mysql query to establish booked walks still available

Question

I need some support with a query. Basically I have a database for a Walks company. The tables are as follows, Guides - Walks - Participants - Programme - Booking and GuideWalk, which is a linking table.

I need to do a query that would give output data that would tell me for each walk, how many places there are, how many have booked and how many are available.

Any help would be appreciated.

The tables are as follows

CREATE TABLE `Participant` (
  `part_id` int(11) NOT NULL AUTO_INCREMENT,
  `part_sname` varchar(20) NOT NULL,
  `part_fname` varchar(15) NOT NULL,
  `part_phone` varchar(20) DEFAULT NULL,
  `part_email` varchar(20) DEFAULT NULL,
  `part_status` varchar(45) NOT NULL,
  PRIMARY KEY (`part_id`)
) ENGINE=InnoDB AUTO_INCREMENT=73 DEFAULT CHARSET=latin1;

CREATE TABLE `Booking` (
  `bookprog_id` int(11) NOT NULL AUTO_INCREMENT,
  `part_id` int(11) NOT NULL,
  PRIMARY KEY (`bookprog_id`,`part_id`),
  KEY `part_id_idx` (`part_id`),
  CONSTRAINT `bookprog` FOREIGN KEY (`bookprog_id`) REFERENCES `Programme` (`prog_id`) ON DELETE NO ACTION ON UPDATE NO ACTION,
  CONSTRAINT `part_id` FOREIGN KEY (`part_id`) REFERENCES `Participant` (`part_id`) ON DELETE NO ACTION ON UPDATE NO ACTION
) ENGINE=InnoDB AUTO_INCREMENT=10 DEFAULT CHARSET=latin1;

CREATE TABLE `Guide` (
  `guide_id` int(11) NOT NULL AUTO_INCREMENT,
  `guide_fname` varchar(20) NOT NULL,
  `guide_sname` varchar(20) NOT NULL,
  `guide_address1` varchar(20) NOT NULL,
  `guide_address2` varchar(20) DEFAULT NULL,
  `guide_town` varchar(20) NOT NULL,
  `guide_pcode` varchar(15) NOT NULL,
  `guide_email` varchar(20) NOT NULL,
  `guide_mphone` varchar(15) NOT NULL,
  PRIMARY KEY (`guide_id`)
) ENGINE=InnoDB AUTO_INCREMENT=274 DEFAULT CHARSET=latin1;

CREATE TABLE `GuideWalk` (
  `guide_id` int(11) NOT NULL,
  `prog_id` int(11) NOT NULL,
  PRIMARY KEY (`guide_id`,`prog_id`),
  KEY `prog_id_idx` (`prog_id`),
  CONSTRAINT `prog_id` FOREIGN KEY (`prog_id`) REFERENCES `Programme` (`prog_id`) ON DELETE NO ACTION ON UPDATE NO ACTION,
  CONSTRAINT `guide_id` FOREIGN KEY (`guide_id`) REFERENCES `Guide` (`guide_id`) ON DELETE NO ACTION ON UPDATE NO ACTION
) ENGINE=InnoDB DEFAULT CHARSET=latin1;

CREATE TABLE `Programme` (
  `prog_id` int(11) NOT NULL,
  `pwalk_id` int(11) NOT NULL,
  `prog_date` date NOT NULL,
  `prog_stime` varchar(9) NOT NULL,
  `prog_max` int(11) NOT NULL,
  PRIMARY KEY (`prog_id`),
  KEY `pwalk_id_idx` (`pwalk_id`),
  CONSTRAINT `pwalk_id` FOREIGN KEY (`pwalk_id`) REFERENCES `Walk` (`walk_id`) ON DELETE NO ACTION ON UPDATE NO ACTION
) ENGINE=InnoDB DEFAULT CHARSET=latin1;

CREATE TABLE `Walk` (
  `walk_id` int(11) NOT NULL AUTO_INCREMENT,
  `walk_title` varchar(15) NOT NULL,
  `walk_dur` int(11) NOT NULL,
  `walk_spoint` varchar(20) NOT NULL,
  `walk_fpoint` varchar(20) NOT NULL,
  `walk_desc` text NOT NULL,
  `walkleader_1` varchar(20) NOT NULL,
  `walkleader_2` varchar(20) DEFAULT NULL,
  PRIMARY KEY (`walk_id`)
) ENGINE=InnoDB AUTO_INCREMENT=8 DEFAULT CHARSET=latin1;

I have managed to output every day and how many places are available with this query

SELECT prog_max AS 'Places', prog_date AS 'Day'
FROM  Programme
WHERE prog_date BETWEEN '2014-07-20' AND '2014-08-03'

I really haven't a clue on how to progress from here

Solution

SELECT walk_title, DAYNAME(prog_date) AS 'Day', DATE_FORMAT(prog_date, '%D %M %Y') AS 'Date', 
prog_stime AS 'Start Time', walk_spoint AS 'Meet At', walkleader_1 AS 'Guide 1', 
walkleader_2 AS 'Guide 2', Booking.bookprog_id AS 'Booked', ABS(Programme.prog_id -Programme.prog_max) AS'Available', 
prog_max AS 'Places'
FROM Walk, Programme
INNER JOIN Booking
ON bookprog_id=prog_id
LIMIT 10

Answer 1

SELECT walk_title，DAYNAME（prog_date）AS'Day'，DATE_FORMAT（prog_date，'％D％M％Y'）AS'Date'，prog_stime AS'Start Time'，walk_spoint AS'Meet At'，walkleader_1 AS'Guide 1' ，walkleader_2 AS'指南2'，Booking.bookprog_id AS'已预订'，ABS（Programme.prog_id -Programme.prog_max）AS'Available'，prog_max AS'Places'from Walk，计划内联接预订bookprog_id = prog_id LIMIT 10