Проблемы с внутренним соединением SQL

Question

У меня есть установка WordPress, и я создаю настраиваемый плагин. В моем плагине хранится хорошая часть usermeta. Я использую внутреннее соединение для объединения данных из таблицы «users» и «usermeta», после чего я возвращаю результат обратно на свой PHP-скрипт. Вот что мой запрос выглядит следующим образом:

select 
# Users table stuff 
users.ID, 
users.user_email, 
users.display_name, 

# Meta stuff 

first_name.meta_value as first_name, 
last_name.meta_value as last_name, 
phone_number.meta_value as phone_number, 
country.meta_value as country, 
years_of_experience.meta_value as years_of_experience, 
highest_degree_obtained.meta_value as highest_degree_obtained, 
availability_for_work.meta_value as availability_for_work, 
english_proficiency.meta_value as english_proficiency, 
disciplines.meta_value as disciplines, 
profile_picture.meta_value as profile_picture, 
resume.meta_value as resume, 
description.meta_value as description, 
hourly_rate.meta_value as hourly_rate, 
satisfaction_rating.meta_value as satisfaction_rating, 
invited_projects.meta_value as invited_projects, 
completed_project_count.meta_value as completed_project_count 

# The table we're selecting from 

from tsd_retro_users as users 

# Join in our usermeta table for each individual meta value 

inner join tsd_retro_usermeta first_name on users.ID = first_name.user_id 
inner join tsd_retro_usermeta last_name on users.ID = last_name.user_id 
inner join tsd_retro_usermeta phone_number on users.ID = phone_number.user_id 
inner join tsd_retro_usermeta country on users.ID = country.user_id 
inner join tsd_retro_usermeta years_of_experience on users.ID = years_of_experience.user_id 
inner join tsd_retro_usermeta highest_degree_obtained on users.ID = highest_degree_obtained.user_id 
inner join tsd_retro_usermeta availability_for_work on users.ID = availability_for_work.user_id 
inner join tsd_retro_usermeta english_proficiency on users.ID = english_proficiency.user_id 
inner join tsd_retro_usermeta disciplines on users.ID = disciplines.user_id 
inner join tsd_retro_usermeta profile_picture on users.ID = profile_picture.user_id 
inner join tsd_retro_usermeta resume on users.ID = resume.user_id 
inner join tsd_retro_usermeta description on users.ID = description.user_id 
inner join tsd_retro_usermeta hourly_rate on users.ID = hourly_rate.user_id 
inner join tsd_retro_usermeta satisfaction_rating on users.ID = satisfaction_rating.user_id 
inner join tsd_retro_usermeta invited_projects on users.ID = invited_projects.user_id 
inner join tsd_retro_usermeta completed_project_count on users.ID = completed_project_count.user_id 


# Define our select stipulations 

where 
(users.ID = 20) 
and 
(first_name.meta_key = 'first_name') 
and 
(last_name.meta_key = 'last_name') 
and 
(phone_number.meta_key = 'phone_number') 
and 
(country.meta_key = 'country') 
and 
(years_of_experience.meta_key = 'years_of_experience') 
and 
(highest_degree_obtained.meta_key = 'highest_degree_obtained') 
and 
(availability_for_work.meta_key = 'availability_for_work') 
and 
(english_proficiency.meta_key = 'english_proficiency') 
and 
(disciplines.meta_key = 'disciplines') 
and 
(profile_picture.meta_key = 'profile_picture') 
and 
(resume.meta_key = 'resume') 
and 
(description.meta_key = 'description') 
and 
(hourly_rate.meta_key = 'hourly_rate') 
and 
(satisfaction_rating.meta_key = 'satisfaction_rating') 
and 
(invited_projects.meta_key = 'invited_projects') 
and 
(completed_project_count.meta_key = 'completed_project_count') 

Как вы можете видеть, я внутреннее соединение таблицы usermeta для каждого значения, что я пытаюсь получить из базы данных. Кажется, что все работает нормально, но после некоторого количества внутренних объединений запрос, похоже, замедляется при сканировании. Прямо сейчас, указанный выше запрос занимает около секунды в среднем, и у меня только около двадцати пользователей в моей таблице пользователей.

