Mysql退出时出现了错误提示？

Question

我有两个看起来像这样的表

contacts
-------------------------------------------------------
| id | fname | lname | email | address | uid | uniqid |
-------------------------------------------------------

numbers
---------------------
| id | pnumber | cid |
---------------------

这是我的Sql代码

SELECT * FROM contacts LEFT JOIN numbers ON contacts.uniqid = numbers.cid 

uniqid与cid相同，我的问题是当我尝试左连接这两个表时会返回此 注意：对不起，必须使用图片，因为当我尝试使用文字时，它会损坏

我想在同一列中获取所有pnumber，而无需重复。有什么办法可以做到这一点，还是我应该使用php？

预期结果

id, fname, lname, email, address, uid, uniqid, id, pnumber,pnumber,pnumber, cid

Answer 1

将@Dao public interface MyDao { @Query("SELECT * FROM movie_table ORDER BY genre") public ListMovies[] findMovieByGenre(String Genre); } 与group_concat()一起使用

distinct

Answer 2

您可以使用GROUP BY语句和GROUP_CONCAT函数

SELECT 
  contacts.* 
  GROUP_CONCAT(numbers.pnumber)
FROM contacts 
LEFT JOIN numbers ON contacts.uniqid = numbers.cid 
GROUP BY contacts.id, contacts.fname, contacts.lname, contacts.email, contacts.address, contacts.uid, contacts.uniqid

Answer 3

您可以使用DISTINCT过滤重复项：

SELECT DISTINCT pnumber FROM contacts LEFT JOIN numbers ON contacts.uniqid = numbers.cid 

Answer 4

尝试-

SELECT c.*, GROUP_CONCAT(n. pnumber SEPARATOR ', ') FROM contacts as c LEFT JOIN numbers as n ON c.uniqid = n.cid 
    GROUP BY c.id, c.fname, c.lname, c.email, c.address, c.uid, c.uniqid