Select user having qualifying data on multiple rows in the wp_usermeta table(选择在 wp_usermeta 表中的多行上具有合格数据的用户)
问题描述
我正在尝试查找具有所有四个限定值的 user_id
- 每个值都位于数据库表的不同行中.
I am trying to find the user_id
which has all four qualifying values -- each in a different row of the database table.
我要查询的表是 wp_usermeta
:
The table that I am querying is wp_usermeta
:
Field Type Null Key Default Extra
---------------------------------------------------------------------------
umeta_id bigint(20) unsigned PRI auto_increment
user_id bigint(20) unsigned IND 0
meta_key varchar(255) Yes IND NULL
meta_value longtext Yes NULL
我写了一个 MySQL 查询,但它似乎没有工作,因为结果是空的.
I have written a MySQL query but it doesn't seem to be working, because the result is empty.
$result = mysql_query(
"SELECT user_id
FROM wp_usermeta
WHERE
(meta_key = 'first_name' AND meta_value = '$us_name') AND
(meta_key = 'yearofpassing' AND meta_value = '$us_yearselect') AND
(meta_key = 'u_city' AND meta_value = '$us_reg') AND
(meta_key = 'us_course' AND meta_value = '$us_course')"
);
如何返回与所有这四行相关的 user_id
?
How do I return the user_id
that relates to all four of these rows?
推荐答案
我会使用这个查询:
SELECT
user_id
FROM
wp_usermeta
WHERE
(meta_key = 'first_name' AND meta_value = '$us_name') OR
(meta_key = 'yearofpassing' AND meta_value = '$us_yearselect') OR
(meta_key = 'u_city' AND meta_value = '$us_reg') OR
(meta_key = 'us_course' AND meta_value = '$us_course')
GROUP BY
user_id
HAVING
COUNT(DISTINCT meta_key)=4
这将选择满足所有四个条件的所有 user_id
.
this will select all user_id
that meets all four conditions.
这篇关于选择在 wp_usermeta 表中的多行上具有合格数据的用户的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持编程学习网!
本文标题为:选择在 wp_usermeta 表中的多行上具有合格数据的用户


基础教程推荐
- YouTube API v3 点赞视频,但计数器不增加 2022-01-01
- PHP 类:全局变量作为类中的属性 2021-01-01
- 如何替换eregi() 2022-01-01
- PHP PDO MySQL 查询 LIKE ->多个关键词 2021-01-01
- 如何在 Laravel 5.3 注册中添加动态下拉列表列? 2021-01-01
- 有什么方法可以用编码 UTF-8 而不是 Unicode 返回 PHP`json_encode`? 2021-01-01
- Cron Jobs 调用带有变量的 PHP 脚本 2022-01-01
- 在PHP中根据W3C规范Unicode 2022-01-01
- 学说 dbal querybuilder 作为准备好的语句 2022-01-01
- 如何在 Laravel 中使用 React Router? 2022-01-01