Question

只有来自path的订单是保留者时，此查询的结果才有意义。但是，在我的Left Inner Join中，结果是混乱的。我想我可以在结果集中创建一个新列，就像结果的索引一样，然后作为整个sql的最后一个子句添加一个ORDER BY idx。

这是我的查询而不是试图保留最终订单：

SELECT path.*, network_link.v0prt
    FROM (SELECT *                              // Need order preserved from this one
        FROM shortest_path_shooting_star(
        'SELECT gid as id, source::integer,
        target::integer, distance::double precision as cost,
        x1, y1, x2, y2, rule, to_cost FROM
            network_link as net ORDER BY net.gid', 9, 1, false, false)) as path
    LEFT OUTER JOIN 
    (SELECT DISTINCT gid, v0prt FROM network_link) as network_link 
    ON (network_link.gid=path.edge_id);

任何见解都会很棒。

感谢。我尝试添加索引值和ORDER BY（不起作用）。

SELECT path.*, network_link.v0prt
    FROM (SELECT incr(0) as idx, * 
        FROM shortest_path_shooting_star(
        'SELECT gid as id, source::integer,
        target::integer, distance::double precision as cost,
        x1, y1, x2, y2, rule, to_cost FROM
            network_link as net ORDER BY net.gid', 9, 1, false, false)) as path
    LEFT OUTER JOIN 
    (SELECT DISTINCT gid, v0prt FROM network_link) as network_link 
    ON (network_link.gid=path.edge_id)
ORDER BY idx;

Answer 1

要保留shortest_path_shooting_star返回的顺序（如果不提供其他方式），您可以使用窗口函数row_number来跟踪原始顺序，并ORDER BY结果：

SELECT path.*, network_link.v0prt
    FROM (SELECT row_number() OVER() AS row_number, *
        FROM shortest_path_shooting_star(
        'SELECT gid as id, source::integer,
        target::integer, distance::double precision as cost,
        x1, y1, x2, y2, rule, to_cost FROM
            network_link as net ORDER BY net.gid', 9, 1, false, false)) as path
    LEFT OUTER JOIN 
    (SELECT DISTINCT gid, v0prt FROM network_link) as network_link 
    ON (network_link.gid=path.edge_id)
ORDER BY path.row_number;

<强>更新

从PostgreSQL版本9.4及更新版本开始，更好的方法是使用WITH ORDINALITY：

SELECT path.*, network_link.v0prt
FROM shortest_path_shooting_star(
        'SELECT gid as id, source::integer,
        target::integer, distance::double precision as cost,
        x1, y1, x2, y2, rule, to_cost FROM
            network_link as net ORDER BY net.gid', 9, 1, false, false)
) WITH ORDINALITY AS path
LEFT OUTER JOIN 
    (SELECT DISTINCT gid, v0prt FROM network_link) as network_link 
    ON (network_link.gid=path.edge_id)
ORDER BY path.ordinality;

在结果集PostgreSQL中创建自动增量列

1 个答案: