Php 相关下拉列表 $fruit_name = $_POST['fruit_name']; #Connect to MySQL #Connect to database $result = mysql_query("SELECT colour FROM fruit WHERE fruit_name = '$fruit_name'"); echo "<select name='colour' id='colour' style='font-family: Calibri;font-size: 10pt;'>"; while($row = mysql_fetch_assoc($result)) { echo "<option value = ''>" . $row['colour'] . "</option>"; } echo "</select>"; mysql_free_result($result); //Closes specified connection ?>

我是PHP新手。。我需要你的帮助

我有两个相关的下拉列表：

        $fruit_name = $_POST['fruit_name'];
#Connect to MySQL
#Connect to database

$result = mysql_query("SELECT colour FROM fruit WHERE fruit_name = '$fruit_name'");

echo "<select name='colour' id='colour' style='font-family: Calibri;font-size: 10pt;'>";

while($row = mysql_fetch_assoc($result))
{
echo "<option value = ''>" . $row['colour'] . "</option>";
}
echo "</select>";

mysql_free_result($result);

//Closes specified connection
?>

下拉列表1：手动插入值

        $fruit_name = $_POST['fruit_name'];
#Connect to MySQL
#Connect to database

$result = mysql_query("SELECT colour FROM fruit WHERE fruit_name = '$fruit_name'");

echo "<select name='colour' id='colour' style='font-family: Calibri;font-size: 10pt;'>";

while($row = mysql_fetch_assoc($result))
{
echo "<option value = ''>" . $row['colour'] . "</option>";
}
echo "</select>";

mysql_free_result($result);

//Closes specified connection
?>

下拉列表2：根据下拉列表1中选择的条件从数据库值附加值

        $fruit_name = $_POST['fruit_name'];
#Connect to MySQL
#Connect to database

$result = mysql_query("SELECT colour FROM fruit WHERE fruit_name = '$fruit_name'");

echo "<select name='colour' id='colour' style='font-family: Calibri;font-size: 10pt;'>";

while($row = mysql_fetch_assoc($result))
{
echo "<option value = ''>" . $row['colour'] . "</option>";
}
echo "</select>";

mysql_free_result($result);

//Closes specified connection
?>

然后，所选的两个值将以另一种形式显示在文本框中

我的问题是：

1第二个下拉列表中的值无法显示

        $fruit_name = $_POST['fruit_name'];
#Connect to MySQL
#Connect to database

$result = mysql_query("SELECT colour FROM fruit WHERE fruit_name = '$fruit_name'");

echo "<select name='colour' id='colour' style='font-family: Calibri;font-size: 10pt;'>";

while($row = mysql_fetch_assoc($result))
{
echo "<option value = ''>" . $row['colour'] . "</option>";
}
echo "</select>";

mysql_free_result($result);

//Closes specified connection
?>

2第一个下拉列表中的值可以传递到其他形式，但第二个不能

        $fruit_name = $_POST['fruit_name'];
#Connect to MySQL
#Connect to database

$result = mysql_query("SELECT colour FROM fruit WHERE fruit_name = '$fruit_name'");

echo "<select name='colour' id='colour' style='font-family: Calibri;font-size: 10pt;'>";

while($row = mysql_fetch_assoc($result))
{
echo "<option value = ''>" . $row['colour'] . "</option>";
}
echo "</select>";

mysql_free_result($result);

//Closes specified connection
?>

请引导我

我不知道如何在这里共享我的代码

form1.php

    <?php
    // put the code to connect to your database here


    $fruit_name = $_POST['value'];  // this will contain the value selected from first dropdown

    $result = mysql_query("SELECT colour FROM fruit WHERE fruit_name = '$fruit_name'");

echo "<select name='colour' id='colour' style='font-family: Calibri;font-size: 10pt;'>";

    while($row = mysql_fetch_assoc($result))
    {
        echo "<option value = '".$row['colour']."'>" . $row['colour'] . "</option>";
    }
    echo "</select>";

mysql_free_result($result);
    ?>

//第一个下拉列表

<select name="fruit_name" id="fruit_name" style="font-family: Calibri;font-size: 10pt;" onchange="loadXMLDoc(this.value); ">
   <option value="0">-- please choose --</option>
   <option value="Pineapple">Pineapple</option>
   <option value="Apple">Apple</option> 

