Question

我可以看到查询返回结果，但我似乎无法将它们放入html下拉框中。此外，下拉框中的条目与查询返回的条目一样多，但它们都是白色空格。但是，页面源显示正确的选项值，例如

<option value="3 John"></option>

<option value="Jude"></option>

<option value="Revelation"></option>

有人可以帮帮我吗？为什么他们实际上没有显示在下拉框中？

<html>
<?php
    //Connect to the database
    $mysqli = new mysqli("localhost", "root", "", "bible");

    //Return an error if we have a connection issue
    if ($mysqli->connect_error) {
        die('Connect Error (' . $mysqli->connect_errno . ') '
                . $mysqli->connect_error);
        }

    //Query the database for the results we want
    $query = $mysqli->query("select distinct bname as Name from kjv limit 1");

    //Create an array  of objects for each returned row
    while($array[] = $query->fetch_object());

    array_pop($array);

    //Print out the array results
    print_r($array);
    ?>
    <h3>Dropdown Demo Starts Here</h3>
    <select name="the_name">
    <?php foreach($array as $option) : ?>
        <option value="<?php echo $option->Name; ?>"></option>
    </select>
        <?php endforeach; ?>

Answer 1

试试这个

<select name="the_name">
<?php foreach($array as $option) : ?>
        <option value="<?php echo $option['Name']; ?>"><?php echo $option['Name']; ?></option>
<?php endforeach; ?>
</select>

Answer 2

执行查询后，使用while循环添加选项以选择

$query = $mysqli->query("select distinct bname as Name from kjv limit 1"); ?>

<select>
    <?php while($option = $query->fetch_object()){ ?>
        <option><?php echo $option->Name; ?></option>
    <?php } ?>
</select>

不确定array_pop在代码中做了什么

Answer 3

AS TIM WAX表示这是解决方案

$query = $mysqli->query("select distinct bname as Name from kjv limit 1"); ?>

<select>
    <?php while($option = $query->fetch_object()){ ?>
        <option><?php echo $option->Name; ?></option>
    <?php } ?>
</select>

Answer 4

<select name="the_name">
<?php foreach($array as $option) : ?>
        <option value="<?php echo $option->Name; ?>"></option>
<?php endforeach; ?>
</select>

您以一种一次又一次创建<select>标记的方式结束了循环。改变它，然后再试一次。我对.php了解不多，但在显示您的下拉框时可能会出现问题。

Answer 5

这是我的..我是初学者，但它对我有用，

    $query = $mysqli->query("SELECT * FROM  `student_type_db`"); //table of student type

    echo "<select>";
    while($row = $query->fetch_array()){
         echo "<option>";
         echo $row['student_type'] . " - " . $row['student_description'];
         echo "</option>"; 
    }
    echo "</select>";

   // student type = 1 | student description = regular
   // output : 1 - regular

使用PHP中的mysql的数组数据填充html <select> </select>

5 个答案: