Question

为什么我得到一个白色的行而不是值？

<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

我怀疑这是一个数据库连接问题，我成功测试了。

请考虑以下事项：

旁注：确保您的设置正确，包括数据库名称和列名称。

数据库连接文件：（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>

最后一点，不推荐使用mysql_*个函数。请考虑将mysqli_*与预准备语句或PDO一起使用。

修改

MySQLi_ *版本

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

数据库连接文件：（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>

php查询结果在html选项中

2 个答案:

修改

MySQLi_ *版本