子查询计数

时间：2016-05-02 05:30:08

标签： php mysql

尝试JOIN 2个表根据第一个表格(student_schedule)计算种族， 学生可能出现超过1次的情况。表2 (bday)只让学生显示1 time by ID student's ethnicity。我正在使用LEFT JOIN，因为有时我可能在bday表中有一名失踪的学生，他们将不会被计算在内（因为他们的种族没有被宣布）。

 SELECT bday.Ethnicity, ROUND(COUNT(DISTINCT student_schedule.ID)/(SELECT  
 COUNT(DISTINCT student_schedule.ID) FROM student_schedule
 WHERE student_schedule.Course LIKE 'AS%')*100,2) AS "% of AS Population",  
 (SELECT COUNT(DISTINCT student_schedule.ID) FROM student_schedule AS 
 "Total Student Population") 
 FROM student_schedule LEFT JOIN bday ON student_schedule.ID=bday.ID WHERE    
 student_schedule.Course LIKE 'AS%' GROUP BY bday.Ethnicity
 ORDER BY COUNT(DISTINCT student_schedule.ID) DESC

结果是3列（种族，AS人口百分比，学生总人口数）。 enter image description here

为了比较课程Like 'AS%'的学生百分比， 我想补充一个专栏，给我一个按种族分组的整个学校的种族。换句话说，32％的白人学生参加AS而白人学校占30％。数据来自表2（bday），其中列出了每个学生的种族。它应该像

SELECT COUNT(bday.Ethnicity) 
FROM student_schedule
LEFT JOIN bday ON student_schedule.ID=bday.ID.

我坚持使用GROUP function所需的COUNT来分类(White, Black,...).

理想情况下，我的结果将是......

enter image description here

任何帮助都将不胜感激。

3 个答案:

答案 0 :(得分：1)

您可以尝试这样的事情：

<?php
 //NOTE: I AM USING PDO FOR SIMPLICITY... BUT ANY OTHER DBAL WORKS AS WELL... 
//DATABASE CONNECTION CONFIGURATION:
defined("HOST")     or define("HOST",   "localhost");           //REPLACE WITH YOUR DB-HOST
defined("DBASE")    or define("DBASE",  "_TEST_");              //REPLACE WITH YOUR DB NAME
defined("USER")     or define("USER",   "root");                //REPLACE WITH YOUR DB-USER
defined("PASS")     or define("PASS",   "root");                //REPLACE WITH YOUR DB-PASS



try {
    $dbh            = new PDO('mysql:host='.HOST.';dbname='. DBASE,USER,PASS);
    $dbh->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);


    $sql            = 'SELECT COUNT(SS.ID) AS "Global_Student_Population" FROM student_schedule AS SS ';
    $stmt           = $dbh->prepare($sql);
    $stmt->execute();
    $result         = $stmt->fetchAll(PDO::FETCH_COLUMN);
    $intAllStudents = (int)$result[0];

    $sql            = " SELECT BD.Ethnicity, {$intAllStudents}  AS Global_Student_Population,
                        COUNT(DISTINCT SS.ID) AS Ethnic_Student_Population,
                        ROUND(COUNT(DISTINCT SS.ID)*100/{$intAllStudents}) AS '%_of_AS_Population'
                        FROM student_schedule AS SS
                        LEFT JOIN bday AS BD ON SS.ID=BD.ID
                        WHERE SS.Course LIKE 'AS%' GROUP BY BD.Ethnicity";

    $stmt           = $dbh->prepare($sql);
    $stmt->execute();
    $result         = $stmt->fetchAll();

      var_dump($result);
      var_dump($intAllStudents);

    //GARBAGE COLLECTION
    $dbh        = null;
}catch(PDOException $e){
    echo $e->getMessage();
}

我的var_dump（）显示如下内容：

        array (size=3)
      0 => 
        array (size=8)
          'Ethnicity' => string 'Black' (length=5)
          0 => string 'Black' (length=5)
          'Global_Student_Population' => string '10' (length=2)
          1 => string '10' (length=2)
          'Ethnic_Student_Population' => string '3' (length=1)
          2 => string '3' (length=1)
          '%_of_AS_Population' => string '30' (length=2)
          3 => string '30' (length=2)
      1 => 
        array (size=8)
          'Ethnicity' => string 'Hispanic' (length=8)
          0 => string 'Hispanic' (length=8)
          'Global_Student_Population' => string '10' (length=2)
          1 => string '10' (length=2)
          'Ethnic_Student_Population' => string '1' (length=1)
          2 => string '1' (length=1)
          '%_of_AS_Population' => string '10' (length=2)
          3 => string '10' (length=2)
      2 => 
        array (size=8)
          'Ethnicity' => string 'White' (length=5)
          0 => string 'White' (length=5)
          'Global_Student_Population' => string '10' (length=2)
          1 => string '10' (length=2)
          'Ethnic_Student_Population' => string '2' (length=1)
          2 => string '2' (length=1)
          '%_of_AS_Population' => string '20' (length=2)
          3 => string '20' (length=2)

