PDO选择并计数“有差异”以显示在下拉菜单中

Question

I have this mysql select working great. It returns the proper data. I can't seem to get the context correct to place the actual count of the custnum so it will appear on the end of the dropdown option select. This statement returns the proper location names

$statement = $pdo->prepare("SELECT locationname FROM location WHERE locationname IN (SELECT locationname FROM location_user WHERE custnum= :custnum GROUP BY locationname HAVING COUNT( DISTINCT email) < 6 )");
  $statement->execute(array(':custnum' => $session->custnum));

 while($row = $statement->fetch(PDO::FETCH_ASSOC)){
echo'<option value="'.$row['locationname'].'">'.$row['locationname'].'('. $row['COUNT(total)'] .')</option>';    
}

Here's one of my attempts to grab the total for each custnum

$statement = $pdo->prepare("SELECT locationname, COUNT(custnum) AS total FROM location WHERE  locationname IN (SELECT locationname FROM location_user WHERE custnum= :custnum GROUP BY locationname HAVING COUNT( DISTINCT email) < 6 )");
$statement->execute(array(':custnum' => $session->custnum));

while($row = $statement->fetch(PDO::FETCH_ASSOC)){
echo'<option value="'.$row['locationname'].'">'.$row['locationname'].'('. $row['total'] .')</option>';    
}

Here's my tables

       table location                            table location_user
custnum   |   locationname             custnum   |    locationname  |   email    |  userlevel
   1           location1                 1            location1       1me@you.com       3
   1           location2                 1            location1       1me@you.com       1
                                         1            location1       2me@you.com       2
                                         1            location1       3me@you.com       2
                                         1            location1       4me@you.com       2
                                         1            location1       5me@you.com       2
                                         1            location2       1me@you.com       2
                                         1            location2       1me@you.com       3

The first select returns

 location1()
 location2()

The second select returns

 location1(2)

I actually need the count of the distinct email which the query is doing and returning only the locationnames of the distinct email in the table less then 6 but how do I get the actual number of distinct emails for each locationname.

This select will retrieve the total for DISTINCT email, but how do I combine the two into one for my while loop?

$statement2 = $pdo->prepare("SELECT COUNT(email) AS total FROM location_user WHERE custnum= :custnum GROUP BY locationname HAVING COUNT( DISTINCT email) < 6");
$statement2->execute(array(':custnum' => $session->custnum));

Here's the working version from the help of Peter and a little prodding from Tin.

 $statement = $pdo->prepare("SELECT l.locationname, COUNT(DISTINCT lu.email) AS total 
 FROM location l LEFT JOIN location_user lu ON l.locationname = lu.locationname AND l.custnum = lu.custnum WHERE l.custnum = :custnum GROUP BY l.locationname HAVING COUNT(DISTINCT lu.email)  < 5 ");
 $statement->execute(array(':custnum' => $session->custnum));

 while($row = $statement->fetch(PDO::FETCH_ASSOC)){
 echo'<option value="'.$row['locationname'].'">'.$row['locationname'].'('. $row['total'] .')</option>';    
 }

Here's another version that I'm working on to skip the user that adds the locations to the table. This user will always have a userlevel > 2. The uselevel is placed in the location_user table only as a value between 1-9. So I still need the location name but I don't want their location included in the count. I just realized that I could actually go a better route because the only email that I want to count will have a userlevel of 2. I was using the distinct email to filter out the userlevel of 1. I'll give it a go. The below version drops my locations that arn't in the location_user table but it's returning the proper count.

     SELECT l.locationname, COUNT(lu.userlevel) AS total 
     FROM location l LEFT JOIN location_user lu
     ON l.locationname = lu.locationname
     AND l.custnum = lu.custnum
     WHERE l.custnum = :custnum
     AND lu.userlevel = 2
     GROUP BY l.locationname
     HAVING COUNT(lu.userlevel) < 6

Answer 1

UPDATE2: based on your comments. Try it this way

SELECT l.locationname, COUNT(DISTINCT lu.email) AS total 
  FROM location l LEFT JOIN location_user lu
    ON l.locationname = lu.locationname
   AND l.custnum = lu.custnum
   AND lu.userlevel < 3 -- consider only users with user level < 3
 WHERE l.custnum = ?
 GROUP BY l.locationname 
HAVING COUNT(DISTINCT lu.email) < 6

Sample output:

| LOCATIONNAME | TOTAL |
|--------------|-------|
|    location1 |     5 |
|    location2 |     1 |
|    location3 |     0 |

Here is SQLFiddle demo

Answer 2

您实际上不需要从表location查询，因为您已经从表location_user获得了locationname字段

SELECT locationname, count(DISTINCT email) as total FROM location_user WHERE custnum = :custnum GROUP BY locationname HAVING count(DISTINCT email) < 6