0

我有可以工作的 PHP 编码，但我没有运气将它转换为 HTML 表单。有什么建议吗？ 

<?php
$db_host =  "localhost";
$db_username  =  "combsb_combsb";
$db_pass =  "pat60086";
$db_name =  "combsb_sample";

@mysql_connect ("$db_host", "$db_username", "$db_pass") or die ("Could not connect to MySQL");
@mysql_select_db("$db_name") or die ("No Database");

Echo"Successful Connection";


$sql = "SELECT compname FROM Crew";
$result = mysql_query($sql);

echo "<select name='compname'>";
while ($row = mysql_fetch_array($result)) {
echo "<option value='" . $row['compname'] . "'>" . $row['compname'] . "</option>";
}
echo "</select>";


?>
4

1 回答 1

0

首先，我会确保在您的 PHP 周围添加 HTML 样板。然后我认为你的意思是 mysql_fetch_row 而不是 mysql_fetch_array。您的最终文件应类似于：

<!DOCTYPE html>
<html>
  <head></head>
  <body>
<?php
// *Cut out for brevity, remember to paste back in* 
$result = mysql_query($sql);
?>

    <form action="" method="POST">

<?php
echo "<select name='compname'>";

while ($row = mysql_fetch_row($result)) {
  echo "<option value='" . $row['compname'] . "'>" . $row['compname'] . "</option>";
}

echo "</select>";
?>

    </form>
  </body>
</html>

下次发布您的输出/错误，或者如果这不起作用

于 2013-06-05T18:29:06.420 回答