Question

我很难找到如何在LIKE％var％startment中执行Python变量：％s或（{url}）的答案。

百分号会导致我出现问题...语法错误

以下是我正在尝试查询的SQLite数据库的示例转储：

PRAGMA foreign_keys=OFF;
BEGIN TRANSACTION;
CREATE TABLE IF NOT EXISTS "user_cats" (
    `category`  TEXT,
    `uploader`  TEXT,
    `folder`    TEXT,
    `playlists` TEXT,
    `playlist_image`    TEXT,
    `save_directory`    TEXT
);
INSERT INTO user_cats VALUES('Comics','ComicsExplained','DC Rebirth','[''New to DC Comics? Start here!'']',NULL,'%USERPROFILE%\Videos\Online Videos\Comics\ComicsExplained\');
INSERT INTO user_cats VALUES('Comics','Comicstorian',NULL,NULL,NULL,NULL);
INSERT INTO user_cats VALUES('Video Games','IGN','Daily Fix','[''Daily Fix'']',NULL,'%USERPROFILE%\Videos\Online Videos\Video Games\IGN\Daily Fix\');
INSERT INTO user_cats VALUES('Comics','Marvel Entertainment','Marvel Top 10','[''Marvel Top 10'']',NULL,'%USERPROFILE%\Videos\Online Videos\Comics\Marvel Entertainment\');
INSERT INTO user_cats VALUES('','ScrewAttack!',NULL,NULL,NULL,NULL);
COMMIT;

我想做的是使用youtube-dl检索上传者和播放列表标题的公式（我可能会变成一个对象方法）。

然后使用数据库将文件路由到不同的保存位置，并且 - 可能保存其他信息。但是，我遇到SELECT语句和百分号的问题。错误取决于我如何编辑语句 - 各种语法或意外的字符等。（它可能没有帮助键入我得到的每一个错误 - 这是变量赋值的问题 - 而不是语句。

如果我使用简单的查询SQLite（在Python之外）...

SELECT save_directory FROM user_cats WHERE uploader='IGN' AND playlists LIKE '%ail%';

宾果。的工作原理。

def save_dir (uploader, playlist):
    query = "SELECT save_directory FROM user_cats WHERE uploader=({ul}) AND playlists LIKE '%%({pl})%%'.format(ul = uploader, pl = playlist)".format(ul=uploader, pl=playlist))

    c.execute(query)
    all_rows = c.fetchall()
    print('1):', all_rows)

Answer 1

你可以这样格式化：

>>> uploader, playlist = 'someUploader', 'somePlaylist'
>>> query ='SELECT save_directory FROM user_cats WHERE uploader={0:s} AND playlists LIKE %{1:s}'
>>> query.format(uploader, playlist)
'SELECT save_directory FROM user_cats WHERE uploader=someUploader AND playlists LIKE %somePlaylist'

甚至这样：

>>> query ='SELECT save_directory FROM user_cats WHERE uploader=%(ul)s AND playlists LIKE %(pl)s'
>>> query % {'ul': 'someUploader', 'pl': '%somePlaylist%'}
'SELECT save_directory FROM user_cats WHERE uploader=someUploader AND playlists LIKE %somePlaylist%'

Answer 2

它并不漂亮，但我最终通过将字符串添加到一起来查询工作，而不是尝试替换值。

def save_dir（上传者，播放列表）：

save_dir('IGN','Daily Fix')

1): [('%USERPROFILE%\\Videos\\Online Videos\\Video Games\\IGN\\Daily Fix',)] [Finished in 0.116s]

输出......

{{1}}

最后-___-

如何在SQL LIKE语句中使用Python变量？

2 个答案: