Here is the mySQL I got
id terms
1 a
2 c
3 a
4 b
5 b
6 a
7 a
8 b
9 b
10 b
I want to get an alphabetized list sorted by count as follows
terms count
a 4
b 5
c 1
What mySQL statement do I need for that?
I believe something like this will work:
SELECT terms, COUNT( id) AS count
FROM table
GROUP BY terms
ORDER BY terms DESC
Read : GROUP BY (Transact-SQL)
Groups a selected set of rows into a set of summary rows by the values of one or more columns or expressions in SQL. One row is returned for each group. Aggregate functions in the SELECT clause list provide information about each group instead of individual rows.
You just need to apply group by clause for getting result
select terms, count (id) as count from table
group by terms
order by terms
I had a very similar need for a used record store to display artists in stock alphabetically with their count in parenthesis e.g.:
Smokey Robinson and The Miracles (2) | Sonic Youth (2) | Spoon (3) | Steely Dan (1) | Stevie Wonder (2) | Sufjan Stevens (1) |
Note that I used SELECT DISTINCT when pulling from my table "records". Here are the relevant code snippets:
//QUERY
$arttool = mysql_query("SELECT DISTINCT * FROM records GROUP BY artist ORDER BY artist ASC");
//OUTPUT LOOP START
while($row = mysql_fetch_array($arttool)){
//CAPTURE ARTIST IN CURRENT LOOP POSITION
$current=$row['Artist'];
//CAPTURING THE NUMBER OF ALBUMS IN STOCK BY CURRENT ARTIST
$artcount = mysql_num_rows(mysql_query("SELECT * FROM records WHERE artist = '$current'"));
//ECHO OUT.
echo $current . "($artcount)";
The actual code in my site is more complicated, but this is the bare bones of it. Hope that helps...
Related
I am trying to search two tables, match the results and then concatenate the answer... Only finding results >= today's date. This will then give the user the option to delete the selected from the DB. So...
Table 1 called Prog_name
id prog_name
1 Breakfast
2 Mid Morning
3 Afternoon
Table 2 called talk_ups
id date_tx prog_name (prog_name value = prog_name.id)
1 2017-06-30 2
2 2017-07-03 1
3 2017-07-01 3
The result I am after is something like: "01-07-2017, Afternoon". But I do also need the talk_ups.id to ensure it only deletes the correct record.
I managed to figure out how to get the name to match the talk_ups.prog_name value:
'$sql. = "SELECT talk_ups.prog_name, prog_name.id as progID, prog_name.prog_name as theName FROM prog_name, talk_ups WHERE talk_ups.prog_name = prog_name.id";'
But I can't figure out how to do the two searches and end up with the right result and how to separate out the results to then concatenate them.
You can use JOIN with WHERE condition, e.g.:
SELECT pn.id, pn.prog_name, tu.date_tx
FROM prog_name pn JOIN talk_ups tu ON pn.id = tu.prog_name
WHERE tu.date_tx > NOW();
I have a table with various categories, and multiple entries for each category.
Word | Category
------------------
Apple | Food
Orange | Food
Grapes | Food
Mango | Food
I wish to retrieve 3 random rows for the category 'food', for which I run the following query,
$query = "SELECT * FROM table WHERE category='food' ORDER BY RAND() LIMIT 3"
$fetch_row = mysqli_query($db_connect, $query);
while ($row = mysqli_fetch_array($fetch_row)) {
array_push($words, $row['word']);
}
However, when I print the contents of the array $words, they tend to repeat sometimes (not on all runs), for example;
Apple, Orange, Apple
i.e. Its not always unique. I want to select random, yet unique words for a given category. What am I doing wrong? I've tried going through other related answers, but I keep messing something up. I've also tried the following query;
SELECT * FROM table WHERE category='food' GROUP BY category ORDER BY RAND() LIMIT 3
But this still gives repetitions occasionally.
Since word column have same values, do GROUP BY word like below:-
SELECT * FROM table WHERE category='food' GROUP BY word LIMIT 3
I've 4 table for a newsletter. Newsletters, Subscribers, Subscriber Groups and Selected Subscriber Groups. I've choose subscriber groups in campaign edit area, and its save selected groups to tbl_newsletter_groups table like;
tbl_newsletters
NID title details
1 text 1 content 1
2 text 2 content 2
tbl_subscriber_groups
GID group_name
5 group 1
6 group 2
tbl_subscribers
SID GID email name
10 5 sub1#mail.com sub1 name
11 6 sub1#mail.com sub1 name
tbl_newsletter_groups
NGID NID GID
15 1 6
16 1 6
17 1 6
I want to show total selected subscriber count when I list newsletters in my page. My soulution works fine, Im looking for simple and clearly statement, there any faster way available like in single newsletter list statement?
Here my own count style (yes I know its too bad and long way);
$subGID = array();
$list = $myconn->query("SELECT * FROM tbl_newsletters");
while($listRs = $list->fetch_assoc()){
$grps = $myconn->query("SELECT * FROM tbl_newsletter_groups WHERE NID=". $listRs['NID'] ."");
while($grpsRs = $grps->fetch_asscoc()){
$subGID[] = $grpsRs['GID'];
} $grps->free();
$subs = implode(" OR GID=",$subGID);
$count = mysqli_num_rows($myconn->query("SELECT ID FROM tbl_subscribers WHERE GID=". $subs));
echo('Total Selected Subscriber: '.$count);
} $list->free();
Thanks.
