我还没有找到一种方法将两个*.mbtiles文件连接在一起(第一个文件包含从0到16的缩放级别,第二个文件包含缩放级别17)。我使用了不同的sqlite管理器,但不管我如何将database2导出和导入到database1中,我都没有成功-二进制字段总是被严重损坏,以至于无法找回image.png。
有谁知道将两个mbtile文件连接在一起的简单步骤吗?
发布于 2013-07-19 12:55:54
如果两个文件具有相同的元数据,并且tiles
表实际上是表而不是视图,则只需将其中一个表的数据附加到另一个表中:
/* open database1 as main database, then: */
ATTACH 'database2' AS db2;
INSERT INTO tiles SELECT * FROM db2.tiles;
发布于 2014-06-11 13:25:44
在我的例子中,@CL有一个错误。解决方案:
Error: cannot modify tiles because it is a view
所以我的数据库的模式是不同的:
> .schema
CREATE TABLE grid_key (
grid_id TEXT,
key_name TEXT
);
CREATE TABLE grid_utfgrid (
grid_id TEXT,
grid_utfgrid BLOB
);
CREATE TABLE images (
tile_data blob,
tile_id text
);
CREATE TABLE keymap (
key_name TEXT,
key_json TEXT
);
CREATE TABLE map (
zoom_level INTEGER,
tile_column INTEGER,
tile_row INTEGER,
tile_id TEXT,
grid_id TEXT
);
CREATE TABLE metadata (
name text,
value text
);
CREATE VIEW tiles AS
SELECT
map.zoom_level AS zoom_level,
map.tile_column AS tile_column,
map.tile_row AS tile_row,
images.tile_data AS tile_data
FROM map
JOIN images ON images.tile_id = map.tile_id;
CREATE VIEW grids AS
SELECT
map.zoom_level AS zoom_level,
map.tile_column AS tile_column,
map.tile_row AS tile_row,
grid_utfgrid.grid_utfgrid AS grid
FROM map
JOIN grid_utfgrid ON grid_utfgrid.grid_id = map.grid_id;
CREATE VIEW grid_data AS
SELECT
map.zoom_level AS zoom_level,
map.tile_column AS tile_column,
map.tile_row AS tile_row,
keymap.key_name AS key_name,
keymap.key_json AS key_json
FROM map
JOIN grid_key ON map.grid_id = grid_key.grid_id
JOIN keymap ON grid_key.key_name = keymap.key_name;
CREATE UNIQUE INDEX grid_key_lookup ON grid_key (grid_id, key_name);
CREATE UNIQUE INDEX grid_utfgrid_lookup ON grid_utfgrid (grid_id);
CREATE UNIQUE INDEX images_id ON images (tile_id);
CREATE UNIQUE INDEX keymap_lookup ON keymap (key_name);
CREATE UNIQUE INDEX map_index ON map (zoom_level, tile_column, tile_row);
CREATE UNIQUE INDEX name ON metadata (name);
因此,您可以这样调整解决方案:
首先按照您想要的方式更新元数据,例如:
sqlite> UPDATE metadata SET value = '7' WHERE name = 'minzoom';
然后插入瓷砖:
sqlite> INSERT OR REPLACE INTO images SELECT * from db2.images;
sqlite> INSERT OR REPLACE INTO map SELECT * from db2.map;
https://stackoverflow.com/questions/17746079
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