假设文件 f1,f2,f3 都有标签 TODO
希望在 f4 文件中实现 列出 TODO 标签所在的全部文档
f1
f2
f3
sql
SELECT * FROM blocks WHERE markdown LIKE '%#TODO#%'
sql
{{ SELECT * FROM blocks WHERE type = 'd' AND id IN (
SELECT root_id FROM blocks WHERE markdown LIKE '%#TODO#%'
) }}
sql
{{ SELECT DISTINCT b.* FROM blocks b
JOIN blocks t ON b.id = t.root_id
WHERE t.markdown LIKE '%#TODO#%' AND b.type = 'd' }}
方法 1 返回的是有标签的那一行内容
方法 23 返回的是有标签的文件的全部内容
我需要的是返回有标签的文档的文档链接
请教各位高手
SELECT '[' || content || '](siyuan://blocks/' || id || ')'
AS __1____pre__文档标题,
hpath AS __2__文档路径,
updated AS __3____datetime__更新时间
FROM blocks
WHERE type = 'd'
AND id IN (
SELECT root_id FROM blocks WHERE markdown LIKE '%#TODO#%'
)
AND hpath not LIKE '%TEST%'
ORDER BY hpath
LIMIT 9999;
最好加上 LIMIT，不然可能限制在最大搜索数量内，比如默认 64
使用查询挂件
SELECT '[' || b.content || '](siyuan://blocks/' || b.id || ')'
AS __1____pre__文档标题,
b.hpath AS __2___文档路径,
b.updated AS __3____datetime__更新时间
FROM blocks AS b
WHERE
b.markdown LIKE '%#list#%'
AND b.hpath not LIKE '%TEST%'
ORDER BY
b.hpath
标题也和预期不符。
是否就没有源数据？ @88250
SELECT '[' || content || '](siyuan://blocks/' || id || ')'
AS __1____pre__文档标题,
hpath AS __2__文档路径,
updated AS __3____datetime__更新时间
FROM blocks
WHERE type = 'd'
AND id IN (
SELECT root_id FROM blocks WHERE markdown LIKE '%#TODO#%'
)
AND hpath not LIKE '%TEST%'
ORDER BY hpath
LIMIT 9999;
最好加上 LIMIT，不然可能限制在最大搜索数量内，比如默认 64
