同时查询“照片”表和“书签”表，以便了解带有书签的照片

Question

How can I query for all records in a table called photos, and know which of the resulting photos have been bookmarked by the current user using a single query?

Here are my tables:

--
-- Table structure for table `photos`
--

CREATE TABLE IF NOT EXISTS `photos` (
  `id` int(11) NOT NULL auto_increment,
  `author` bigint(20) NOT NULL COMMENT 'The user''s Facebook ID.',
  `filename` varchar(255) NOT NULL,
  `thumbnail` varchar(255) NOT NULL,
  `post_date` timestamp NOT NULL default CURRENT_TIMESTAMP,
  `description` varchar(140) NOT NULL,
  `finalist` tinyint(4) NOT NULL DEFAULT '0',
  CONSTRAINT user_must_exist FOREIGN KEY (author)
    REFERENCES users(facebook_id)
    ON DELETE CASCADE,
  PRIMARY KEY  (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=1 ;

-- --------------------------------------------------------

--
-- Table structure for table `bookmarks`
--

CREATE TABLE IF NOT EXISTS `bookmarks` (
  `facebook_id` bigint(20) NOT NULL COMMENT 'The author''s Facebook ID.',
  `photo_id` int(11) NOT NULL,
  CONSTRAINT photo_should_exist FOREIGN KEY (photo_id)
    REFERENCES photos(id)
    ON DELETE CASCADE,
  CONSTRAINT user_should_exist FOREIGN KEY (facebook_id)
    REFERENCES users(facebook_id)
    ON DELETE CASCADE,
  UNIQUE KEY `no_duplicates` (`facebook_id`,`photo_id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 COMMENT='The user''s favourite photos.';

I would imagine this query would look something like the following:

SELECT 
    photos.*, 
    bookmarks.photo_id AS bookmark 
FROM photos
LEFT JOIN bookmarks 
    ON bookmarks.photo_id = photos.id 
    AND photos.author = 123456789

However this doesn't work and I receive the following MySQL error:

Unknown column 'photos.id' in 'on clause'

keyur's code worked for me after a minor typo fix which Barmar pointed out.

SELECT photos . * , bookmarks.photo_id AS bookmark
FROM photos
LEFT JOIN bookmarks ON photos.id = bookmarks.photo_id
AND bookmarks.facebook_id = 123456789

Thank you.

Editted for typo.

Editted for second typo.

Answer 1

Here is a sample working code on SQL Fiddle . As pointed out on a comment, you do not have a bookmarks.id column on you bookmarks table and photo.id indeed exists on the photos table. Check again the error you are receiving as seems something is not correct.

Answer 2

you query should be like this.

SELECT 
    photos.*, 
    bookmarks.id AS bookmark 
FROM photos
LEFT JOIN bookmarks 
    ON photos.id = bookmarks.photo_id 
    AND bookmarks.facebook_id = 123456789

according your table structure photos.facebook_id is not exist; it is in bookmarks table