Мой вопрос: каковы производственные последствия внутренних соединений? Есть ли лучший способ выполнить вышеупомянутый запрос и добиться того же результата?

Answer 1

Попробуйте так:

select 
# Users table stuff 
users.ID, 
users.user_email, 
users.display_name, 

# Meta stuff 

MAX(CASE WHEN meta_table.meta_key='first_name' THEN meta_table.first_name ELSE NULL END) as first_name, 
MAX(CASE WHEN meta_table.meta_key='last_name' THEN meta_table.last_name ELSE NULL END) as last_name, 
MAX(CASE WHEN meta_table.meta_key='phone_number' THEN meta_table.phone_number ELSE NULL END) as phone_number, 
MAX(CASE WHEN meta_table.meta_key='country' THEN meta_table.country ELSE NULL END) as country, 
MAX(CASE WHEN meta_table.meta_key='years_of_experience' THEN meta_table.years_of_experience ELSE NULL END) as years_of_experience, 
MAX(CASE WHEN meta_table.meta_key='highest_degree_obtained' THEN meta_table.highest_degree_obtained ELSE NULL END) as highest_degree_obtained, 
MAX(CASE WHEN meta_table.meta_key='availability_for_work' THEN meta_table.availability_for_work ELSE NULL END) as availability_for_work, 
MAX(CASE WHEN meta_table.meta_key='english_proficiency' THEN meta_table.english_proficiency ELSE NULL END) as english_proficiency, 
MAX(CASE WHEN meta_table.meta_key='disciplines' THEN meta_table.disciplines ELSE NULL END) as disciplines, 
MAX(CASE WHEN meta_table.meta_key='profile_picture' THEN meta_table.profile_picture ELSE NULL END) as profile_picture, 
MAX(CASE WHEN meta_table.meta_key='resume' THEN meta_table.resume ELSE NULL END) as resume, 
MAX(CASE WHEN meta_table.meta_key='description' THEN meta_table.description ELSE NULL END) as description, 
MAX(CASE WHEN meta_table.meta_key='hourly_rate' THEN meta_table.hourly_rate ELSE NULL END) as hourly_rate, 
MAX(CASE WHEN meta_table.meta_key='satisfaction_rating' THEN meta_table.satisfaction_rating ELSE NULL END) as satisfaction_rating, 
MAX(CASE WHEN meta_table.meta_key='invited_projects' THEN meta_table.invited_projects ELSE NULL END) as invited_projects, 
MAX(CASE WHEN meta_table.meta_key='completed_project_count' THEN meta_table.completed_project_count ELSE NULL END) as completed_project_count 

# The table we're selecting from 

from tsd_retro_users as users 

# Join in our usermeta table for each individual meta value 

inner join tsd_retro_usermeta as meta_table 
ON users.ID = meta_table.user_id 

# Define our select stipulations 

where 
(users.ID = 20) 
GROUP BY users.ID 

И как только мы используем PHP вы можете сделать так:

$requiredMetaFields = array(
'first_name', 
'last_name', 
'phone_number', 
'country', 
'years_of_experience', 
'highest_degree_obtained', 
'availability_for_work', 
'english_proficiency', 
'disciplines', 
'profile_picture', 
'resume', 
'description', 
'hourly_rate', 
'satisfaction_rating', 
'invited_projects', 
'completed_project_count' 
); 

$query = "select 
    users.ID, 
    users.user_email, 
    users.display_name,"; 

foreach($requiredMetaFields as $metafield) { 
$query .= "MAX(CASE WHEN meta_table.meta_key='$metafield' THEN meta_table.$metafield ELSE NULL END) as $metafield, "; 
} 
$query =substr($query,0,-1); 
$query .= "from tsd_retro_users as users 
inner join tsd_retro_usermeta as meta_table 
ON users.ID = meta_table.user_id 
where 
(users.ID = 20) 
GROUP BY users.ID"; 

отредактированные ошибки синтаксиса