        $fruit_name = $_POST['fruit_name'];
#Connect to MySQL
#Connect to database

$result = mysql_query("SELECT colour FROM fruit WHERE fruit_name = '$fruit_name'");

echo "<select name='colour' id='colour' style='font-family: Calibri;font-size: 10pt;'>";

while($row = mysql_fetch_assoc($result))
{
echo "<option value = ''>" . $row['colour'] . "</option>";
}
echo "</select>";

mysql_free_result($result);

//Closes specified connection
?>

//第二个下拉列表

        $fruit_name = $_POST['fruit_name'];
#Connect to MySQL
#Connect to database

$result = mysql_query("SELECT colour FROM fruit WHERE fruit_name = '$fruit_name'");

echo "<select name='colour' id='colour' style='font-family: Calibri;font-size: 10pt;'>";

while($row = mysql_fetch_assoc($result))
{
echo "<option value = ''>" . $row['colour'] . "</option>";
}
echo "</select>";

mysql_free_result($result);

//Closes specified connection
?>

form2.php

 <?php
//connection

        $fruit_name = $_POST['fruit_name'];
        $colour = $_POST['colour'];
?>
  <label>
  <input type="text" name="fruit_name" id="fruit_name" value = "<?php echo $fruit_name;?>" readonly>
  </label>
  <p>
    <label>
    <input type="text" name="colour" id="colour" value="<?php echo $colour;?>" readonly>
    </label>
  </p>

---

我通常不这样做，但由于我手头有一些空闲时间，我将给出您可以遵循的一般方法：

在标签之间包含以下内容

在下面，粘贴此代码

<script type="text/javascript">
    $(function(){
        $('select#fruit_name').change(function(){
                var selectedVal = $(this).val(); // get the selected value

                $.ajax({    // send ajax request to the php file to process data
                    type:'post',
                    url:'php-page-name.php',
                    data:{'value':selectedVal},
                    success:function(ret)   // display the result from php-page-name.php page
                    {
                        $('div#result').html(ret);
                    }
                });

        });
    });
</script>

现在让我们转到HTML

php-page-name.php页面不要忘记创建此页面，并将其放在与form1.php相同的文件夹中

    <?php
    // put the code to connect to your database here


    $fruit_name = $_POST['value'];  // this will contain the value selected from first dropdown

    $result = mysql_query("SELECT colour FROM fruit WHERE fruit_name = '$fruit_name'");

echo "<select name='colour' id='colour' style='font-family: Calibri;font-size: 10pt;'>";

    while($row = mysql_fetch_assoc($result))
    {
        echo "<option value = '".$row['colour']."'>" . $row['colour'] . "</option>";
    }
    echo "</select>";

mysql_free_result($result);
    ?>

---

PS：我在本例中使用mysql_*函数，因为我假设您也是。但不建议这样做，因为它们很快就会被弃用。您可能想切换到mysqli或PDO，代码在哪里？如果你没有发布相关的代码ajax是你的答案，那么1号将会回答你的问题。这里有一些小东西让你开始请输入一些代码，以便更好地回答我很抱歉。我不知道如何在这里共享我的代码。请编辑您的问题并将代码粘贴到其中。然后使用{}工具将其标记为代码，以便正确格式化。谢谢@asprin。我尝试了你的代码，但我有一个错误-致命错误：无法在写上下文中使用函数返回值。我可以知道错误吗？我已使用你的值编辑了我的答案。过来看。还要确保创建一个名为php-page-name.php的php页面，并将其与第一个drown所在的文件夹放在同一个文件夹中。谢谢@asprin。我使用过它，但我有一个错误警告：mysql\u free\u result期望参数1是resource，mysql\u free\u result$result处的布尔值；代码。我已经编辑了代码。现在，在我选择第一个下拉列表后，第二个下拉列表值不显示。但在我选择第一个下拉列表并提交后，将显示第二个下拉列表的值。在我选择第一个值而不提交后，您能帮助我使其显示吗？

        $fruit_name = $_POST['fruit_name'];
#Connect to MySQL
#Connect to database

$result = mysql_query("SELECT colour FROM fruit WHERE fruit_name = '$fruit_name'");

echo "<select name='colour' id='colour' style='font-family: Calibri;font-size: 10pt;'>";

while($row = mysql_fetch_assoc($result))
{
echo "<option value = ''>" . $row['colour'] . "</option>";
}
echo "</select>";

mysql_free_result($result);

//Closes specified connection
?>