php - 与 mysql 数据库不同的结果作为下拉列表/php

Question

我确定我昨晚发布了此内容，但我找不到它，而且我的帐户显示我已问了 0 个问题。无论如何...

我的 MySQL 数据库中有一个包含英国县的字段。我想将这些列在下拉框中，以便用户可以选择它们，以选择他们想要查看结果的英国地区。

到目前为止我已经有了......但我的下拉框显示为空？

非常感谢任何帮助。

<?php
include 'connect.php';

//set variable
$option = '';

// Get the county names from database - no duplicates
$query = "SELECT DISTINCT tradingCounty FROM offers";

// execute the query, $result will hold all of the Counties in an array
$result = mysqli_query($con,$query);


while($row = mysql_fetch_array($result)) {
    $option .="<option>" . $row['tradingCounty'] . "</option>";
    echo $option;
}

echo "<form id='filter' name='filter' method='post' action='resultssimple.php'>
        <p><label>County</label></p>
        <select name='filter' id='filter'>" . $option . "</select>
        </form>";
?>

Answer 1

你可以尝试一下吗

    echo "<form id='filter' name='filter' method='post' action='resultssimple.php'>
            <p><label>County</label></p>
            <select name='filter' id='filter'>";
                // Get the county names from database - no duplicates
                $query = "SELECT DISTINCT tradingCounty FROM offers";

        // execute the query, $result will hold all of the Counties in an array
        $result = mysqli_query($con,$query);
        while($row = mysql_fetch_array($result)) {
            $option .="<option>" . $row['tradingCounty'] . "</option>";
            echo $option;
        }

    echo "</select>
    </form>";