relational-division

SQL how to search a many to many relationship

由 こ雲淡風輕ζ 提交于 2019-11-28 04:29:34
问题 I have a database with two main tables notes and labels . They have a many-to-many relationship (similar to how stackoverflow.com has questions with labels). What I am wondering is how can I search for a note using multiple labels using SQL? For example if I have a note "test" with three labels "one", "two", and "three" and I have a second note "test2" with labels "one" and "two" what is the SQL query that will find all the notes that are associated with labels "one" and "two"? 回答1: To obtain

Using same column multiple times in WHERE clause

由 廉价感情. 提交于 2019-11-27 14:56:04
问题 I have a following table structure. USERS PROPERTY_VALUE PROPERTY_NAME USER_PROPERTY_MAP I am trying to retrieve user/s from the users table who have matching properties in property_value table. A single user can have multiple properties. The example data here has 2 properties for user '1', but there can be more than 2. I want to use all those user properties in the WHERE clause. This query works if user has a single property but it fails for more than 1 properties: SELECT * FROM users u

PostgreSQL where all in array

由 旧巷老猫 提交于 2019-11-27 08:36:40
What is the easiest and fastest way to achieve a clause where all elements in an array must be matched - not only one when using IN ? After all it should behave like mongodb's $all . Thinking about group conversations where conversation_users is a join table between conversation_id and user_id I have something like this in mind: WHERE (conversations_users.user_id ALL IN (1,2)) UPDATE 16.07.12 Adding more info about schema and case: The join-table is rather simple: Table "public.conversations_users" Column | Type | Modifiers | Storage | Description -----------------+---------+-----------+------

SQL: Get Products from a category but also must be in another set of categories

由 北慕城南 提交于 2019-11-27 07:51:01
问题 I am currently stuck in a situation. The scenario is this. I have products who may be associated with multiple categories. The data structure is shown below: Products Table: product_id name 1 Lemon 2 Kiwis 3 Cheese Product to Categories Table product_id category_id 1 1 1 2 1 3 2 1 2 3 3 2 3 4 Category Table (not required in query however adding it here to help visualize what is happening) category_id name 1 Fruit 2 Yellow 3 Round 4 Dairy What I'm struggling with here is that originally I want

SQL where joined set must contain all values but may contain more

由 ℡╲_俬逩灬. 提交于 2019-11-27 05:33:00
I have three tables offers , sports and the join table offers_sports . class Offer < ActiveRecord::Base has_and_belongs_to_many :sports end class Sport < ActiveRecord::Base has_and_belongs_to_many :offers end I want to select offers that include a given array of sport names. They must contain all of the sports but may have more. Lets say I have these three offers: light: - "Yoga" - "Bodyboarding" medium: - "Yoga" - "Bodyboarding" - "Surfing" all: - "Yoga" - "Bodyboarding" - "Surfing" - "Parasailing" - "Skydiving" Given the array ["Bodyboarding", "Surfing"] I would want to get medium and all

jsonb query with nested objects in an array

由 百般思念 提交于 2019-11-27 02:20:44
问题 I'm using PostgreSQL 9.4 with a table teams containing a jsonb column named json . I am looking for a query where I can get all teams which have the Players 3 , 4 and 7 in their array of players. The table contains two rows with the following json data: First row: { "id": 1, "name": "foobar", "members": { "coach": { "id": 1, "name": "A dude" }, "players": [ { "id": 2, "name": "B dude" }, { "id": 3, "name": "C dude" }, { "id": 4, "name": "D dude" }, { "id": 6, "name": "F dude" }, { "id": 7,

PostgreSQL where all in array

由 会有一股神秘感。 提交于 2019-11-26 12:44:37
问题 What is the easiest and fastest way to achieve a clause where all elements in an array must be matched - not only one when using IN ? After all it should behave like mongodb\'s $all. Thinking about group conversations where conversation_users is a join table between conversation_id and user_id I have something like this in mind: WHERE (conversations_users.user_id ALL IN (1,2)) UPDATE 16.07.12 Adding more info about schema and case: The join-table is rather simple: Table \"public.conversations

Select group of rows that match all items in a list

由 ▼魔方 西西 提交于 2019-11-26 12:22:20
Assume I have two tables: cars – list of cars carname | modelnumber | ... passedtest – contains every test that a car passed: id | carname | testtype | date | ... 1 | carA | A | 2000 | 2 | carB | C | 2000 | 3 | carC | D | 2001 | 4 | carA | C | 2002 | Now, how can I select a car from the passedtest table that passed all tests (A, B, C, D)? I tried the IN statement but it also matches cars that pass even one test. I am looking for a statement to match all values in a list across all rows. MarcinJuraszek How about this? SELECT carname FROM PassedTest GROUP BY carname HAVING COUNT(DISTINCT

SQL where joined set must contain all values but may contain more

由 本秂侑毒 提交于 2019-11-26 11:37:26
问题 I have three tables offers , sports and the join table offers_sports . class Offer < ActiveRecord::Base has_and_belongs_to_many :sports end class Sport < ActiveRecord::Base has_and_belongs_to_many :offers end I want to select offers that include a given array of sport names. They must contain all of the sports but may have more. Lets say I have these three offers: light: - \"Yoga\" - \"Bodyboarding\" medium: - \"Yoga\" - \"Bodyboarding\" - \"Surfing\" all: - \"Yoga\" - \"Bodyboarding\" - \

How to filter SQL results in a has-many-through relation

由 守給你的承諾、 提交于 2019-11-25 23:56:36
问题 Assuming I have the tables student , club , and student_club : student { id name } club { id name } student_club { student_id club_id } I want to know how to find all students in both the soccer (30) and baseball (50) club. While this query doesn\'t work, it\'s the closest thing I have so far: SELECT student.* FROM student INNER JOIN student_club sc ON student.id = sc.student_id LEFT JOIN club c ON c.id = sc.club_id WHERE c.id = 30 AND c.id = 50 回答1: I was curious. And as we all know,
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