Here is database table:
$sql[2] = "SELECT u.* , oi.* , COUNT(oi.user_id) AS count
FROM users u, order_items oi
WHERE u.id=oi.user_id ";
$result3= mysqli_query($conn,$sql[2]) or die(mysqli_error());
if (mysqli_num_rows($result3) > 0) {
while ($record = mysqli_fetch_array($result3)) {
echo $record['count'];
}
}
I want to count how much order have every user. Example: Like Thomas have 3 order, but my code is writing 4, i want to write Thomas (3), Gracian(1). Any idea how to fix it ?
Use this query:
SELECT u.id,
COUNT(oi.user_id) AS orderCount
FROM users u
LEFT JOIN order_items oi
ON u.id = oi.user_id
GROUP BY u.id
The reason we count user_id from the order_items table is because of the edge case where a given user has no orders. In this case, we want to make sure that his count would appear as zero. The COUNT function ignores NULLs, which is what we want.
There is another way to perform the sql query using subqueries:
SELECT id,
email,
address,
name,
(SELECT count(user_id) FROM order_items WHERE user_id = users.id) AS orderCount FROM users;
Related
I have users with rating and I want to show the total amount of rating of each user. I have 2 tables: users and feedback. I have 4 users in USERS table. And in table FEEDBACK (consists of applicant_id and app_experience columns) I have the rating of each user. But, I have several ratings for each person.
$sql4 = "
SELECT SUM(app_experience) as app_experience
, applicant_id
FROM feedback f
JOIN users u
ON u.id=f.applicant_id
GROUP
BY feedback.applicant_id
";
$res4 = mysqli_query($db,$sql4) or die(mysqli_error($db));
This is my output, but it prints only 2 users, because in table FEEDBACK 2 of the users do not have any feedback yet:
foreach ($res4 as $row)
{
echo "".$row['applicant_id']."----".$row['app_experience']."";
echo "<br>";
}
My question is how to output all 4 users and I want to show the total number of rating next to each user.
I am doing something like this, but I do not know where to add the foreach loop above in the code below. Do you have any ideas?:
$sql2 = "SELECT * FROM users";
$res2 = mysqli_query($db,$sql2) or die(mysqli_error($db));
while($row2 = mysqli_fetch_assoc($res2))
{
$id = $row2['id'];
$name = $row2['name'];
}
You just have to modify the below statement :
$sql4 = "SELECT SUM(app_experience) as app_experience, applicant_id FROM feedback INNER JOIN users ON users.id=feedback.applicant_id GROUP BY feedback.applicant_id";
to :
$sql4 = "SELECT COALESCE(SUM(app_experience), 0) as app_experience, applicant_id FROM users LEFT JOIN feedback ON users.id=feedback.applicant_id GROUP BY users.id";
The users LEFT JOIN feedback ... clause will allow the query to return a row even for users who do not have a feedback yet.
The COALESCE(SUM(app_experience), 0) will evaluate to 0 when a user has no feedback yet.
With this solution you don't need to loop.
SELECT users.*, (SELECT SUM(app_experience) FROM feedback WHERE applicant_id = users.id) as feedback FROM users
Selects all users and their feedback in a short SQL
As written, your query will not run. The GROUP BY clause is referencing feedback, but that reference has been changed to f by the alias.
You appear to want a LEFT JOIN:
SELECT u.id, SUM(f.app_experience) as app_experience
FROM feedback f LEFT JOIN
users u
ON u.id = f.applicant_id
GROUP BY u.id;
This will return a NULL value for app_experience. If you want a value (such as 0) use COALESCE(SUM(f.app_experience), 0).
My tables
$sql="SELECT *
FROM addresses
LEFT JOIN users ON address_id = user_id
LEFT JOIN notes ON note_id = user_id
ORDER BY id DESC
LIMIT 1";
This is my SQL query, my task is to show the last records from 3 tables, but the table is blank, I don't know why,thanks in advance people :)
I guess the problem is coming from the ORDER BY id DESC .
Indeed, you have no column so called id.
You should probably remove this clause, in order to make your code work.
