I'm trying to run a MySQL query and use PHP to do the following:
I need to group data in a column (GROUP BY) then count (COUNT) the number of rows in each group. After that, I need to divide the number of rows in a given group by the number of groups to get that groups percentage of popularity.
So if I had a table with the following data:
Version_Number
1.1
1.2
1.1
1.2
1.2
1.1
I need the final output to be:
1.1 50%
1.2 50%
You may need to adjust this for mysql (I normally work in oracle) but:
SELECT
( count(*) / totalCount * 100) AS percentOfVersionedThings
, Version_Number
FROM tableOfVersionedThings
INNER JOIN ( SELECT count(*) as totalCount FROM tableOfVersionedThings ) ON 1=1
GROUP BY Version_Number
If this was Oracle, I'd suggest using analytics, but I'm not sure if there is an equivalent in MySQL. That said, in your case a simple sub-query should solve the problem, and it should be workable on any SQL database.
This is what I ended up doing after trying the above:
$result = mysql_query("SELECT COUNT(*) AS nsites FROM table WHERE auth = '$auth'") or
die(mysql_error());
$return_sites = mysql_fetch_array($result);
if (!$return_sites['nsites']) {
echo "...";
} else {
$esult = mysql_query("SELECT * FROM table WHERE auth = '$auth' GROUP BY theme_version") or die(mysql_error());
while($row = mysql_fetch_array($esult)){
$get_current_version = $row['theme_version'];
$vresult = mysql_query("SELECT COUNT(*) AS nversion FROM table WHERE auth = '$auth' AND theme_version = '$get_current_version'") or die(mysql_error());
$version = mysql_fetch_array($vresult);
$percent = round($version['nversion'] / $return_sites['nsites'] * 100);
echo "...";
}
}
Something like this:
select v."Version_Number", count(v."Version_Number"),count(v."Version_Number")::float / (select count(v2."Version_Number")::float from versions v2 )::float * 100::float
from versions as v
group by v."Version_Number"
Related
I have 2 tables - users and articles.
users:
user_id (int)
name (varchar)
articles:
article_id (int)
user_id (int)
title (varchar)
description (text)
In my application I need to display 20 RANDOM articles on a page.
My query is like this:
SELECT a.title
, a.description
, u.name
FROM articles a
JOIN users u
USING (user_id)
ORDER
BY RAND()
LIMIT 20
A user can have any number of articles in the database.
Now the problem is sometimes out of 20 results, there are like 9-10 articles from one single user.
I want those 20 records on the page to not contain more than 3 (or say 4) articles from a particular user.
Can I achieve this through SQL query. I am using PHP and MySQL.
Thanks for your help.
You could try this?
SELECT * FROM
(
SELECT B.* FROM
(
SELECT A.*, ROW_NUMBER() OVER (PARTITION BY A.USER_ID ORDER BY A.R) USER_ROW_NUMBER
FROM
(
SELECT a.title, a.description, u.name, RND() r FROM articles a
INNER JOIN users u USING (user_id)
) A
) B
WHERE B.USER_ROW_NUMBER<=4
) C
ORDER BY RAND() LIMIT 20
Mmm, intresting I don't think this is possible through a pure sql query.
My best idea would be to have an array of the articles that you'll eventually display query the database and use the standard SELECT * FROM Articles ORDER BY RAND() LIMIT 20
The go through them, making sure that you have indeed got 20 articles and no one has breached the rules of 3/4 per user.
Have another array of users to exclude, perhaps using their user id as an index and value of a count.
As you go through add them to your final array, if you find any user that hits you rule add them to the array.
Keep running the random query, excluding users and articles until you hit your desired amount.
