我无法正确编写正确的 SQL 查询
I cant write the correct SQL query correcty
我在数据库中有 2 个 table,我正在尝试根据某些条件过滤一些用户。
criterias table
+---------+----------------------+--------------------+----------------------+
| user_id | searching_friendship | searching_practice | conversation_subject |
+---------+----------------------+--------------------+----------------------+
| 31 | | | science |
| 26 | on | on | love |
| 32 | on | off | science |
| 34 | | | |
+---------+----------------------+--------------------+----------------------+
user table
+---------+------------+-----------------------------+---------------+-------------+---------------+--------------+------------------+
| user_id | user_name | user_email | user_password | user_gender | user_language | language_lvl | last_login |
+---------+------------+-----------------------------+---------------+-------------+---------------+--------------+------------------+
| 26 | Furkan | furkanakgun@windowslive.com | 123 | | Türkiye | basic | 16.06.2019 11:57 |
| 31 | sfsdf | asdfasdf@hotmail.com | 123 | Male | Afrikaans | basic | 09.06.2019 20:01 |
| 32 | denemeuser | xxx@hotmail.com | 123 | Male | Amharic | intermediate | 16.06.2019 11:57 |
| 33 | Smith | ssdf | 123 | male | ing | upper | NULL |
| 34 | luser | llll@hotmai.com | 123456 | Male | Afrikaans | basic | 16.06.2019 10:32 |
+---------+------------+-----------------------------+---------------+-------------+---------------+--------------+------------------+
我正在尝试匹配与 criterias.What 相同的用户
$userId=$_SESSION['userId'];
$sql="SELECT* FROM criterias WHERE user_id='$userId'";
$query=mysqli_query($conn,$sql);
while($result=mysqli_fetch_assoc($query)){
$friendshipCheck=$result['searching_friendship'];
$pracCheck=$result['searching_practice'];
$conversationSub=$result['conversation_subject'];
}
所以我正在获取当前标准信息,并且我正在尝试与其他用户匹配
SELECT* FROM users,criterias WHERE (users.user_id=criterias.user_id AND users.user_id!='$userId') AND criterias.searching_friendship='$friendshipCheck' OR criterias.searching_practice='$pracCheck'
但是没用。亲爱的朋友,你知道如何解决这个问题吗?
提供预期结果与您所获得的结果将有助于证明最佳解决方案。
在这里,我尝试根据对您尝试在代码中编写的内容的理解来编写查询:
;WITH CTE AS
(
SELECT C.*
FROM Criterias C
WHERE C.user_id = <user_id_input>
)
SELECT *
FROM User U
INNER JOIN Criterias C
ON U.user_id = C.user_id
WHERE EXISTS (
SELECT 1
FROM CTE
WHERE CTE.searching_friendship = C.searching_friendship
OR CTE.searching_practice = C.searching_practice
)
AND U.user_id != <input_user_id> ;
如果您只想获取与您的条件相同的用户,此查询应该有效
SELECT *
FROM users AS u
INNER JOIN criterias AS c ON u.user_id = c.user_id
WHERE u.user_id != '$userId'
AND c.searching_friendship = '$friendshipCheck'
AND c.searching_practice = '$pracCheck'
AND c.conversation_subject = '$conversation_subject'
我在数据库中有 2 个 table,我正在尝试根据某些条件过滤一些用户。
criterias table
+---------+----------------------+--------------------+----------------------+
| user_id | searching_friendship | searching_practice | conversation_subject |
+---------+----------------------+--------------------+----------------------+
| 31 | | | science |
| 26 | on | on | love |
| 32 | on | off | science |
| 34 | | | |
+---------+----------------------+--------------------+----------------------+
user table
+---------+------------+-----------------------------+---------------+-------------+---------------+--------------+------------------+
| user_id | user_name | user_email | user_password | user_gender | user_language | language_lvl | last_login |
+---------+------------+-----------------------------+---------------+-------------+---------------+--------------+------------------+
| 26 | Furkan | furkanakgun@windowslive.com | 123 | | Türkiye | basic | 16.06.2019 11:57 |
| 31 | sfsdf | asdfasdf@hotmail.com | 123 | Male | Afrikaans | basic | 09.06.2019 20:01 |
| 32 | denemeuser | xxx@hotmail.com | 123 | Male | Amharic | intermediate | 16.06.2019 11:57 |
| 33 | Smith | ssdf | 123 | male | ing | upper | NULL |
| 34 | luser | llll@hotmai.com | 123456 | Male | Afrikaans | basic | 16.06.2019 10:32 |
+---------+------------+-----------------------------+---------------+-------------+---------------+--------------+------------------+
我正在尝试匹配与 criterias.What 相同的用户
$userId=$_SESSION['userId'];
$sql="SELECT* FROM criterias WHERE user_id='$userId'";
$query=mysqli_query($conn,$sql);
while($result=mysqli_fetch_assoc($query)){
$friendshipCheck=$result['searching_friendship'];
$pracCheck=$result['searching_practice'];
$conversationSub=$result['conversation_subject'];
}
所以我正在获取当前标准信息,并且我正在尝试与其他用户匹配
SELECT* FROM users,criterias WHERE (users.user_id=criterias.user_id AND users.user_id!='$userId') AND criterias.searching_friendship='$friendshipCheck' OR criterias.searching_practice='$pracCheck'
但是没用。亲爱的朋友,你知道如何解决这个问题吗?
提供预期结果与您所获得的结果将有助于证明最佳解决方案。 在这里,我尝试根据对您尝试在代码中编写的内容的理解来编写查询:
;WITH CTE AS
(
SELECT C.*
FROM Criterias C
WHERE C.user_id = <user_id_input>
)
SELECT *
FROM User U
INNER JOIN Criterias C
ON U.user_id = C.user_id
WHERE EXISTS (
SELECT 1
FROM CTE
WHERE CTE.searching_friendship = C.searching_friendship
OR CTE.searching_practice = C.searching_practice
)
AND U.user_id != <input_user_id> ;
如果您只想获取与您的条件相同的用户,此查询应该有效
SELECT *
FROM users AS u
INNER JOIN criterias AS c ON u.user_id = c.user_id
WHERE u.user_id != '$userId'
AND c.searching_friendship = '$friendshipCheck'
AND c.searching_practice = '$pracCheck'
AND c.conversation_subject = '$conversation_subject'