在postgre中创建newguid()方法

     Postgre不支持newid()方法，需要自己创建，方法：在Postgre中执行以下语句，就完成了newid()的创建。

CREATE or REPLACE FUNCTION "newid"() RETURNS "pg_catalog"."varchar" AS $BODY$ DECLARE   v_seed_value varchar(32); BEGIN select md5( inet_client_addr()::varchar || timeofday() || inet_server_addr()::varchar || to_hex(inet_client_port()) ) into v_seed_value; return (substr(v_seed_value,1,8) || '-' ||         substr(v_seed_value,9,4) || '-' ||         substr(v_seed_value,13,4) || '-' ||         substr(v_seed_value,17,4) || '-' ||         substr(v_seed_value,21,12)); END; $BODY$ LANGUAGE 'plpgsql' VOLATILE SECURITY DEFINER; 

参考自： http://postgresql.1045698.n5.nabble.com/newid-in-postgres-td1879346.html