Question

这是一个假设的问题。如果我有来自3个独立的sql db查询的3个数组，这些数组都与另一个相关。例如......

//db
schools
id | school_name

classes
id | class_name | school_id

students
id | student_name | class_id

我想在这样一个巨大的列表中显示所有内容......

//php
foreach(schools as school){

    echo '<h1>' . $school->school_name . '<h1>';

    foreach(classes as class){

        if($class->school_id == $school->id){

            echo '<h2>' . $class->class_name . '<h2>';

            foreach(students as student){

                if($student->class_id == $class->id){

                    echo '<h3>' . $student->student_name . '<h3>';

                }
            }
        }
    }
}

我必须进行3次数据库调用。有没有办法在单个数据库查询中获取所有这些信息？就像数组中的数组中的数组然后以某种方式循环？或者这是最好的方法吗？

Answer 1

你可以进行一次加入，每次加入一次。你想要一切或任何类型的过滤器吗？

Answer 2

您可以加入这些表格，以获得一个包含展平数据的大数组。循环遍历此数据时，您可以检查上一条记录的ID是否仍与当前记录的ID匹配。如果没有，您可以输出新的标题。但重要的是，结果集已正确排序。

SELECT
  s.id AS school_id,
  s.school_name,
  c.id AS class_id,
  c.class_name,
  st.id AS student_id,
  st.student_name
FROM
  schools s
  INNER JOIN classes c ON c.school_id = s.id
  INNER JOIN students st ON st.class_id = c.id
ORDER BY
  s.id,
  c.id,
  st.id

如果你把它全部放在扁平结构中，你甚至可以把它变成一个嵌套的数组结构：

foreach ($resultset as $row)
{
    $schools[$row->school_id]->school_name = 
        $row->school_name;
    $schools[$row->school_id]->classes[$row->class_id]->class_name = 
        $row->class_name;
    $schools[$row->school_id]->classes[$row->class_id]->students[$row->student_id]->student_name =
        $row->student_name;
}
var_dump($schools);

之后，您仍然可以使用嵌套的for循环来处理数组，但它将以更有效的方式，因为数据已经被整理出来：类已经添加到他们所属的学校，并且学生已添加到正确的课程中。

Answer 3

您可以在一个SQL查询中完成所有操作，可能类似于：

SELECT schools.schoolname, classes.class_name, students.student_name
FROM
    schools INNER JOIN classes ON (schools.id = classes.school_id)
    INNER JOIN students ON (classes.id = students.class_id)
ORDER BY 1, 2;

然后你可以在一个循环中遍历结果集，但是你可能想要添加一些逻辑，以便每次更改时只显示学校名称和类名。

Answer 4

$res = mysql_query('SELECT school_name, class_name, student_name, sc.id AS scid, c.id AS cid, st.id AS stid FROM schools sc LEFT JOIN classes c ON (sc.id = c.school_id) LEFT JOIN students st ON (c.id = st.class_id) ');  
$arr = array();
while ($v = mysql_fetch_assoc($res)) {
 $arr[$v['school_name']][$v['class_name']][$v['stid']] = $v['student_name'];
}

print_r($arr);

Answer 5

<?php

try {
    $pdo = new PDO("mysql:host=127.0.0.1;dbname=school", "username");
} catch (PDOException $e) {
    echo "PDO Connection failed: " . $e->getMessage();
    exit(1);
}

$sql = <<<SQL

    SELECT schools.school_name, classes.class_name, students.student_name
    FROM
        schools INNER JOIN classes ON (schools.id = classes.school_id)
        INNER JOIN students ON (classes.id = students.class_id)
    ORDER BY 1, 2;

SQL;

$result = $pdo->query($sql);
if ($result == false) {
    die("query failed?!");
}

$school = "";
$class = "";
while ($row = $result->fetch(PDO::FETCH_ASSOC)) {
    if ($school != $row['school_name']) {
        $school = $row['school_name'];
        echo "\nSchool:  $school\n\n";
    }
    if ($class != $row['class_name']) {
        $class = $row['class_name'];
        echo "   Class:  $class\n\n";
        echo "      Student list:\n";
    }
    echo "         {$row['student_name']}\n";
}

php foreach in foreach in foreach

5 个答案: