Question

我有这个查询来返回2个日期之间的访问者数量。

$SQLVisit = "SELECT     
            count(score) as counts,
            date FROM persons  
            WHERE date > '".$_SESSION['data1']."' AND date < '".$_SESSION['data2]."' 
            GROUP BY date 
            ORDER BY date asc"; 

$result = mysql_query($SQLVisit);   
$num = mysql_num_rows($result);

这些访客，有些是男性，有些是女性。

我已经有一个数组，可以告诉我访问次数和相应的日期。像这样：[＆＃39; 2012-1-1＆＃39;，50]。我的想法是有一个数组，可以保持日期，访问次数，男性人数和女性人数。像这样：[＆＃39; 2012-1-1＆＃39;，50,35.15]。

有什么想法吗？ ps：我正在使用PHP。

编辑：数组代码

$data[0] = array('day','counts');       
    for ($i=1; $i<($num+1); $i++)
    {
        $data[$i] = array(substr(mysql_result($result, $i-1, "date"), 0, 20),
            (int) mysql_result($result, $i-1, "counts"),);
    }   
    echo json_encode($data);

EDIT2：是的我有一个性别（int）列，男性为1，女性为2

Answer 1

如果您有gender列，请尝试

$SQLVisit = "SELECT  date, gender,   
                     count(score) as counts,
            FROM  persons  
            WHERE date > '".$_SESSION['data1']."' AND date < '".$_SESSION['data2]."' 
            GROUP BY date, gender 
            ORDER BY date asc";

如果您要格式化性别而不是数字，则希望将其显示为male或female

$SQLVisit = "SELECT  date, 
                     IF(gender = 1, 'Male', 'Female') gender,   
                     count(score) as counts,
            FROM  persons  
            WHERE date > '".$_SESSION['data1']."' AND date < '".$_SESSION['data2]."' 
            GROUP BY date, gender 
            ORDER BY date asc";

如果您还想要这样的格式，

date    Male      Female
===========================
'date'     5          6

使用以下查询

SELECT  DATE,
        SUM(CASE WHEN gender = 1 then 1 ELSE 0 END) Male,
        SUM(CASE WHEN gender = 2 then 1 ELSE 0 END) Female
FROM    person
-- WHERE clause
GROUP BY DATE
-- ORDER clause

您的查询容易被SQL Injection攻击，请阅读以下文章

How can I prevent SQL injection in PHP?

Answer 2

您可以使用SUM（IF（））：

$SQLVisit = "SELECT     
            count(score) as counts,
            SUM(IF(gender=1,1,0)) as male,
            SUM(IF(gender=2,1,0)) as female,
            date FROM persons  
            WHERE date > '".$_SESSION['data1']."' AND date < '".$_SESSION['data2]."' 
            GROUP BY date 
            ORDER BY date asc";

Answer 3

你可以这样做：

SELECT     
  SUM(CASE WHEN Gender = 1 THEN 1 ELSE 0) malescounts,
  SUM(CASE WHEN Gender = 2 THEN 1 ELSE 0) femalescounts,
  count(score) as counts
FROM persons  
WHERE date > ...
GROUP BY date;

查询以返回基于性别的访客数量

3 个答案: