2016-08-27 95 views

回答

0

按照下面的代码。

<?php 
$servername = "localhost"; 
$username = "username"; 
$password = "password"; 
$dbname = "myDB"; 

// Create connection 
$conn = new mysqli($servername, $username, $password, $dbname); 
// Check connection 
if ($conn->connect_error) { 
    die("Connection failed: " . $conn->connect_error); 
} 

$sql = "SELECT name FROM state"; 
$result = $conn->query($sql); 

if ($result->num_rows > 0) { 
?> 
<select name="myCountry"> 
<?php 
    // output data of each row 
    while($row = $result->fetch_assoc()) { 
    ?> 
    <option value="<?php echo $row['name']; ?>"><?php echo $row['name']; ?></option> 
    <?php 
    } 
    ?> 
</select> 
<?php 
} 
$conn->close(); 
?> 
+0

这对我来说很好。我希望当我选择我的选择有一个提交按钮带我到另一个页面,但我想保持我的选择,因为我想要去的页面然后将改变这个值。我可以这样做吗? –

+0

。在another-page.php页面中,您可以通过** $ _ POST ['MyCountry'] **变量找到值。 –

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