Let me try some code (it's been a while since I did php)
$finalArray = [];
$userArray = [];
while(count($finalArray) < 20) {
$query = "SELECT * FROM Articles ";
if(count($finalArray) > 0) {
$query = $query . " WHERE articleID NOT IN(".$finalArray.")";
$query = $query . " AND userID NOT IN (".$userArray.filter(>4).")";
}
$query = $query . " ORDER BY Rand()";
$result = mysql_query($query);
foreach($row = mysql_fetch_array($result)) {
if(in_array($finalArray,$row) == false) {
$finalArray[] = $row;
}
if(in_array($userArray,$row[userId]) == false) {
$userArray[$row[userId]] = 1;
}
else {
$userArray[$row[userId]] = $userArray[$row[userId]] + 1;
}
}
I got a little problem, I've got a database, in that database are different names, id, and coins. I want to show people their rank, so your rank has to be 1 if you have the most coins, and 78172 as example when your number 78172 with coins.
I know I can do something like this:
SELECT `naam` , `coins`
FROM `gebruikers`
ORDER BY `coins` DESC
But how can I get the rank you are, in PHP :S ?
You can use a loop and a counter. The first row from MySql is going the first rank,I.e first in the list.
I presume you want something like:
1st - John Doe
2nd - Jane Doe
..
..
right?
See: http://www.if-not-true-then-false.com/2010/php-1st-2nd-3rd-4th-5th-6th-php-add-ordinal-number-suffix
Helped me a while ago.
You could use a new varariable
$i = "1";
pe care o poti folosi in structura ta foreach,while,for,repeat si o incrementezi mereu.
and you use it in structures like foreach,while,for,repeat and increment it
$i++;
this is the simplest way
No code samples above... so here it is in PHP
// Your SQL query above, with limits, in this case it starts from the 11th ranking (0 is the starting index) up to the 20th
$start = 10; // 0-based index
$page_size = 10;
$stmt = $pdo->query("SELECT `naam` , `coins` FROM `gebruikers` ORDER BY `coins` DESC LIMIT {$start}, {$page_size}");
$data = $stmt->fetchAll();
// In your template or whatever you use to output
foreach ($data as $rank => $row) {
// array index is 0-based, so add 1 and where you wanted to started to get rank
echo ($rank + 1 + $start) . ": {$row['naam']}<br />";
}
Note: I'm too lazy to put in a prepared statement, but please look it up and use prepared statements.
If you have a session table, you would pull the records from that, then use those values to get the coin values, and sort descending.
If we assume your Session table is sessions(session_id int not null auto_increment, user_id int not null, session_time,...) and we assume that only users who are logged in would have a session value, then your SQL would look something like this: (Note:I am assuming that you also have a user_id column on your gebruikers table)
SELECT g.*
FROM gebruikers as g, sessions as s WHERE s.user_id = g.user_id
ORDER BY g.coins DESC
You would then use a row iterator to loop through the results and display "1", "2", "3", etc. The short version of which would look like
//Connect to database using whatever method you like, I will assume mysql_connect()
$sql = "SELECT g.* FROM gebruikers as g, sessions as s WHERE s.user_id = g.user_id ORDER BY g.coins DESC";
$result = mysql_query($sql,$con); //Where $con is your mysql_connect() variable;
$i = 0;
while($row = mysql_fetch_assoc($result,$con)){
$row['rank'] = $i;
$i++;
//Whatever else you need to do;
}
EDIT
In messing around with a SQLFiddle found at http://sqlfiddle.com/#!2/8faa9/6
I came accross something that works there; I don't know if it will work when given in php, but I figured I would show it to you either way
SET #rank = 0; SELECT *,(#rank := #rank+1) as rank FROM something order by coins DESC
EDIT 2
This works in a php query from a file.
SELECT #rank:=#rank as rank,
g.*
FROM
(SELECT #rank:=0) as z,
gebruikers as g
ORDER BY coins DESC
If you want to get the rank of one specific user, you can do that in mysql directly by counting the number of users that have more coins that the user you want to rank:
SELECT COUNT(*)
FROM `gebruikers`
WHERE `coins` > (SELECT `coins` FROM `gebruikers` WHERE `naam` = :some_name)
(assuming a search by name)
Now the rank will be the count returned + 1.
