按性别和种族搜索和统计

Question

我只想知道任何有关性别和种族的想法。数据应显示如下：

-----------------------------
course   | Malay  | chinese | 
         | M | F  |  M | F  |
-----------------------------
science  | 0 | 1  |  0 | 0  |
-----------------------------
business | 0 | 0  | 1  | 0  |

//这是表结构

$sql = "CREATE TABLE student (
id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY, 
name VARCHAR(30) NOT NULL,
gender VARCHAR(30) NOT NULL,
race VARCHAR(50),
session VARCHAR(50),
course VARCHAR(50),
)";

示例数据

id |名字|性别|比赛|会议|当然

98773552 | nurul |女性|马来语| 2014年6月|科学计算机

5273544 |李|男性|中文| june2012 |商业研究

我已有的代码：

<?php
include "connect_db.php";
if ((isset($_POST['submit'])) AND ($_POST['search'] <> "")) {

    $search = $_POST['search'];

    $sql = "SELECT gender, count(gender) AS cnt FROM student WHERE session LIKE '%$search%' AND gender IN ('Male', 'Female') GROUP BY gender" or die(mysqli_error(
        ''
    ));
    $result = $conn->query($sql);

    while ($row = fetch_assoc($result)) {
        $dataSet->addPoint(new Point($row['gender'], $row['cnt']));

    }

}

Answer 1

作为初步想法，您希望生成一个包含所有可能组合的结果集。

理想情况下，你会有一个性别表，一个种族表和一个课程表，然后你可以交叉加入那些以获得所有组合，然后离开加入学生表来获得计数。

如果你没有使用子查询来获取可能的值，那么就会出现这样的情况： -

$sql = "SELECT c.course,
                r.race,
                g.gender,
                COUNT(s.id)
        FROM
        (
            SELECT DISTINCT race
            FROM student
        ) r
        CROSS JOIN
        (
            SELECT DISTINCT gender
            FROM student
            WHERE gender IN ('Male', 'Female')
        ) g
        CROSS JOIN
        (
            SELECT DISTINCT course
            FROM student
        ) c
        LEFT OUTER JOIN student s
        ON r.race = s.race
        AND g.gender = s.gender
        AND c.course = s.course
        AND session LIKE '%".$conn->escape($search)."%' 
        GROUP BY c.course, r.race, g.gender" or die(mysqli_error(''));

然后只是简单的php将它放在一个表中，当课程改变时推出一个新的表格行。

修改

以下是完全未经测试的，但希望能让您了解如何使用2个标题行回显表： -

<?php

include "connect_db.php";
if ((isset($_POST['submit'])) AND ($_POST['search'] <> "")) 
{

    $search = $_POST['search'];

$sql = "SELECT c.course,
                r.race,
                g.gender,
                COUNT(s.id) AS res_cnt
        FROM
        (
            SELECT DISTINCT race
            FROM student
        ) r
        CROSS JOIN
        (
            SELECT DISTINCT gender
            FROM student
            WHERE gender IN ('Male', 'Female')
        ) g
        CROSS JOIN
        (
            SELECT DISTINCT course
            FROM student
        ) c
        LEFT OUTER JOIN student s
        ON r.race = s.race
        AND g.gender = s.gender
        AND c.course = s.course
        AND session LIKE '%".$conn->escape($search)."%' 
        GROUP BY c.course, r.race, g.gender" or die(mysqli_error(
        ''
    ));
    $result = $conn->query($sql);

    echo "<table>";
    $header1 = "<tr><td rowspan='2'>course</td>";
    $header2 = "<tr>";
    $out_row = "<tr>";
    $prev_course = '';
    $prev_race = '';
    $first_row = true;
    while ($row = $conn->fetch_assoc($result)) 
    {
        if ($prev_course != $row['course'] and $prev_course != '')
        {
            if ($first_row)
            {
                echo $header1."</tr>";
                echo $header2."</tr>";
                $first_row = false;
            }
            echo $out_row."</tr>";
            $out_row = "<tr>";
            $prev_course = $row['course'];
        }
        if ($first_row)
        {
            if ($prev_race != $row['race'])
            {
                $header1 .= "<td colspan='2'>".$row['race']."</td>";
                $prev_race = $row['race'];
            }
            $header2 .= "<td>".$row['gender']."</td>";
        }
        $out_row .= "<td>".$row['res_cnt']."</td>";
    }
    if ($prev_course != $row['course'] and $prev_course != '')
    {
        if ($first_row)
        {
            echo $header1."</tr>";
            echo $header2."</tr>";
            $first_row = false;
        }
        echo $out_row."</tr>";
        $out_row = "<tr>";
        $prev_course = $row['course'];
    }
    echo "</table>";

}