If you want to take the last records anyway, you can put an ORDER BY address_id DESC which will do the job !
The code directly edited :
$sql="SELECT *
FROM addresses
LEFT JOIN users ON address_id = user_id
LEFT JOIN notes ON note_id = user_id
ORDER BY adress_id DESC
LIMIT 1";
This may work:
SELECT a.address_id, u.user_id, n.note_id
FROM addresses a
LEFT JOIN users_addresses ua ON ua.ua_address_id = a.address_id
LEFT JOIN users u ON u.user_id = ua.ua_user_id
LEFT JOIN notes n ON n.note_user_id = u.user_id
ORDER BY a.address_id DESC
LIMIT 1
Here is the query to get all data from all the tables, not sure what do you mean last records from 3 tables, I can see four tables there:
SELECT *
FROM `addresses`
LEFT JOIN `users_addresses` ON `users_addresses`.`ua_address_id` = `addresses`.`address_id`
LEFT JOIN `users` ON `users`.`user_id` = `users_addresses`.`ua_user_id`
LEFT JOIN `notes` ON `notes`.`note_user_id` = `users`.`user_id`;
I have two tables: publick_feed and users
I want to SELECT all from public_feed and also SELECT a three columns from users whose id is the same of user_id in public_feed
and assign the rows returned from public_feed to the column in users table ( correspondent)
I try this:
<?php
$sql = "
SELECT * FROM public_feed
WHERE user_id IN
(SELECT id FROM users) AND
(SELECT Firstname,Lastname,Avatar FROM users WHERE id IN(SELECT user_id FROM public_feed))
";
$query = mysqli_query($dbc_conn,$sql);
if(mysqli_num_rows($query) > 0){
while($row = mysqli_fetch_assoc($query)){
//echo rows with correspondent details from the users table
echo $row['user_id'];
}
}
<?
Please any help will be much appreciated.
Thank you.
Or version with left join in case if there is no user in public_feed, and you still want to fetch user data
SELECT
u.*, f.*
FROM
public_feed f LEFT JOIN
users u ON f.user_id = u.id;
Because author asked for explanation, here it is:
First we are going to use table name alias to make query shorter
public_feed f
and
users u
we are saying that want to refer to tables with an alias. Of course * means that we want to select all columns
SELECT users.*, public_feed.*
is equal to
SELECT u.*, f.*
Of course you can use any other letters as an alias
Next we are saying that public_feed.user_id must be equal to users.id. But when public feed entry does not exists just display columns with null values. This is why we are using LEFT JOIN instead of INNER JOIN. In general JOINS are used to fetch related data from more than one related tables.
ON keyword is saying values from which columns in the tables must be equal to satisfy the request
I think doing a join would be cleaner than using a complicated subquery:
SELECT u.Firstname,
u.Lastname,
u.Avatar,
COALESCE(pf.User_id, 'NA'),
COALESCE(pf.Post, 'NA'),
COALESCE(pf.Date, 'NA')
FROM users u
LEFT JOIN public_feed pf
ON u.Id = pf.User_id
I chose a LEFT JOIN of users against public_feed on the assumption that every feed will have an entry in the users table, but not necessarily vice-versa. For those users who have no feed entries, NA would appear in those columns and that user would appear in only a single record.
i need help getting data from different tables and insert into other different table Here are the Queries
"SELECT commentID, date, comment, subject, parentID, aBUserID FROM comments WHERE status = 'APPROVED'"
"SELECT topicID, subForumID, aBUserID, lastPostID, views, replies, startDate FROM topic WHERE status = 'APPROVED' AND topicID = $parentid";
// $parentID need to be matched from above query parentID,
"SELECT userName FROM users WHERE aBUserID = $cmtaBUserID";
// $cmtaBUserID = aBUserID from first query
"SELECT userName FROM users WHERE aBUserID = $topicaBUserID";
//$topicaBUserID = aBUserID from second query
Last 2 queries are from same table but using different where clause
i used different inner join left join from solutions posted here but non of these worked for me stuck since last 2 weeks please help
PS data from all above Queries will be inserted to a single table i need these to be combined so i can have them all in one place
If you want to perform the operation in same query use 'OR'
"SELECT userName FROM users WHERE aBUserID = $cmtaBUserID OR aBUserID = $topicaBUserID";
Please try this
SELECT userName from users where aBUserID IN(SELECT aBUserID FROM comments WHERE status = 'APPROVED')
Couldn't test it but Maybe this is what you are looking for.