Or you do SELECT COUNT(*) + 1 in mysql...
I am trying to calculate how much a user has earned so it reflects on the users home page so they know how much their referrals have earned.
This is the code I have.
$get_ref_stats = $db->query("SELECT * FROM `members` WHERE `referral` = '".$user_info['username']."'");
$total_cash = 0;
while($ref_stats = $get_ref_stats->fetch_assoc()){
$get_ref_cash = $db->query("SELECT * FROM `completed` WHERE `user` = '".$ref_stats['username']."' UNION SELECT * FROM `completed_repeat` WHERE `user` = '".$ref_stats['username']."'");
$countr_cash = $get_ref_cash->fetch_assoc();
$total_cash += $countr_cash['cash'];
$countr_c_rate = $setting_info['ref_rate'] * 0.01;
$total_cash = $total_cash * $countr_c_rate;
}
It worked fine when I just had
$get_ref_cash = $db->query("SELECT * FROM `completed` WHERE `user` = '".$ref_stats['username']."'");
but as soon as I added in the UNION it no longer calculated correctly.
For example, there is 1 entry in completed and 1 entry in completed_repeat both of these entries have a cash entry of 0.75. The variable for $countr_c_rate is 0.10 so $total_cash should equal 0.15 but instead it displays as 0.075 with and without the UNION it acts as if it is not counting from the other table as well.
I hope this makes sense as I wasn't sure how to explain the issue, but I am very unsure what I have done wrong here.
In your second query instead of UNION you should use UNION ALL since UNION eliminates duplicates in the resultset. That is why you get 0.075 instead of 0.15.
Now, instead of hitting your database multiple times from client code you better calculate your cash total in one query.
It might be inaccurate without seeing your table structures and sample data but this query might look like this
SELECT SUM(cash) cash_total
FROM
(
SELECT c.cash
FROM completed c JOIN members m
ON c.user = m.username
WHERE m.referral = ?
UNION ALL
SELECT r.cash
FROM completed_repeat r JOIN members m
ON r.user = m.username
WHERE m.referral = ?
) q
Without prepared statements your php code then might look like
$sql = "SELECT SUM(cash) cash_total
FROM
(
SELECT c.cash
FROM completed c JOIN members m
ON c.user = m.username
WHERE m.referral = '$user_info['username']'
UNION ALL
SELECT r.cash
FROM completed_repeat r JOIN members m
ON r.user = m.username
WHERE m.referral = '$user_info['username']'
) q";
$result = $db->query($sql);
if(!$result) {
die($db->error()); // TODO: better error handling
}
if ($row = $result->fetch_assoc()) {
$total_cash = $row['cash_total'] * $setting_info['ref_rate'];
}
On a side note: make use of prepared statements in mysqli instead of building queries with concatenation. It's vulnerable for sql-injections.
With $countr_cash = $get_ref_cash->fetch_assoc(); you only fetch the first row of your result. However, if you use UNION, you get in your case two rows.
Therefore, you need to iterate over all rows in order to get all values.
Ok, So there is only one row in members table. You are iterating only once on the members table. Then you are trying to get rows using UNION clause which will result in two rows and not one. Then you are just getting the cash column of the first row and adding it to the $total_cash variable.
What you need to do is iterate over the results obtained by executing the UNION query and add the $total_cash variable. That would give you the required result.
