Question

我有一个选择国家和选择城市，我想在城市选择中显示不同的值，具体取决于国家选择的选择。

像：值= countries.selection

SELECT CITY FROM cities WHERE Country = Value

$ myrow [城市]

请参阅下面的代码（这会为数据库中的选择带来值并填充它们。

可能有更好的方法来编码我的代码，我也非常感谢任何使其更容易和更有效的建议。

此外，如果在Javascrip或jQuery中更好或更容易，我也会接受这方面的建议。

非常感谢

<html>
<body>


<select name="taskOption" id="drop_drop">
    <?php
    $db = pg_connect('host=localhost dbname=test user=myuser password=mypass');

    $query = "SELECT country FROM countries";

    $result = pg_query($query);
    if (!$result) {
        echo "Problem with query " . $query . "<br/>";
        echo pg_last_error();
        exit();
    }

    printf ("<option value=Select>Select a Country</option>");
    while($myrow = pg_fetch_assoc($result)) {

     printf ("<option value=$myrow[country]>$myrow[country]</option>");

}

    ?>

</select>


<select name="taskOption" id="drop_drop">
    <?php
    $db = pg_connect('host=localhost dbname=test user=myuser password=mypass');

    $query = "SELECT city FROM cities";

    $result = pg_query($query);
    if (!$result) {
        echo "Problem with query " . $query . "<br/>";
        echo pg_last_error();
        exit();
    }
    printf ("<option value=Select>Select a City</option>");
    while($myrow = pg_fetch_assoc($result)) {

     printf ("<option value=$myrow[city]>$myrow[city]</option>");

}

    ?>

</select>

</body>
</html>

Answer 1

您必须在国家/地区更改时提出请求，将国家/地区的值传递给服务器以查找其城市。

input = gets.chomp.to_i
  case input
  when 1

请注意 <select name="taskOption" id="countries"> <?php $db = pg_connect('host=localhost dbname=test user=myuser password=mypass'); $query = "SELECT country FROM countries"; $result = pg_query($query); if (!$result) { echo "Problem with query " . $query . "<br/>"; echo pg_last_error(); exit(); } printf ("<option value=Select>Select a Country</option>"); while($myrow = pg_fetch_assoc($result)) { printf ("<option value=$myrow[country]>$myrow[country]</option>"); } ?> </select> <select name="taskOption" id="cities"> <?php $db = pg_connect('host=localhost dbname=test user=myuser password=mypass'); // Note query change here. $selectedCountry = $_GET['country']; $query = "SELECT city FROM cities where country_id = " + $selectedCountry; $result = pg_query($query); if (!$result) { echo "Problem with query " . $query . "<br/>"; echo pg_last_error(); exit(); } printf ("<option value=Select>Select a City</option>"); while($myrow = pg_fetch_assoc($result)) { printf ("<option value=$myrow[city]>$myrow[city]</option>"); } ?> </select> <script type='text/javascript'> // Note this new script. $(document).ready(function(){ $('#countries').on('change', function(){ window.location += '?country=' + $(this).val(); }); }); </script>元素中的id属性，并注意可能的SQL注入。

PHP：如何根据另一个Select的选择更改Select的值

1 个答案: