PHP / MySQL：在列中显示与特定值匹配的所有行

Question

解决了
Sloan Trasher的回答非常有效。 我添加了另一组if语句，用于检查subcat_id并在依据之前打印正确的标题。

if($curr_sub_cat != $row['subcat_id']) {
  if($curr_sub_cat != '') {
    echo "</div>\n";
  }
  $curr_sub_cat = $row['subcat_id'];
  if($curr_sub_cat == '1') {
    echo "<h1 class='anchor' id='smartphones'>Smartphones</h1>";
    echo "<div class='flexed-boxes'>\n";
  }
  if($curr_sub_cat == '2') {
    echo "<h1 class='anchor' id='tablets'>Tablets</h1>";
    echo "<div class='flexed-boxes'>\n";
  }
}

原始问题
我有一个数据库表，看起来像这样：

dev_id  dev_name         cat_id  subcat_id

1       Apple iPhone 6   1       1
2       Apple iPhone 6s  1       1
3       Apple iPad Air   1       2

dev_id是主键，每个品牌的cat_id都不相同（Apple = 1，Samsung = 2，等等）， subcat_id指示设备是手机（1）还是平板电脑（2）。

我想在标题为Smartphones的一个列表中回显分配给subcat_id == 1的一个品牌的所有设备。 列subcat_id == 2的所有行都需要进入带有标题Tablets的列表的下方。 这两个列表都需要按主键升序排序。

像这样使用PDO::FETCH_GROUP和PDO::FETCH_COLUMN ：

 $sql = "SELECT subcat_id,dev_id FROM devices WHERE cat_id='$cat_id' ORDER BY dev_id";
 $data = $db->query($sql)->fetchAll(PDO::FETCH_GROUP|PDO::FETCH_COLUMN);

用所有（在本例中为Apple）设备很好地打印阵列：

Array
(
[0] => Array
    (
        [0] => 1
        [1] => 2
        [2] => 3
        [3] => 4
        [4] => 5
        [5] => 6
        [6] => 7
        [7] => 8
        [8] => 9
        [9] => 10
        [10] => 11
        [11] => 12
    )

[1] => Array
    (
        [0] => 13
        [1] => 14
        [2] => 15
        [3] => 16
        [4] => 17
        [5] => 18
    )
)

如果我实现一个for循环，如下所示：

 $cat_name_lower=strtolower($cat_name);
 $dev_name_ul=str_replace(array("$cat_name "," "),array("","_"),$row_all_dev['dev_name']);

 foreach ($data as $dev => $dev_data)
 {
   echo "<div class='flexed-boxes'>
        ";
     foreach ($dev_data as $row)
     {
       echo "<a class='singlebox trpl' href='/devices/".$cat_name_lower."/smartphone_details.php?dev_name=".$dev_name_ul."'>
              <img class='singlebox-media' src='".$cat_name_lower."/img/".$row_all_dev['dev_img']."' alt='".$row_all_dev['dev_name']."'>
              <span class='singlebox-header'>".$row_all_dev['dev_name']."</span>
            </a>
            ";
     }
   echo "</div>";
 }

我得到表中的第一个条目（iPhone 6，在我的实际数据库中，dev_id为6）打印了18 张图片以进行说明 。 但是，我希望这些行像这样显示 （图片的第二行包含iPhone 6、6 Plus和6s，它们恰好具有相同的图片）。
对我来说，似乎正确地循环了for循环，但是内部数组指针未提前一个？ 我要去哪里错了？

这个问题是一个跟进到这一个

Answer 1

尝试这个：

$sql = "";
$sql .= "SELECT\n";
$sql .= "   subcat_id,\n";
$sql .= "   dev_id,\n";
$sql .= "   dev_img,\n";    //  Not sure if that column exists in the table???
$sql .= "   dev_name\n";
$sql .= "FROM devices\n";
$sql .= "WHERE cat_id='$cat_id'\n";
$sql .= "ORDER BY subcat_id, dev_id";

$data = $db->query($sql)->fetchAll();

$cat_name_lower = strtolower($cat_name);

echo "<div class='flexed-boxes'>\n";
$curr_sub_cat = '';
foreach($data as $row_no => $row) {
    $dev_name_ul = str_replace(array("$cat_name "," "),array("","_"),$row['dev_name']);

    if($curr_sub_cat != $row['subcat_id']) {
        if($curr_sub_cat != '') {
            echo "</div>\n";
            echo "<div class='flexed-boxes'>\n";
        }
        $curr_sub_cat = $row['subcat_id'];
    }
    echo "<a class='singlebox trpl' href='/devices/".$cat_name_lower."/smartphone_details.php?dev_name=".$dev_name_ul."'>
              <img class='singlebox-media' src='".$cat_name_lower."/img/".$row["dev_img"]."' alt='".$row['dev_name']."'>
              <span class='singlebox-header'>".$row['dev_name']."</span>
            </a>\n";
}
echo "</div>\n";
//  echo "<p>\$data:<pre>".print_r($data,true)."</pre></p>\n";

Answer 2

试试这个，并在您的问题中显示结果。 这还不是答案，但是会提供给您答案所需的信息。

$sql = "";
$sql .= "SELECT\n";
$sql .= "   subcat_id,\n";
$sql .= "   dev_id,\n";
$sql .= "   dev_name\n";
$sql .= "FROM devices\n";
$sql .= "WHERE cat_id='$cat_id'\n";
$sql .= "ORDER BY subcat_id, dev_id";

$data = $db->query($sql)->fetchAll();

echo "<p>\$data:<pre>".print_r($data,true)."</pre></p>\n";