resultaten filteren op twee geselecteerde opties

SELECT c.gender, COUNT(*) AS 'count'
FROM ads a 
INNER JOIN 
    (SELECT fieldvalue, fieldtitle AS country FROM field_values) b 
        ON b.fieldvalue = a.ad_country
INNER JOIN
    (SELECT fieldvalue, fieldtitle AS gender FROM field_values) c
        ON c.fieldvalue = a.ad_gender
GROUP BY c.gender

U kunt de volgende filters toevoegen vóór GROUP BY :

• Jaar:WHERE YEAR(a.ad_birthday) = '2012'
• Land:WHERE b.country = 'Albania'

Zie het in actie .

Nadat u uw mysql_ . heeft geconverteerd functies naar mysqli_ of PDO (omdat het verouderd is ), kunt u het resultaat eenvoudig per geslacht weergeven:

| GENDER | COUNT |
------------------
| Female |     2 |
|   Male |     1 |

Update 1

Om deze code te implementeren, kun je zoiets proberen (niet getest):

$link = mysqli_connect("localhost", "user_name", "password", "stock");

if (mysqli_connect_error()) {
die('Connect Error (' . mysqli_connect_errno() . ') ' . mysqli_connect_error());
}

$stmt = mysqli_prepare($link, "SELECT c.gender, COUNT(*) AS 'count' FROM ads a 
INNER JOIN (SELECT fieldvalue, fieldtitle AS country FROM field_values) b ON b.fieldvalue = a.ad_country
INNER JOIN (SELECT fieldvalue, fieldtitle AS gender FROM field_values) c ON c.fieldvalue = a.ad_gender
WHERE b.country = ? AND (YEAR(a.ad_birthday) = ? OR YEAR(a.ad_birthday) <> NULL) GROUP BY c.gender");

mysqli_bind_param($stmt, 'ss', $country, $year) or die(mysqli_error($dbh));

$result = mysqli_stmt_execute($stmt) or die(mysqli_error($link));

while($row = mysqli_fetch_assoc($result)) {
echo "There are " . $row[count] . ' ' . $row[gender] . "<br />\n";
}

mysqli_close($link);

Update 2

Aangezien u mysqli . niet kunt gebruiken om de een of andere reden zou de onderstaande code moeten werken. Houd er rekening mee dat het ervan uitgaat dat het land niet leeg is.

$query = "SELECT c.gender, COUNT(*) AS 'count' FROM ads a 
    INNER JOIN (SELECT fieldvalue, fieldtitle AS country FROM field_values) b ON b.fieldvalue = a.ad_country
    INNER JOIN (SELECT fieldvalue, fieldtitle AS gender FROM field_values) c ON c.fieldvalue = a.ad_gender
    WHERE b.country = " . mysql_real_escape_string($country);

if(isset($year)) $query .= " AND YEAR(a.ad_birthday) = " . mysql_real_escape_string($year);

$query .= ' GROUP BY c.gender';

$sql2 = mysql_query($query);
while($row = mysql_fetch_assoc($sql2)) {
    echo "There are " . $row[count] . ' ' . $row[gender] . "<br />\n";
}