在postgresql中使用xml？

Question

我有一個xml文檔。把它放到posgresql表中。 這就是我所說的。

create or replace function bytea_import(p_path text, p_result out bytea) 
        language plpgsql as $$ 
declare 
    l_oid oid; 
    r record; 
begin 
    p_result := ''; 
    select lo_import(p_path) into l_oid; 
    for r in (select data 
      from pg_largeobject 
      where loid = l_oid 
      order by pageno) loop 
    p_result = p_result || r.data; 
    end loop; 
    perform lo_unlink(l_oid); 
end;$$; 


insert into mydocs(docform,content) 
values (3, convert_from(bytea_import('D:/html/ex08.xml'), 'utf-8')); 

我需要更改文檔。例如

<?xml version="1.0"?> 
<list_of_items> 
<item id="1"><first/>first</item> 
<item id="2">second <sub_item>subsecond 1</sub_item></item> 
<item id="3">third</item> 
<item id="4"><last/>last</item> 
</list_of_items> 

必須從該文件取出文本並投入其他表：

first投入表1第2列

second投入表第2欄2

subsecond放入表2第3欄

third放入第4列o ˚F表2

last投產表2

加布的5列做一個很好的答案，但我有1點人的問題。 如何製作任何種類的xml。 如何刪除行像

<?xml version="1.0"?> 
<list_of_items> 
</list_of_items> 

，並選擇任何文本，不xpath搜索？

Answer 1

您贏得了當天最難的問題。獲得upvote吧！

用下面的查詢：

WITH x(col) AS (
    SELECT '<?xml version="1.0"?> 
    <list_of_items> 
    <item id="1"><first/>first</item> 
    <item id="2">second <sub_item>subsecond 1</sub_item></item> 
    <item id="3">third</item> 
    <item id="4"><last/>last</item> 
    </list_of_items> 
    '::xml 
), interpreted as (
    SELECT xpath('.//item/text()', col) AS val, 
      xpath('.//sub_item/text()', col) AS val2 
    FROM x 
) 
SELECT val[1] as col1, val[2] as col2, val2[1] as col3, val[3] as col4, val[4] as col5 
FROM interpreted 
; 

你可以得到：

col1 | col2 | col3  | col4 | col5 
-------+---------+-------------+-------+------ 
first | second | subsecond 1 | third | last 
(1 row)