php&mysql:将group by用于类别

我建议只使用一个简单的查询来获取所有行,并按类别ID排序。只有当类别的值与前一行的值不同时,才输出该类别。

<?php

$stmt = $pdo-> query("SELECT * FROM `myTable` ORDER BY categoryID");

$current_cat = null;
while ($row = $stmt->fetch()) {
  if ($row["categoryID"] != $current_cat) {
    $current_cat = $row["categoryID"];
    echo "Category #{$current_cat}\n";
  }
  echo $row["productName"] . "\n";
}

?>

这应该有效:

$categories = array();
$result= mysql_query("SELECT category_id, product_name  FROM `table` GROUP BY `catagory_id` ORDER BY `catagory_id`");
while($row = mysql_fetch_assoc($result)){
    $categories[$row['category_id']][] = $row['product_name'];
}

// any type of outout you like
foreach($categories as $key => $category){
    echo $key.'<br/>';
    foreach($category as $item){ 
        echo $item.'<br/>';
    }
}

你可以自己设计的输出。只需将所有内容添加到一个多维数组中,类别ID作为第一级键。

编辑 :结果数组可能如下所示:

$categories = array(
    'cateogy_id_1' => array(
        1 => 'item_1',
        2 => 'item_2',
        ...
    ),
    'cateogy_id_2' => array(
        1 => 'item_1',
        2 => 'item_2',
        ...
    ),
    ....
);

你想要的是按类别排序,而不是按组排序。

SELECT * FROM MyTable
ORDER BY category_id, product_id

当您遍历列表时,只要在类别_id更改时输出一个新的头。

这种方法不需要按类别ID对数据进行排序。

您可以使用PHP将每个类别分组在一个嵌套数组中。 之后,数据可以很容易地显示在表中,传递到图形/图表库等。

<?php
$stmt = $pdo-> query("SELECT * FROM myTable ORDER BY categoryID");


$categories = []; //the array to hold the restructured data

//here we group the rows from different categories together
while ($row = $stmt->fetch())
{
    //i'm sure most of you will see that these two lines can be performed in one step
    $cat = $row["categoryID"]; //category id of current row
    $categories[$cat][] = $row; //push row into correct array
}

//and here we output the result
foreach($categories as $current_cat => $catgory_rows)
{
    echo "Category #{$current_cat}\n";

    foreach($catgory_rows as $row)
    {
        echo $row["productName"] . "\n";
    }
}
?>

最简单的方法可能是提取类别列表,迭代并提取其附加产品。(即几个问题。)

例如(伪代码):

$result = fetch_assoc("SELECT category_id FROM categories");
foreach($result as $row)
{
  echo $row['category_id'];
  $result2 = fetch_assoc("SELECT product_id FROM products");
  foreach($result2 as $row2)
  {
    echo $row2['product_id'];
  }
}

如果您想在一个查询中完成这一切,可以执行如下操作:

$result = fetch_assoc("SELECT product_id, category_id FROM products p JOIN categories c ON p.category_id = c.category_id ORDER BY c.category_id ASC");
$last = null;
foreach($result as $row)
{
  # Output the category whenever it changes
  if($row['category_id'] != last)
  {
    echo $row['category_id'];
    $last = $row['category_id'];
  }
  echo $row['item_id'];
}

然后,您可以对结果集进行迭代,并在类别名称发生更改时将其拉出。

在从数据库中获取SQL之前,可能有一种更复杂、更优雅的方法来编写它来完成所有工作,但我并不那么聪明。;)

注意:示例使用伪代码。我使用的mysql抽象类的工作方式如下:

$db->query("SELECT stuff");
$db->multi_result(); // returns 2d-associative array

那时我可以 foreach($db->multi_result() as $row) 太懒了,太棒了。

如果您愿意,我可以发布抽象层的代码。

$stmt = $dbConnection->prepare("SELECT exam_id, COUNT(report_id) FROM bug_report GROUP BY exam_id; ");
        //$stmt->bind_param('s',$exam_id); 
        $stmt->execute();
        $extract = $stmt->get_result();
        $count = $extract->num_rows;
        if($count){
            while($rows = $extract->fetch_assoc()){
                $exam_id = $rows["exam_id"];
                switch($exam_id){
                    case "jee_questions":
                        $jeeBugCount = $rows["COUNT(report_id)"];
                        break;
                    case "gate_questions":
                        $gateBugCount = $rows["COUNT(report_id)"];
                        break;
                }
            }   
        }