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SQL根据初始表的ID联接另一个表

[英]SQL Joining another table based on the ID of the initial table

I really just can't seem to grasb the sytanx of SQL Joins, etc ... so I'm needing some help with this (which I think is quite simple) 我真的真的似乎无法掌握SQL Joins的共生关系,等等...所以我需要一些帮助(我认为这很简单)

I'm querying bid_tag as follows: 我正在查询bid_tag ,如下所示:

SELECT paid_date, term, pet_id FROM bid_tag WHERE active = 1

And I need to use the pet_id to then grab some information from another table wp_postmeta , where the table is actually in a meta_key meta_value structure (WordPress) ... 我需要使用pet_id从另一个表wp_postmeta获取一些信息,该表实际上位于meta_key meta_value结构(WordPress)中。

So I need to grab the meta_value of meta_key "bid_name", for example... amongst other values. 因此,我需要获取meta_valuemeta_key ”的meta_value,例如...以及其他值。

TABLE
id | meta_key | meta_value
1    bid_name    Max

That ID is the same ID that I need to connect the initial table to... ID是一样的ID ,我需要初始表连接到...

I really appreciate it! 我真的很感激!

Join on both the id and the meta_key to select your desired value. 加入idmeta_key来选择所需的值。 If you want values that go with other keys in the meta table as well, one way is to join multiple times: 如果您还希望值与元表中的其他键一起使用,则一种方法是多次联接:

SELECT bt.paid_date, bt.term, bt.pet_id,
    bn.meta_value as bidname,
    bo.meta_value as bidother
FROM bid_tag bt
INNER JOIN wp_postmeta bn on bn.id = bt.pet_id and bn.meta_key = 'bid_name'
INNER JOIN wp_postmeta bo on bo.id = bt.pet_id and bo.meta_key = 'bid_other'
WHERE bt.active = 1

If the values won't necessarily be present in the meta table, use a LEFT OUTER JOIN instead and it will return null for whatever is missing. 如果值不一定在元表中出现,请改用LEFT OUTER JOIN ,如果缺少任何内容,它将返回null

Here's how your query would go: 这是查询的方式:

SELECT B.meta_key, B.meta_value, A.pet_id
FROM bid_tag A
JOIN wp_postmeta B
ON B.id = A.pet_id;

there're some ways to use join: 有一些使用联接的方法:
1)use inner join or equally just join : 1)使用inner join或相等地只是join

select 
  --some columns
from bid_tag
join wp_postmeta on
  bid_tag.pet_id = wp_postmeta.id
where
  wp_postmeta.meta_key = 'bid_name'

The result will have only that records which pet_id exists in wp_postmeta . 其结果将只具有哪些记录pet_id存在于wp_postmeta The other will be omitted. 另一个将被省略。 And if wp_postmeta have a couple of records with the same id (and exists pet_id equal this id ) then you'll get a couple records with this pet_id ( BUT with different meta_key and meta_value ). 如果wp_postmeta具有相同几个记录id (和存在pet_id等于这个id ),那么你会得到这个pet_id一对夫妇记录( 具有不同的meta_keymeta_value )。
2)use left join : 2)使用left join

select 
  --some columns
from bid_tag
left join wp_postmeta on
  bid_tag.pet_id = wp_postmeta.id
where
  wp_postmeta.meta_key = 'bid_name'

In this case behavior will be the same except one thing - even if there aren't any id for a pet_id the result will contain record from wp_postmeta with this pet_id . 在这种情况下,除了一件事外,行为将是相同的-即使pet_id没有任何id ,结果也将包含来自wp_postmeta带有此pet_id记录。 But meta_key and meta_value in this case will be null . 但是在这种情况下, meta_keymeta_value将为null

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