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我目前正在从我的数据库中提取我需要的信息，但希望显示的信息根据下拉结果进行更改。我做了一些研究，并认为最好的方法是使用表格，但不确定它是如何工作的。

这是我下面的代码：

      <section id="compare">

    <select>
      <option>Select Gym</option>       
      <option>fitness first</option>
      <option>anytime fitness</option>
      <option>the gym</option>
      <option>its leisure ltd</option>
      <option>the armoury</option>
    </select>
    <select>
      <option>Select Gym</option>       
      <option>fitness first</option>
      <option>anytime fitness</option>
      <option>the gym</option>
      <option>its leisure ltd</option>
      <option>the armoury</option>      
    </select>

    <section id="left"> 
    <?php

    mysql_select_db("gyms", $con);
    $result = mysql_query("SELECT * FROM gym WHERE id='1'") or die ('Error: '.mysql_error ());

        while($row = mysql_fetch_array($result)){
          echo "<h1>" . $row['name'] . "</h1>";
          echo "<p><h6>type</h6>" . $row['type'] . "</p>";
          echo "<p><h6>price</h6>" . $row['price'] . "</p>";
          echo "<p><h6>hours</h6>" . $row['hours'] . "</p>";
          echo "<p><h6>parking</h6>" . $row['parking'] . "</p>";
          echo "<p><h6>facilities</h6>" . $row['facilities'] . "</p>";
          } 
    ?>      
    </section>

    <section id="right">
    <?php
    $result = mysql_query("SELECT * FROM gym WHERE id='2'") or die ('Error: '.mysql_error ());

        while($row = mysql_fetch_array($result2)){
          echo "<h1>" . $row['name'] . "</h1>";
          echo "<p><h6>type</h6>" . $row['type'] . "</p>";
          echo "<p><h6>price</h6>" . $row['price'] . "</p>";
          echo "<p><h6>hours</h6>" . $row['hours'] . "</p>";
          echo "<p><h6>parking</h6>" . $row['parking'] . "</p>";
          echo "<p><h6>facilities</h6>" . $row['facilities'] . "</p>";
          } 

    mysql_close($con);
    ?>
    </section>

  </section>
4

2 回答 2

1

您是绝对正确的，您需要使用实现 POST 方法的表单，该方法将所选值发布到表单中，与数据库中包含的“id”字段匹配。

这是一个例子：

<form action="compare.php" method="post">
    <select name="gyms-1">
      <option value="0">Select Gym</option>         
      <option value="1">fitness first</option>
        </select>
    <select name="gyms-2">
      <option value="0">Select Gym</option>         
      <option value="1">fitness first</option>      
    </select>
    <input name="send" id="send" type="submit" value="compare" />
</form>

下面是实现 post 方法的 PHP 代码：

<?php
    $gyms=$_POST['gyms-1'];
    $jimmy=$_POST['gyms-2'];

    mysql_select_db("gyms", $con);
    $result = mysql_query("SELECT * FROM gym WHERE id='$gyms'") or die ('Error: '.mysql_error ());

        while($row = mysql_fetch_array($result)){
          echo "<h1>" . $row['COLUMN NAME'] . "</h1>";
          echo "<p><h6>type</h6>" . $row['COLUMN NAME'] . "</p>";
          } 

    $result2 = mysql_query("SELECT * FROM gym WHERE id='$jimmy'") or die ('Error: '.mysql_error ());

        while($row = mysql_fetch_array($result2)){
          echo "<h1>" . $row['COLUMN NAME'] . "</h1>";
          echo "<p><h6>type</h6>" . $row['COLUMN NAME'] . "</p>";
          } 

    mysql_close($con);
?>

这应该允许您根据用户选择的值在数据库中显示多行。有什么问题给我留言！

于 2013-02-10T19:52:19.793 回答
0

用户 jquery .change。例如沿途：

   $("#left").change(function(){ 
    $("#right").html("[post your data]");
   });
于 2013-02-10T19:36:22.277 回答