在 MySQL 中使用 COUNT 时如何返回 0 而不是 null

Question

I am using this query to return return a list of songs stored in $sTable along with a COUNT of their total projects which are stored in $sTable2.

 /*
     * SQL queries
     * Get data to display
     */
    $sQuery = "
        SELECT SQL_CALC_FOUND_ROWS ".str_replace(" , ", " ", implode(", ", $aColumns))."

    FROM $sTable b 
    LEFT JOIN (
   SELECT COUNT(*) AS projects_count, a.songs_id

   FROM $sTable2 a
   GROUP BY a.songs_id
) bb ON bb.songs_id = b.songsID


        $sWhere
        $sOrder
        $sLimit
    ";
    $rResult = mysql_query( $sQuery, $gaSql['link'] ) or die(mysql_error());

'projects_count' is put into an array along with the columns in '$sTable', this is then spat out via JSON and displayed in a table on page.

This is working perfectly apart from when a song has no projects linked to it. It of course returns NULL.

All I want is for any null values to be returned as a '0'.

I have tried the COUNT(), COUNT(IFNULL (project_id,0) and using COUNT(DISTINCT)...

And also:-

SELECT COALESCE(COUNT(*),0) AS projects_count, a.songs_id

All without success.

Any ideas?

Answer 1

Use the COALESCE() function. COALESCE() takes at least 2 arguments, calculated in order, and returns the first non-null argument. So COALESCE(null, 0) would return 0 , and COALESCE(null, null, null, null, 1) would return 1 . Here's MySQL's documentation about COALESCE() .

In re-reading your query, you should be able to get the results you want like this:

SELECT <all the fields you want>, b.songsID, COUNT(*) AS projects_count
FROM $sTable b
LEFT OUTER JOIN $sTable2 bb ON bb.songs_id = b.songsID
$sWhere
GROUP BY b.songsID
$sOrder
$sLimit

Like I said, this should work, but something about it doesn't feel quite right.

Answer 2

COALESCE() returns the first non-null argument. So if you say COALESCE(count(...),0) it will return the count(...) if it's not null, or it will return 0 if the count(...) is null

Answer 3

You don't have to do the join with a subquery. The following should work just fine without the COALESCE etc:

SELECT ".str_replace(" , ", " ", implode(", ", $aColumns)).", 
SUM(b.songsID is not null) as countprojects
FROM $sTable b 
LEFT JOIN $sTable2 a ON a.songs_id=b.songsID
GROUP BY ".str_replace(" , ", " ", implode(", ", $aColumns))."

This will return what you ask for in countprojects.

The way it works: The LEFT JOIN just makes certain you get all data. You can't use COUNT because that would return 1 for the NULL rows. But, if you just use the fact that a boolean TRUE evaluates to 1, and a boolean FALSE evaluates to 0, you can SUM over those results.

Answer 4

Simply add this line in your code after SELECT

IF(projects_count IS NULL, 0, projects_count) As projects_countList

Like This:

$sQuery = "
    SELECT SQL_CALC_FOUND_ROWS ".str_replace(" , ", " ", implode(", ", $aColumns)).",
     IF(projects_countIS NULL, 0, projects_count) As projects_countList

FROM $sTable b 
LEFT JOIN (SELECT COUNT(*) AS projects_count, a.songs_id FROM $sTable2 a GROUP BY a.songs_id) bb ON bb.songs_id = b.songsID
    $sWhere
    $sOrder
    $sLimit
";
$rResult = mysql_query( $sQuery, $gaSql['link'] ) or die(mysql_error());

To return 0 instead of null in MySQL
USE

SELECT id, IF(age IS NULL, 0, age) FROM tblUser

USE with count() having join 2 tables
like

SELECT 
    tblA.tblA_Id,
    tblA.Name,
    tblC.SongCount,
    IF(tblC.SongCount IS NULL, 0, tblC.SongCount) As noOfSong
  FROM  tblA
    LEFT JOIN
    (   
        SELECT 
            ArtistId,count(*) AS SongCount 
        FROM 
            tblB  
        GROUP BY 
            ArtistId
    ) AS tblC
    ON 
        tblA.tblA_Id = NoOfSong.ArtistId

And Result is

tblA_Id    Name     SongCount   noOfSong
-----------------------------------------------------
7          HSP      NULL        0
6          MANI     NULL        0
5          MEET     1           1
4          user     NULL        0
3          jaani    2           2
2          ammy     NULL        0
1          neha     2           2 

Answer 5

SELECT blahblahblah, IFNULL(bb.projects_count, 0)
FROM $sTable b
etc...