这是我的测试表定义：          -          - 表bday的表结构          - 

    CREATE TABLE `bday` (
    `ID` int(11) NOT NULL,
      `Ethnicity` varchar(255) NOT NULL
    ) ENGINE=InnoDB AUTO_INCREMENT=11 DEFAULT CHARSET=utf8;

    --
    -- Dumping data for table `bday`
    --

    INSERT INTO `bday` (`ID`, `Ethnicity`) VALUES
    (1, 'Black'),
    (2, 'Black'),
    (3, 'Black'),
    (4, 'White'),
    (5, 'Hispanic'),
    (6, 'White'),
    (7, 'Asian'),
    (8, 'Hispanic'),
    (9, 'White'),
    (10, 'Black');

    --
    -- Indexes for dumped tables
    --

    --
    -- Indexes for table `bday`
    --
    ALTER TABLE `bday`
      ADD PRIMARY KEY (`ID`);

    --
    -- AUTO_INCREMENT for dumped tables
    --

    --
    -- AUTO_INCREMENT for table `bday`
    --
    ALTER TABLE `bday`
      MODIFY `ID` int(11) NOT NULL AUTO_INCREMENT,AUTO_INCREMENT=11;




    --
    -- Table structure for table `student_schedule`
    --

    CREATE TABLE `student_schedule` (
    `ID` int(11) unsigned NOT NULL,
      `Course` varchar(255) NOT NULL
    ) ENGINE=InnoDB AUTO_INCREMENT=11 DEFAULT CHARSET=utf8;

    --
    -- Dumping data for table `student_schedule`
    --

    INSERT INTO `student_schedule` (`ID`, `Course`) VALUES
    (1, 'AS'),
    (2, 'AS'),
    (3, 'EN'),
    (4, 'EN'),
    (5, 'AS'),
    (6, 'AS'),
    (7, 'EN'),
    (8, 'EN'),
    (9, 'AS'),
    (10, 'AS');

    --
    -- Indexes for dumped tables
    --

    --
    -- Indexes for table `student_schedule`
    --
    ALTER TABLE `student_schedule`
      ADD PRIMARY KEY (`ID`);

    --
    -- AUTO_INCREMENT for dumped tables
    --

    --
    -- AUTO_INCREMENT for table `student_schedule`
    --
    ALTER TABLE `student_schedule`
      MODIFY `ID` int(11) unsigned NOT NULL AUTO_INCREMENT,AUTO_INCREMENT=11;

答案 1 :(得分：1)

你的SQL有点乱，让我们先重新构建它，使它更清晰可读：

SELECT bday.Ethnicity
        , ROUND(
            COUNT(*)/(SELECT COUNT(DISTINCT ID) FROM student_schedule
                                                WHERE Course LIKE 'AS%')
            *100,2) AS "% of AS Population"
FROM bday INNER JOIN student_schedule on bday.ID=student_schedule.ID
WHERE student_schedule.Course LIKE 'AS%'
GROUP BY bday.Ethnicity

请注意，我在这里使用INNER JOIN更好，如果有一些学生没有Ethnicity宣布你将获得有意义的百分比（并且可以在{{1}中过滤掉}子句）

现在我们要添加表WHERE中每个(count)的学生Ethnicity，为此，我们只需向{{1}添加bday list：

Sub-Query

请注意，我将子查询的表SELECT的{​​{1}}设置为SELECT bday.Ethnicity , ROUND( COUNT(*)/(SELECT COUNT(DISTINCT ID) FROM student_schedule WHERE Course LIKE 'AS%') *100,2) AS "% of AS Population" , ROUND( (SELECT COUNT(*) FROM bday a WHERE a.Ethnicity=bday.Ethnicity) /(SELECT COUNT(*) FROM bday) *100,2) AS "Ethnicity of School" FROM bday INNER JOIN student_schedule on bday.ID=student_schedule.ID WHERE student_schedule.Course LIKE 'AS%' GROUP BY bday.Ethnicity ，以避免在{{1}中使用它时与外部alias表冲突子查询中的子句。

答案 2 :(得分：0)

有时你必须在挂断电话时从头开始。我决定从全局COUNTS开始，然后使用CASE语句来执行WHERE计数。我终于按照我需要的方式工作了。我想出的解决方案是：

SELECT bday.Ethnicity, ROUND(COUNT(DISTINCT CASE WHEN 
student_schedule.Course LIKE 'AS%' THEN student_schedule.ID END)/(SELECT 
COUNT(DISTINCT student_schedule.ID) FROM student_schedule WHERE 
student_schedule.Course LIKE 'AS%')*100,2) AS '% of AS Population', 
ROUND(COUNT(DISTINCT student_schedule.ID)/(SELECT COUNT(DISTINCT 
student_schedule.ID) FROM student_schedule)*100,2) AS '% of Student 
Population'
FROM student_schedule INNER JOIN bday ON student_schedule.ID=bday.ID
GROUP BY bday.Ethnicity
ORDER BY COUNT(DISTINCT student_schedule.ID) DESC

结果是： enter image description here

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