php查询在html选项中的结果

Question

为什么我得到的是白色行而不是值？

<select name="select">
    <?php                    
    $connessione = mysql_connect('localhost' , 'root', '') or die("Impossibile connettersi: " .mysql_error());
    mysql_select_db("musica",$connessione);
    $query = mysql_query("SELECT * FROM artisti_preferiti");
    while($row = mysql_fetch_array($query))
    {
        ?><option value="<?php echo $row['nome']; ?>"> <?php echo $row['cognome'];?></option> 
    <?php }?>
</select>

Answer 1

我怀疑这是一个DB连接问题，在这里，我成功地测试了。

请考虑以下几点：

Sidenote:确保设置是正确的，包括DB名称和colum名称。

DB连接文件：(db_connect.php)

<?php
$mysql_hostname = 'xxx';
$mysql_username = 'xxx';
$mysql_password = 'xxx';

$mysqli = mysql_connect("$mysql_hostname", "$mysql_username", "$mysql_password");

if($mysqli->connect_errno > 0) {
  die('Connection failed [' . $mysqli->connect_error . ']');
}

?>

PHP (example.php)

<select name="select">
<?php

include 'db_connect.php';

    mysql_select_db("musica",$mysqli);
    $query = mysql_query("SELECT * FROM artisti_preferiti",$mysqli);
    while($row = mysql_fetch_array($query))
    {
        ?><option value="<?php echo $row['nome']; ?>"> <?php echo $row['cognome'];?></option> 
    <?php }?>
</select>

对最后一个注意事项，mysql_*函数是不推荐的。请考虑将mysqli_*与准备好的语句或PDO结合使用。

编辑

MySQLi_*版本

Sidenote:最好使用列名而不是SELECT * --即：SELECT nome, cognome

DB连接文件：(db_connect_mysqli.php)

<?php
$mysql_hostname = 'xxx';
$mysql_username = 'xxx';
$mysql_password = 'xxx';
$mysql_dbname = 'xxx';

$mysqli = new mysqli("$mysql_hostname", "$mysql_username", "$mysql_password","$mysql_dbname");

if($mysqli->connect_errno > 0) {
  die('Connection failed [' . $mysqli->connect_error . ']');
}

?>

PHP (example_mysqli.php)

<select name="select">
    <?php

include 'db_connect_mysqli.php';

$query = $mysqli->query("SELECT * FROM artisti_preferiti");


    while($row = mysqli_fetch_array($query))
    {
        ?><option value="<?php echo $row['nome']; ?>"> <?php echo $row['cognome'];?></option> 
    <?php }?>
</select>

Answer 2

我更改了您的脚本以便我们可以调试

<?php
$lnk = mysql_connect('localhost' , 'root', '') or die("Impossibile connettersi: " .mysql_error());
mysql_select_db("musica",$lnk);
$q = 'SELECT nome,cognome FROM artisti_preferiti';
$result = mysql_query($q,$lnk);
$numRows = mysql_num_rows($result);
$rows = array();
while($row = mysql_fetch_array($result)) $rows[$row['nome']] = $row['cognome'];
/* show us this!! */
print_r($rows);
?>
<select name="select"><?php
  foreach ($rows as $nome => $cognome) {
    echo "<option value='$nome'>$cognome</option>";
  }
?></select>