SELECT c.commentID, c.date, c.comment, c.subject, c.parentID, c.aBUserID,
t.topicID, t.subForumID, t.aBUserID, t.lastPostID, t.views, t.replies, t.startDate,
u.userName
FROM
comments c
left outer join topic t on t.topicID = c.parentID
left outer join users u on u.aBUserID = c.aBUserID and u.aBUserID = t.aBUserID
WHERE
c.status = 'APPROVED' and t.status = 'APPROVED';
try this:
SELECT
comment.[commentID],
comment.[date],
comment.[comment],
comment.[subject],
comment.[parentID],
comment.[aBUserID],
commentuser.[userName],
topic.[topicID],
topic.[subForumID],
topic.[aBUserID],
topic.[lastPostID],
topic.[views],
topic.[replies],
topic.[startDate],
topic.[userName]
FROM comments comment
LEFT OUTER JOIN users commentuser
ON commentuser.aBUserID = comment.[aBUserID]
LEFT OUTER JOIN
(
SELECT
t.[topicID],
t.[subForumID],
t.[aBUserID],
t.[lastPostID],
t.[views],
t.[replies],
t.[startDate],
u2.[userName] --user from users table joined to topics table
FROM topic t
LEFT OUTER JOIN users u
ON u.aBUserID = t.[aBUserID]
WHERE t.[status] = 'APPROVED'
) topic
ON topic.topicID = comment.parentID
WHERE comment.[status] = 'APPROVED'
I am making a cookie based favorite system and need to join data from two tables based on the unique user id stored in the cookie so I can tell what items that use has marked as favorites.
I know I need to do a JOIN but have not used them much and dont really have my head around them yet.
Existing query that selects the items from the db:
$query = mysql_query("SELECT *, UNIX_TIMESTAMP(`date`) AS `date` FROM songs WHERE date >= DATE_SUB( NOW( ) , INTERVAL 2 WEEK ) ORDER BY date DESC");
My favorites table is setup as: ID FAVORITE USERID where ID is the primary key, FAVORITE is the song ID from table songs and USERID is a hash stored in a cookie.
I need to join in all the rows from the favorites table where the USERID field matches the cookie hash variable.
I also need to gather the total number of rows in favorites that match the song id so I can display a count of the number of people who set the item as favorite so I can display how many people like it. But maybe need to do that as a separate query?
This should do it, I would imagine:
$user_id = intval($_COOKIE['user_id']);
$query = mysql_query(sprintf("
SELECT *
FROM songs s
INNER JOIN favorites f
ON f.favorite = s.id
WHERE f.userid = %s
", $user_id));
You should probably read up on the different types of joins.
And then to get the total amount of rows returned, you can just call mysql_num_rows on the result:
$favorite_song_count = mysql_num_rows($query);
EDIT: To select all songs but note which are favorited, you would do this:
$query = mysql_query(sprintf("
SELECT s.*, f.id as favorite_id
FROM songs s
LEFT JOIN favorites f
ON f.favorite = s.id AND f.userid = %s
", $user_id));
By switching it from an INNER JOIN to a LEFT JOIN we are selecting all songs even if they don't have a corresponding record in the favorites table. Any songs that are favorites of the user_id provided will have a non-NULL value for favorite_id.
You can have logical (and, or, ...) operators in join conditions:
select t1.*
from t1
join t2 on t1.id = t2.fid and t2.foo = 'blah'
If you are also querying the total number of times each song has been "favorited" then you need a group by construct also, like this way:
select *, count(f.id)
from songs as s
left join favorites as f on s.id = f.favorite and f.userid = <hash>
group by s.id