我创建了一个网页，它使用JQuery将表单的内容重定向到另一个网页，使用PHP连接到数据库以找到一些内容并将其放回第一页。
一切都很好(感谢stack overflow的追随者的帮助:-))，但现在我想要以下内容:我正在询问一个城市的邮政编码，如果我很幸运，这个邮政编码是唯一的(只有一个城市有它)，但它也发生了，邮政编码是相同的几个城市，所以我想在这种情况下显示一个列表框供用户选择他/她的城市。有人知道怎么做吗?
my code:
home。
<!DOCTYPE html>
<html>
<head>
<script src="http://code.jquery.com/jquery-latest.js"></script>
</head>
<body>
<form action="/" id="myform">
<input type="text" name="postal_code" id="postal_code" placeholder="Search..." />
<input type="submit" value="Search" />
</form>
<!-- the result of the search will be rendered inside this div -->
<div id="result"></div>
<script>
$('#myform').submit(function() {
var url = 'target.php';
var postal_code = $('#postal_code').val();
$.post( url, { postal_code: postal_code },
function( data ) {
$( "#result" ).empty().append( data );
}
);
return false;
});
target.php
<?php
try
{
$pdo_options[PDO::ATTR_ERRMODE] = PDO::ERRMODE_EXCEPTION;
$bdd = new PDO('mysql:host=localhost;dbname=mydatabase', 'root', '', $pdo_options);
$response = $bdd->prepare('SELECT city FROM city_list where postal_code = ?');
$response->execute(array($_POST['postal_code']));
echo '<ul>';
while ($data = $response->fetch())
{
?>
<br/>The city you entered the postal code is : <?php echo $data['city'];
}
$response->closeCursor();
}
catch (Exception $e)
{
die('Error : ' . $e->getMessage());
}
?>
这是我需要的代码。我只需要从Jules的代码中做一些非常小的改变，使它ok(因为一个未知的原因，他的答案对他来说是完美的，但不适合我:-))
<?php
try {
//Get the postal code:
$postcode = $_POST['code_postal'];
//Make MySQL connection
mysql_connect("localhost", "root", "") or die (mysql_error());
//Select the database
mysql_select_db("site_artisans_amélioré");
//Do your query based on the postcode...
$query = "SELECT ville FROM liste_communes_code_postaux where code_postal = '" . mysql_real_escape_string($postcode) . "'";
//Return the response in a variable
$data = mysql_query($query) or die (mysql_error());
//echo "Num rows: " . mysql_num_rows($data);
//Check how many rows the query returned. If more than 1 that means several cities
//exist for one postcode, so you should show a listbox.
//If not, just return the ville name
if (mysql_num_rows($data) > 1) { ?>
<select name="cities">
<?php while ($row = mysql_fetch_assoc($data)) { ?>
<option value="<?php echo $row['ville']?>"><?php echo $row['ville']?></option>
<?php } ?>
</select>
<?php }
else {
$row = mysql_fetch_assoc($data);
echo $row['ville'];
}
}
catch (Exception $e) {
die("Error : " . $e->getMessage());
}
?>
我不确定你使用的数据库查询库，所以我会在伪代码和mysql_query..
target.php
<?php
try {
//Get the postal code:
$postcode = $_POST['postal_code'];
//Make MySQL connection
mysql_connect("localhost", "username", "password") or die (mysql_error());
//Select the database
mysql_select_db("mydatabase");
//Do your query based on the postcode...
$query = "SELECT city FROM city_list where postal_code = '" . mysql_real_escape_string($postcode) . "'";
//Return the response in a variable
$data = mysql_query($query);
//Check how many rows the query returned. If more than 1 that means several cities
//exist for one postcode, so you should show a listbox.
//If not, just return the city name
if (mysql_num_rows($data) > 1) { ?>
<select name="cities" multiple="multiple">
<? while ($row = mysql_fetch_assoc($data)) { ?>
<option value="<?=$row['city']?>"><?=$row['city']?></option>
<? } ?>
</select>
<? }
else {
$row = mysql_fetch_assoc($data);
echo $row['city'];
}
}
catch (Exception $e) {
die("Error : " . $e->getMessage());
}
?>
我希望你明白我的意思，你可以自己完成。