在MySQL中的两个表中选择查询

Question

I am running this query

select * from user_meta JOIN user ON user_meta.userid=user.userid where user_meta.userid=9

But this is not resulting what i wanted, its returning multiple lines.

My one table looks like this name user_meta :

umeta_id     userid      meta_key       meta_value
 1              9         mobile        123324
 2              9         address       some address
 3              9         city          some city
 4              9         country       some country
 5              9         occupation    some details
 6              9         website       someurl
 7              9         mobile        123324
 8              9         address       some address
 9              9         city          some city
 10             10        country       some country
 11             10        occupation    some details
 12             10        website       someurl

Another table looks like this name user :

userid          username      fullname      email              role
  9             someuser       john Doe    123324@gmail.com    admin

How can i make it select the query so that all the values related to userid 9 can be fetch from both the tables and make it look like this

Desired output:

userid          username      fullname      email              role      Mobile     address     city     country    occupation     website
  9             someuser       john Doe    123324@gmail.com    admin    123123     someaddres    Somecity    somecountry    some details    someurl

Thank you! (In Advance!)

Answer 1

SELECT 
    *
FROM
    user
        INNER JOIN
    user_meta ON user_meta.userid = user.userid
WHERE
    user.userid = 9

The above answer was with respect to your initial requirement. But as per your new requirement it can not be done with simple query, it needs to be dynamic and here I have created the demo for you, you can use this

http://sqlfiddle.com/#!2/39861/2

Answer 2

SELECT * FROM user_meta um, user u WHERE um.userid = u.userid

Answer 3

select 
    user_meta.umeta_id,
    user_meta.userid,
    user_meta.meta_key,
    user_meta.meta_value,
    user.userid,
    user.username,
    user.fullname,
    user.email,
    user.role
from
    user_meta
        LEFT JOIN
    user ON user_meta.userid = user.userid
where
    user_meta.userid = 9

Answer 4

simple as this:

select * from user_meta, user where user_meta.userid=9 
and user_meta.userid=user.userid

Answer 5

The only real way to reduce it to a single line is using an aggregate function

for example, grouping the details in a field

SELECT user.userid, user.username, user.fullname, user.email, user.role, GROUP_CONCAT(CONCAT_WS('-', user_meta.umeta_id, user_meta.meta_key, user_meta.meta_value))
FROM user_meta 
INNER JOIN user ON user_meta.userid=user.userid 
WHERE user_meta.userid=9
GROUP BY user.userid, user.username, user.fullname, user.email, user.role

Answer 6

If you want only user_meta table all values means use this.

select meta.* from user_meta AS meta JOIN user AS us ON meta.userid=us.userid where meta.userid=9

If you want only user table all values means use this.

select us.* from user_meta AS meta JOIN user AS us ON meta.userid=us.userid where meta.userid=9

If you want both means mention like this

select 
    meta.umeta_id,
    meta.userid, // only one userid is enough
    meta.meta_key,
    meta.meta_value,
    us.username,
    us.fullname,
    us.email,
    us.role from user_meta JOIN
    user ON meta.userid = us.userid

Answer 7

SELECT * FROM USERS U INNER JOIN Meta_Users M ON U.userid = M.userid
WHERE U.userid = 9

Answer 8

To give you the output you want (but this isn't flexible, so won't cope without changes if you add more details that you want to output):-

SELECT user.userid, 
        user.username, 
        user.fullname, 
        user.email, 
        user.role, 
        mobile_meta.meta_value AS `Mobile`, 
        address_metameta_value AS `Address`, 
        city_metameta_value AS `City`, 
        country_metameta_value AS `Country`, 
        occupation_metameta_value AS `Occupation`, 
        website_metameta_value AS `Website`
FROM user
LEFT OUTER JOIN user_meta AS mobile_meta ON mobile_meta.userid=user.userid AND mobile_meta.meta_key = 'mobile'
LEFT OUTER JOIN user_meta AS address_meta ON address_meta.userid=user.userid AND address_meta.meta_key = 'address'
LEFT OUTER JOIN user_meta AS city_meta ON city_meta.userid=user.userid AND city_meta.meta_key = 'city'
LEFT OUTER JOIN user_meta AS country_meta ON country_meta.userid=user.userid AND country_meta.meta_key = 'country'
LEFT OUTER JOIN user_meta AS occupation_meta ON occupation_meta.userid=user.userid AND occupation_meta.meta_key = 'occupation'
LEFT OUTER JOIN user_meta AS website_meta ON website_meta.userid=user.userid AND website_meta.meta_key = 'website'
WHERE user.userid=9