Question

我想显示此结果

Bodypart1，Bodypart2

表Excersise Image
表体部 Image

我的代码：

$sql = "
SELECT x.*
     , b.bodypart_name bodypart
  FROM exercises x
     , equipments e
     , bodyparts b
 WHERE x.exercise_bodypart = b.bodypart_id
 ";

$statement = $pdo->prepare($sql);
$statement->execute();
$exercises = $statement->fetchAll();

?>

<table>
<th>Bodypart</th>
</tr>
</thead>
<?php foreach($exercises as $exercise): ?>

echo $exercise['bodypart'];


?></span></td>
</tr>
<?php endforeach; ?>
</table>

Answer 1

$statement->fetchAll(PDO::FETCH_ASSOC) //将结果提取到关联数组中。

你也忘记了开场<td><span>

Answer 2

首先，您的数据库结构有点偏。

#Your excercise_bodypart field contains multiple keys, remove the field.
ALTER TABLE test_answer_override
DROP COLUMN exercise_bodypart;

#Build a bridge table which will establish a many to many relationship between exercises and bodyparts.
CREATE TABLE exercise_body_part
    exercise_id INT NOT NULL,
    bodypart_id INT NOT NULL,
    PRIMARY KEY (exercise_id,bodypart_id),
    KEY exercise_id (exercise_id),
    KEY bodypart_id (bodypart_id),
    CONSTRAINT fk_exercise_bodypart FOREIGN KEY (`bodypart_id`) REFERENCES `bodyparts`(`bodypart_id`) ON UPDATE CASCADE ON DELETE CASCADE,
    CONSTRAINT fk_bodypart_exercise FOREIGN KEY (`exercise_id`) REFERENCES `exercise`(`exercise_id`) ON UPDATE CASCADE ON DELETE CASCADE;

#Insert the rows needed that were removed with the deletion of exercise_bodypart field in step one above.
INSERT INTO excercise_body_part(exercise_id,bodypart_id) VALUES(5,8),(5,9);

你在这里缺少的是一个加入。出于这个问题的目的，尽管没有这样规定，我将假设你真的想要的不只是列出身体部分，但也会列出练习。现在让我们来解决这个问题：

SELECT 
  exercises.*,
  bodyparts.bodypart_name AS 'bodypart' 
FROM bodyparts 
JOIN exercise_body_part ON bodyparts.bodypart_id=exercise_body_part.bodypart_id
JOIN excercise ON exercise_bodypart.exercise_id=exercise.exercise_id
ORDER BY bodypart;

就你的HTML而言，看起来你有一个错误，你不匹配你的PHP标签。也许这就是你要找的东西？

<table>
  <tr>
    <th>Body Part</th>
    <th>Exercises</th>
  <tr>
<?php 
$currentBodypart="";
foreach($exercises as $exercise){
  if($currentBodypart!=$exercise['bodypart']){
    $currentBodypart=$exercise['bodypart'];
    //create a new row and add the first column with the bodypart name ?>
  <tr>
  <td><?=$exercise['bodypart'];?></td><td>
  <?php } else { ?>
  <table border="1">
    <tr>
        <th>Title</th>
        <th>Reps</th>
    </tr>
    <tr>
        <td><?=$exercise['exercise_title'];?></td>
        <td><?=$exercise['exercise_reps'];?></td>
    </tr>
  </table>
  <?php 
  }
  if($currentBodypart!=$exercise['bodypart']){
    //close of the cell containing exercises and end the new row ?>
  </td></tr>
  <?php }
}?>
</table>

没有PHP的HTML可能看起来像这样：

<html>
  <body>
    <table border="1">
      <tr>
        <th>Bodypart</th>
        <th>Exercises</th>
      <tr>
      <tr>
        <td>Bodypart 1</td>
        <td>
          <table>
            <tr>
                <th>Title</th>
                <th>Reps</th>
            </tr>
            <tr>
                <td>Exercise 1</td>
                <td>12</td>
            </tr>
          </table>
        </td>
      </tr>
    </table>
  </body>
</html>

将您的任务分解成更小的部分，您将获得成功。首先确保查询正在运行。然后确保PHP正常工作，并使用var_dump（）函数从数据库中获取数据。然后，尝试使用简单的HTML示例来使模板正确。最后，通过PHP将从MySQL中提取的数据插入到您的页面中。

如何显示/回显与多个id相关的名称

2 个答案: