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Possible to optimize/merge this kind of query scenario?

Posted on 2013-01-07
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Last Modified: 2013-01-08
Hi,
I have 3 queries and they are doing the exact same thing other than one difference (The TextSubscribed column).
Seems odd to do 3 different ones when all I'm trying to do is know how many of each.
Can the code be combined to query once and somehow break it out to 3 different variables instead of 3 separate requests? I'm just trying to find out how many 'Yes, No and Pendings there are.

$result = mysql_query("SELECT * FROM subscribers WHERE ContestId='$contest_id' AND TextSubscribed = 'Yes'");
$subscribed = mysql_num_rows($result);
mysql_free_result($result);

$result = mysql_query("SELECT * FROM subscribers WHERE ContestId='$contest_id' AND TextSubscribed = 'Pending'");
$pending = mysql_num_rows($result);
mysql_free_result($result);

$result = mysql_query("SELECT * FROM subscribers WHERE ContestId='$contest_id' AND TextSubscribed = 'No'");
$unsubscribed = mysql_num_rows($result);
mysql_free_result($result);

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*I'm very much a newbie so if there is a much better way, please don't hesitate to point it out. Thanks.
0
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Question by:tjyoung
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11 Comments
 
LVL 11

Expert Comment

by:Guru Ji
ID: 38752327
You can just do one statement to show the count of all Yes, No and Pending like this

$result = mysql_query("SELECT Count(CASE WHEN TextSubscribed = 'Yes' THEN 1 END) as NumOfYes, Count(CASE WHEN TextSubscribed = 'Pending' THEN 1 END) as NumOfPending, Count(CASE WHEN TextSubscribed = 'No' THEN 1 END) as NumOfNo FROM subscribers WHERE ContestId='$contest_id'");
$unsubscribed = mysql_num_rows($result);
mysql_free_result($result);

Open in new window

0
 
LVL 13

Expert Comment

by:darren-w-
ID: 38752546
$result = mysql_query("SELECT * FROM subscribers WHERE ContestId='$contest_id' AND TextSubscribed  IN ('Yes'","Pending","No");
$subscribed = mysql_num_rows($result);
mysql_free_result($result);
0
 
LVL 1

Author Comment

by:tjyoung
ID: 38752886
I think it must be the way I worded my issue: the solutions seem to give me a solitary 'how many of X' but they don't give me: how many Yes, how many No and how many Pendings.
I'm trying to figure out how many of each of the 3 possible answers.
0
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LVL 13

Expert Comment

by:darren-w-
ID: 38752928
Select subscribers , count(subscibers) where ContestId='$contest_id' AND TextSubscribed  IN ('Yes'","Pending","No");
0
 
LVL 1

Author Comment

by:tjyoung
ID: 38752966
@darren-w-
Within my original question, if I was to run those 3 queries, I would have ended up with a count for how many of each (numbers are just for example):
$subscribed = 8
$unsubscribed = 4
$pending = 2

Problem was it seemed kind of redundant to run 3 queries on the same table one after the other to get a value for each

I'm trying to figure out how many of EACH "Yes, No and Pending" there are and assign each to a variable so I can output to my page.

Kind of like:
"Last week you had <php echo $subscribed> subscribers, <?php echo $unsubscribed?> people unsubscribed and there are still <?php echo $pending?> people who are pending.
0
 
LVL 11

Accepted Solution

by:
Guru Ji earned 500 total points
ID: 38753005
<?php
$result = mysql_query("SELECT Count(CASE WHEN TextSubscribed = 'Yes' THEN 1 END) as NumOfYes, Count(CASE WHEN TextSubscribed = 'Pending' THEN 1 END) as NumOfPending, Count(CASE WHEN TextSubscribed = 'No' THEN 1 END) as NumOfNo FROM subscribers WHERE ContestId='$contest_id'");

if (!$result) {
    echo 'Could not find any data: ' . mysql_error();
    exit;
}
$row = mysql_fetch_row($result);

echo $row[0]; // Number of subscribed
echo $row[1]; // Number of pending
echo $row[2]; // Number of unsubscribed

?>
Now you have all three Yes, Pending and No in a php variable to be used anywhere in your code.

Hope thats what you wanted
0
 
LVL 13

Expert Comment

by:darren-w-
ID: 38753013
What does the query produce

Not tested but expect something like
Array(
Yes=>4,
No=>6,
Pending=>7
);
0
 
LVL 1

Author Comment

by:tjyoung
ID: 38753947
@darren-w-
I appreciate your efforts but the code simply failed. I noticed a number of quotation errors that I think I fixed but nevertheless couldn't get anything meaningful out of it. Maybe it was just me.
0
 
LVL 1

Author Closing Comment

by:tjyoung
ID: 38753949
Worked out of the box. Thanks!
0
 
LVL 13

Expert Comment

by:darren-w-
ID: 38754137
Hi,

This is the corrected code:
SELECT TextSubscribed, COUNT( * ) AS count
FROM subscribers
WHERE  `ContestId` = $contest_id
AND  `TextSubscribed` 
IN (
 "Yes",  "pending",  "No"
)

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Dump of db:

CREATE TABLE IF NOT EXISTS `subscribers` (
  `id` int(11) unsigned NOT NULL AUTO_INCREMENT,
  `ContestId` int(11) unsigned NOT NULL,
  `TextSubscribed` varchar(30) NOT NULL,
  PRIMARY KEY (`id`),
  UNIQUE KEY `id` (`id`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=7 ;


INSERT INTO `subscribers` (`id`, `ContestId`, `TextSubscribed`) VALUES
(1, 3, 'yes'),
(2, 3, 'pending'),
(3, 1, 'no'),
(4, 1, 'pending'),
(5, 2, 'pending'),
(6, 1, 'no');

Open in new window


with contest id = 1

TextSubscribed  count
no            2
pending   1
0
 
LVL 13

Expert Comment

by:darren-w-
ID: 38754145
write2mohit, are you able to explain how yours differs?, Was on my mobile last night so was not able to test it out
0

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