The search term you want is "set-based logic".
Your thinking is sound: you need everything from tbl_newsletters, then you need to count results from tbl_subscribers, but in order to get those you need information from tbl_newsletter_groups.
In SQL, that's an indication you want a join. You've already discovered the conditions you need, you just don't know the syntax. A reference manual can help there.
Now you'll have a bunch of records, which you need to smash into a smaller number of records. You need aggregation functions and a GROUP BY clause.
So here's the final query:
SELECT n.NID, n.title, n.details, COUNT(s.SID)
FROM tbl_newsletters AS n
JOIN tbl_newsletter_groups AS g ON n.NID = g.NID
JOIN tbl_subscribers AS s ON g.GID = s.GID
GROUP BY n.NID
I've got two tables:
content:
id access
1 3
2 5
3 9
viewlevels:
id group
1 [10,12,15]
2 [8,12,11]
3 [9,10,5]
The access field in content is related with the id field in viewlevels.
I select the rows in viewlevels depending on the current user group. So for example, if group is = 10, my query will select rows 1 and 3. If the group is 12, it will select rows 1 and 2, etc. I'm using the following query:
$query="SELECT id FROM #__viewlevels WHERE rules LIKE '%$group%'";
My challenge is to count the number of rows for column id in table content where the access matches with the selected id's from the above query on table viewlevels.
I tried the following code but it is returning the error: undefined variable: nartigos
$query="SELECT count(id) FROM #__content WHERE access IN (SELECT id FROM #__viewlevels WHERE rules LIKE '%$group%')";
if ($stmt = mysqli_prepare($link, $query)) {
/* execute query */
mysqli_stmt_execute($stmt);
/* store result */
mysqli_stmt_store_result($stmt);
$nartigos=mysqli_stmt_num_rows($stmt);
/* close statement */
mysqli_stmt_close($stmt);
};
echo "Nº de artigos " .$nartigos;
First of all, you really should normalize your data. Consider having a many-to-many join table for viewLevels instead of having all groups in one row. That might look like this:
access_id group_id
1 10
1 12
1 15
2 8
2 11
2 12
3 5
3 9
3 10
That would make your query as simple as
SELECT c.id AS `content_id`, COUNT(v.access_id) AS `content_count`
FROM content AS c INNER JOIN viewLevels AS v
ON c.access_id = v.access_id
WHERE v.group_id = ?
GROUP BY c.id
Here ? is the group id you are querying against.
Without normalization (which again I STRONGLY recommend you do), you would still use a join, but it would look like this:
SELECT c.id AS `content_id`, COUNT(v.access_id) AS `content_count`
FROM content AS c INNER JOIN viewLevels AS v
ON c.access_id = v.access_id
WHERE v.group LIKE '%?%'
GROUP BY c.id
you need to "join" your tables.
the sql command cant query two tables seperately.
when you "join" 2 tables in your sql, think of it as making one virtual/temporary table in the air, of the 2 tables which you can then query.
this is quite a good intro http://www.sitepoint.com/understanding-sql-joins-mysql-database/
I've created a game with a highscore table in MySQL.
I have a "My scores" button that needs to retrieve the users scores, e.g.:
10. John 395
42. John 340
90. John 10
How should I go out retrieving the rank (10th, 42th, 90th) of each score of the user?
I could pull all the scores from the database and iterate through them but that doesn't seem like a good solution.
Let me try to expand:
I retrieve all MY scores from the database. E.g. 10 scores. I want to display these 10 scores however I won't know what the rank of these scores is compared to the other scores in my database! (10th, 16th, etc) ..Hope that makes more sense...
Thanks
For the position in the total list you either need to build up a list every time you want this overview, or use a stored procedure to build a list for a given moment. You could 'cache' a list on a given interval. Or maybe update a list when some one played a game that would change the top 100.
As #WhiteElephant suggested, you'd be making the table every time you want the data.
#stefandoorn suggest to not use the optimized count of sql, i think this is not efficient enough for these kind of computations.
A simple SQL query would do it for you. For example, if you want the 10th score, you could use:
SELECT name, score FROM highscores ORDER BY score DESC LIMIT 1 OFFSET 9
The offset will always be the position required - 1.
If you want to have the associated rank as a column beside the score, you could do the following:
SELECT #rownum:=#rownum+1 position, name, score FROM (SELECT #rownum:=0) r, highscores ORDER BY score DESC
This doesn't work with an offset (the position number will always start at 1). The result would be something like the following:
+----------+-------------+-------+
| position | name | score |
+----------+-------------+-------+
| 1 | Player 1 | 27681 |
| 2 | Player 2 | 14982 |
+----------+-------------+-------+
But I think the best solution is to just loop through the returned values with an index and use the index to keep track of the position.
SELECT
id, name, score,(select count(*) FROM highscores AS higherscores WHERE higherscores.score>currentscores.score)+1 AS rank
FROM
highscores AS currentscores
WHERE name="john"
;
Just use ORDER BYscoreDESC in the end of your query to sort them in reversed order (from high to low). When iterating and showing it in PHP you can use a count:
Query e.g.: SELECT * FROM scores ORDER BY score DESC
$count = 1;
$sql = 'SELECT * FROM scores ORDER BY score DESC';
$result = mysql_query($sql) or die(mysql_error());
while($fetch = mysql_fetch_object($result)) {
echo $count . ' ' . $fetch->score . '<br />';
$count++;
}