我有一个从umbraco(4.7)数据库加载一些信息的批处理。
我需要提取Link to document
节点属性...
网站中的Api就像umbraco.library.NiceUrl(nodeid)
一样...但我有一个控制台应用程序无法访问umbraco的东西......
1)我在哪里可以找到这个"漂亮的网址"数据库中的节点? 2)如果没有(或太复杂),如何使用控制台应用程序配置umbraco API(4.7)库?
答案 0 :(得分:2)
我尝试了上述答案,但是他们需要mysql(需要sql server),clr函数,或者只包含有关当前文档的url信息的信息。对于导航路径深处的儿童和文档,我想要包含完整路径。以下适用于我
; WITH PathXml AS (
/* -- This just gives nodes with their 'urlName' property
-- not in recycle bin,
-- level > 1 which excludes the top level documents which are not included in the url */
SELECT
nodeId,
cast([xml] as xml).query('data(//@urlName[1])').value('.', 'varchar(max)') AS Path
FROM cmsContentXml x
JOIN umbracoNode n ON x.nodeId = n.id AND n.trashed = 0 AND n.level > 1
)
SELECT
un.id,
un.path,
'/' +
/* IsNull after the leading '/'. This will handle the top level document */
IsNull((SELECT
pl.Path + '/' /* Ok to end with a / */
FROM PathXml pl
/* e.g. ',-1,1071,1072,1189,' LIKE '%,1072,%' */
WHERE ',' + un.path + ',' LIKE '%,' + CAST(pl.nodeId AS VARCHAR(MAX)) + ',%'
/* order by the position of ',1072,' in ',-1,1071,1072,1189,' */
ORDER BY CHARINDEX(',' + CAST(pl.nodeId AS VARCHAR(MAX)) + ',',
',' + un.path + ',')
FOR XML PATH('')),
'') AS Url,
un.text PageName
FROM umbracoNode un
WHERE nodeObjectType = 'C66BA18E-EAF3-4CFF-8A22-41B16D66A972' /* "Document" https://our.umbraco.org/apidocs/csharp/api/Umbraco.Core.Constants.ObjectTypes.html#Umbraco_Core_Constants_ObjectTypes_Document */
AND trashed = 0
ORDER BY 3 /* Url */
答案 1 :(得分:1)
SELECT
cast([xml] as xml).query('data(//@urlName[1])').value('.', 'varchar(20)') AS PATH
FROM cmsContentXml where nodeid = 19802
我的2美分值
答案 2 :(得分:0)
select
GROUP_CONCAT(EXTRACTVALUE(xml, "//@urlName") separator '/') as PATH
from
cmscontentxml x inner join
umbraconode n on FIND_IN_SET(x.NODEID,n.PATH)>0
where n.ID=19802 -- your id
group by n.ID;