$get_ref_stats = $db->query("SELECT * FROM `members` WHERE `referral` = '".$user_info['username']."'");
$total_cash = 0;
while($ref_stats = $get_ref_stats->fetch_assoc()){
$get_ref_cash = $db->query("SELECT * FROM `completed` WHERE `user` = '".$ref_stats['username']."' UNION SELECT * FROM `completed_repeat` WHERE `user` = '".$ref_stats['username']."'");
while($countr_cash = $get_ref_cash->fetch_assoc()){
$total_cash += $countr_cash['cash'];
}
$countr_c_rate = $setting_info['ref_rate'] * 0.01;
$total_cash = $total_cash * $countr_c_rate;
}
Hello I would like to query multiple identical tables in my db which has different prefixes and than display the results randomly but somehow I need to track the origin of the item and I couldn't figure out how
I do the query like this because I don't have access to information_schema
$query = "SHOW TABLES FROM mydb WHERE RIGHT( tables_in_mydb, 5 ) = 'table'";
$res = mysql_query($query);
$num = mysql_num_rows($res);
while($row = mysql_fetch_row($res)) {
$numbers = explode('_', $row[0]);
if($num > 0) {
$q = "SELECT `this`, `that`, `something` FROM ".$numbers[0]."_idetinticaltables"; // :)
$r = mysql_query($q);
while($c = mysql_fetch_array($r)) {
/*display the results randomly with an identifier where the come from*/
}
}
}
You could use ORDER BY RAND() to randomly sort it
The following might work:
Get the list of the tables you're interested in. You already do that.
Create a UNION of multiple SELECT statements. Each SELECT statement differs for the table being selected from and you add a column set to the name of the table (so you can identify it later):
(SELECT *, TABLENAME = 'first_name_of_table' FROM first_name_of_table ...)
UNION
(SELECT *, TABLENAME = 'second_name_of_table' FROM second_name_of_table ...)
UNION
...
ORDER BY RAND() LIMIT 10;
Because it is a UNION you can randomize the whole order then. See How can i optimize MySQL's ORDER BY RAND() function? because it is not that trivial to do well, the example above is only to have an ORDER BY and LIMIT clause placed there. With many entries in your tables, it will kill your server.
$aa=array()
while($c = mysql_fetch_array($r))
{
/*display the results randomly with an identifier where the come from*/
$aa[]=$c;
}
echo $aa; // print "Array"
I've been looking for this for a while but with no success.
I am trying to implement a recomendation bar, for example like in youtube, when you are seeing a video it shows the list or recommended videos on the right.
At this moment I am using this method:
$offset_result = mysql_query( " SELECT FLOOR(RAND() * COUNT(*)) AS `offset` FROM `$tablename` ");
$offset_row = mysql_fetch_object($offset_result );
$offset = $offset_row->offset;
$result_rand = mysql_query( " SELECT * FROM `$tablename` LIMIT $offset, 9 " );
This works fine, but sometimes doesn't show any result, and the problem is also that its not completely random, because it shows for example the first ID as 200, so the next result will be id 201 and then 202 and so.
I would like to know if there is a way to show this 9 randon results, for example 1º result id 500, 2º result id 10, 3º result id 788, etc etc?
Thank you
Not entirely sure this answers what you are looking for, but try:
$result_rand = mysql_query("SELECT * FROM " . $tablename . " ORDER BY RAND() LIMIT 9");
You can use php rand() function to create 5 numbers and save them in an array:
http://php.net/manual/en/function.rand.php
<?php
$rand_array = array();
for($i=1;$i<5;$i++) {
$rand_array[$i] = rand(0,500);
}
?>
and after that create a query with every int with a foreach loop and work with your data.
<?php
foreach ($rand_array as $integer) {
$q = "SELECT * from $tablename WHERE id='$integer';";
}
?>
Does this helps?
First you should use mysqli_ functions instead of mysql_ because the latter is deprecated. Second use order by rand() to get random rows:
$rand_result = mysqli_query( "SELECT * FROM $tablename ORDER BY RAND() LIMIT 9;" );
UNTESTED:
SELECT id, #rownum:=#rownum+1 AS rownum, name
FROM users u,
(SELECT #rownum:=0) r
THis will give a unique number to each row in sequence. Now if you create a temp table with 9 random numbers between 1 and count(*) of your table and JOIN those two together...
Not sure about performance but seems like it might be faster than Rand and order by since I only need 9 random numbers