/docs/current/static/plpgsql-statements.html#PLPGSQL-STATMENTS-DIAGNOSTICS
如果有任何行受到影响,则found
为 true。 row_count
为您提供受影响的行数。
IF FOUND THEN
GET DIAGNOSTICS integer_var = ROW_COUNT;
END IF;