从数据库表中填充选择下拉列表答案

【问题标题】：Populate select drop down from a database table从数据库表中填充选择下拉列表
【发布时间】：2010-09-17 00:12:33
【问题描述】：

我有一张表（“场所”），其中存储了志愿者可以工作的所有可能场所，每个志愿者被分配到每个场所工作。

我想从场地表中创建一个选择下拉列表。

现在我可以显示每个志愿者分配的地点，但我希望它显示下拉框，并且地点已经在列表中选择。

<form action="upd.php?id=7">
<select name="venue_id">
<?php //some sort of loop goes here
print '<option value="'.$row['venue_id'].'">'.$row['venue_name'].'</option>';
//end loop here ?>
</select>
<input type="submit" value="submit" name="submit">
</form>

例如id为7的志愿者，分配给venue_id 4

<form action="upd.php?id=7">
<select name="venue_id">
    <option value="1">Bagpipe Competition</option>
    <option value="2">Band Assistance</option>
    <option value="3">Beer/Wine Pouring</option>
    <option value="4" selected>Brochure Distribution</option>
    <option value="5">Childrens Area</option>
    <option value="6">Cleanup</option>
    <option value="7">Cultural Center Display</option>
    <option value="8">Festival Merch</option>
</select>
<input type="submit" value="submit" name="submit">
</form>

Brochure Distribution option will already be selected when it displays the drop down list, because in the volunteers_2009 table, column venue_id is 4.

我知道它将采用 for 或 while 循环的形式从场所表中提取场所列表

我的查询是：

$query = "SELECT volunteers_2009.id, volunteers_2009.comments, volunteers_2009.choice1, volunteers_2009.choice2, volunteers_2009.choice3, volunteers_2009.lname, volunteers_2009.fname, volunteers_2009.venue_id, venues.venue_name FROM volunteers_2009 AS volunteers_2009 LEFT OUTER JOIN venues ON (volunteers_2009.venue_id = venues.id) ORDER by $order $sort";

如何使用场地表中的场地（volunteers_2009.venue_id、venues.id）填充选择下拉框并让它预先选择场地在列表中？

【问题讨论】：

标签： php mysql

【解决方案1】：

$query = "SELECT volunteers_2009.id, volunteers_2009.comments, volunteers_2009.choice1, volunteers_2009.choice2, volunteers_2009.choice3, volunteers_2009.lname, volunteers_2009.fname, volunteers_2009.venue_id, venues.venue_name FROM volunteers_2009 AS volunteers_2009 LEFT OUTER JOIN venues ON (volunteers_2009.venue_id = venues.id) ORDER by $order $sort";

$res = mysql_query($query);
echo "<select name = 'venue'>";
while (($row = mysql_fetch_row($res)) != null)
{
    echo "<option value = '{$row['venue_id']}'";
    if ($selected_venue_id == $row['venue_id'])
        echo "selected = 'selected'";
    echo ">{$row['venue_name']}</option>";
}
echo "</select>";

【讨论】：

【解决方案2】：

<!DOCTYPE html>
<html>
<head>
    <title>table binding</title>
    <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>

</head>
<body>
    <div id="mydiv" style="width:100px;height:100px;background-color:yellow">

        <select id="myselect"></select>
    </div>

</body>
</html>


<?php
include('dbconnection.php');

$sql = "SHOW TABLES FROM $dbname";
$result = mysqli_query($conn,$sql);

if (!$result) {
    echo "DB Error, could not list tables\n";
    echo 'MySQL Error: ' . mysqli_error();
    exit;
}

while ($row = mysqli_fetch_row($result)) {
    echo "<script>
    var z = document.createElement('option');
    z.setAttribute('value', '".$row[0]."');
    var t = document.createTextNode('".$row[0]."');
    z.appendChild(t);
    document.getElementById('myselect').appendChild(z);</script>";

}



?>

【讨论】：

【解决方案3】：

        <?php 
        $query = "SELECT * from blogcategory";
        //$res = mysql_query($query);
        $rows = $db->query($query);
        echo "<select name = 'venue'>";
        // while (($row = mysql_fetch_row($res)) != null)
        while ($record = $db->fetch_array($rows)) 
        {
            echo "<option value = '{$record['CategoryId']}'";
            if ($CategoryId == $record['CategoryId'])
                echo "selected = 'selected'";
            echo ">{$record['CategoryName']}</option>";
        }
        echo "</select>";
        ?>

【讨论】：

【解决方案4】：

假设您有一系列场地...我个人不喜欢将 sql 与其他魔法混为一谈。

function displayDropDown($items, $name, $label, $default='') {
  if (count($items)) {
    echo '<select name="' . $name . '">';
    echo '<option value="">' . $label . '</option>';
    echo '<option value="">----------</option>';
    foreach($items as $item) {
      $selected = ($item['id'] == $default) ? ' selected="selected" : '';
      echo <option value="' . $item['id'] . '"' . $selected . '>' . $item['name'] . '</option>';
    }
    echo '</select>';
  } else {
    echo 'There are no venues';
  }
}

【讨论】：

我有一个表（venues），里面存储了所有的场地（id,venue_name），venues.id 存储在volunteers_2009表中（volunteers_2009.venue_id）