Question

因此，当客户端上的用户从下拉列表中选择一个项目时，应该显示一个包含查询结果的新div容器。我遇到的问题是div容器按预期弹出，但没有包含查询搜索结果的标记。

兴趣点

data.php

<?php
 if(!isset($_GET['name'])) die('A vitamin name is required.');
require '../vitamins.php';

$pdo = new PDO('mysql:host=localhost;dbname=vitamins', 'root', 'root');

$table = new Vitamins($pdo);

$param = $_GET['name'];

$data = $table->getData($param);
?>

<div class="row">
<?php while($data): ?>
    <?php $i = 0;
    while($i < 3){ ?>
        <div class="column">
            <h1><?php print $data['name']; ?></h1>
            <p>serving</p>
            <p><?php print $data['grams']; ?>G = <?php print         $data['percentage']; ?>%</p>
        </div>
        <?$i++;
    }?>
        <br/>
<?php endWhile; ?>

查询方法vitamin.php

<?php
class Vitamins {

protected $db;

public function __construct(PDO $db) {
    $this->db = $db;
}

public function getAll()
{
    return $this->db->query('SELECT * FROM vitamins');
}

public function getData($name) {
  $nameEscaped = $this->db->quote($name);
  $sql = "SELECT 'v'.'name', 'f'.'name', 'p'.'grams', 'p'.'percentage',   'i'.'path' " .
      "FROM 'vitamins' 'v' JOIN 'percentage' 'p' ON 'v'.'id' = 'p'.'vitamin_id' " .
        "JOIN 'food' 'f' ON 'f'.'id' = 'p'.'food_id' " .
        "JOIN 'images' 'i' ON 'i'.'id' = 'f'.'image_id' " .
      "WHERE 'v'.'name' = {$nameEscaped}";
  return $this->db->query($sql);
  }
}

index.php中的

脚本

<script>
$(function() {
  var trans = $('.transbox');
  var lower = $('.containerL');
  $('.dropdown li').click(function(e) {
      e.preventDefault();
        var name = $(e.target).html();

      $.get('./vitamins/data.php', { name: name }, function(data) {
          console.log(data);

          trans.html(data);
          trans.show();
      });


        lower.hide();
        console.log('clicked');
        console.log(trans, lower);
  });
});
</script>

jquery没有从服务器返回标记

0 个答案: