Question

我正在尝试从数据库中的数据中填充一个下拉框，但它似乎不起作用。它传递了一个错误，说数据库中没有数据供它读取，但有。下面是我用来使其工作的脚本。

    <?php
@ini_set('display_errors', 'on');
 echo "<h1>Register</h1>";
if ($_SERVER['REQUEST_METHOD'] == 'POST'){

$errors = array();
if (empty($_POST['firstname'])){
$errors[] = 'Your forgot to enter your first name.';
}else{
  $firstname = trim($_POST['firstname']);
  }
if (empty($_POST['lastname'])){
$errors[] = 'Your forgot to enter your last name.';
}else{
  $lastname = trim($_POST['lastname']);
}
if (empty($_POST['username'])){
$errors[] = 'Your forgot to enter your  username.';
}else{
  $username = trim($_POST['username']);
}
if (!empty($_POST['password1'])) {
if ($_POST['password1'] != $_POST ['password2']) {
  $errors[] = 'Your password did not match the confirmed password!';
 }else{
   $password = trim($_POST['password1']);
  }
} else {
  $errors[] = 'You forgot to enter your password!';
}
if (empty($_POST['birthdate'])){
$errors[] = 'Your forgot to enter your  birthdate.';
}else{
  $birthdate = trim($_POST['birthdate']);
}
if (empty($_POST['gamespyid'])){
$errors[] = 'Your forgot to enter your  gamespy id.';
}else{
  $gamespyid = trim($_POST['gamespyid']);
}
  if (empty($errors)) {
     if (is_file('admin/mysqli_connect.php')) { echo '<p>The connection file is there.</p>';require('admin/mysqli_connect.php'); }

else {
  echo '<p>The connection file is not there</p>';
  }
     $q="INSERT INTO Users (firstname, lastname, username, password1, birthdate, gamespyid, base) VALUES ('$firstname', '$lastname', '$username', md5('$password1'), '$birthdate', '$gamespyid', '$base')";
$r = mysql_query($dbc, $q);
if ($r){
 echo'<p>You are now registered</p>';
}else{
  echo'<p>You have not been registered</p>';
                                            }
  } else {
    echo 'Error<br> <p>The following errors have occured:<br/>';
    foreach ($error as $msg) {
      echo " - $msg<br/>\n";
    }
    echo '</p><p>Please try again.</p><p><br/></p>';
  }   //if no errors

 }     //submit

?>
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 3.2//EN">
<html>
<head>
  <title></title>
</head>

<body>
  <form action="http://www.virtual-aviation.org/gatewayaviation/index.php?p=register" method='POST'>
    <table summary="REgform">
      <tr>
        <td>First Name:</td>

        <td><input type='text' name='firstname' value='<?php echo $firstname; ?>'></td>
      </tr>

      <tr>
        <td>Last Name:</td>

        <td><input type='text' name='lastname'value='<?php echo $lastname; ?>'></td>
      </tr>

      <tr>
        <td>Username:</td>

        <td><input type='text' name='username'value='<?php echo $username; ?>'></td>
      </tr>

      <tr>
        <td>Password:</td>

        <td><input type='password' name='password1'></td>
      </tr>

      <tr>
        <td>Repeat Password:</td>

        <td><input type='password' name='password2'></td>
      </tr>

      <tr>
        <td>Birthdate:</td>

        <td><input type='text  ' name='birthdate'value='<?php echo $birthdate; ?>'></td>
      </tr>

      <tr>
        <td>Gamespy Id:</td>

        <td><input type='text' name='gamespyid'value='<?php echo $gamespyid; ?>'></td>
      </tr>
       <td>Base:</td>
      <tr>
       <td><select name="base" size="1">
          <option>
            Select One
          </option>
         <?php
         $qf = "SELECT  airport_id, CONCAT_WS(' ', airport_name, airport_code) FROM airports ORDER BY airport_code ASC";
         $rf = mysqli_query ($dbc, $qf);

         if (mysqli_num_rows($rf) > 0) {
         while ($row = mysql_fetch_array ($rf, MYSQLI_NUM)) {
         echo "<option value=\"$row[0]\"";
         if (isset($_POST['existing']) && ($_POST['existing'] == $row[0]) ) echo 'selected="selected"'; echo ">$row[1]</option>\n";
         }
   } else {
   echo '<option>Please a new airport first.</option>';
    }
    //mysqli_close($dbc);

         ?>
        </select></td>
      </tr>
    </table>
   <input type='submit' name='submit' value='Register'/>
  </form>
  <?php
  echo '<p>debugger<p>' . $rf ;
  ?>
</body>
</html>

以下是我查看源代码时遇到的一些错误;错误出现在选项框内容的位置。

警告：mysqli_query（）要求参数1为mysqli，null为 /home5/virtua15/public_html/gatewayaviation/pages/register.inc.php 182

警告：mysqli_num_rows（）要求参数1为mysqli_result，在 /home5/virtua15/public_html/gatewayaviation/pages/register.inc.php 行中给出null 184

Please a new airport first.

Answer 1

使用mysql_ *或mysqli_ *函数，但不能同时使用它们。我希望你选择后者并了解准备好的陈述：http://php.net/manual/en/mysqli.quickstart.prepared-statements.php

问题是你的$dbc变量很可能是mysql资源或null ，而不是mysqli资源，因此当你将它传递给mysqli_ *函数时它会失败。有多种方法可以检查并确保您拥有有效的资源：

$mysqli = new mysqli("example.com", "user", "password", "database");
if ($mysqli->connect_errno) {
    echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error;
}

要解决您的问题，您需要重新构建代码以使用仅 mysqli，并确保$dbc是有效的mysqli资源。

Answer 2

在您的示例的第123行，您有$rf = mysqli_query ($dbc, $qf);参数＃1是$dbc，即MySQL连接对象。错误消息指出连接对象是null，这是真的 - 它没有在脚本中的任何位置声明。

Answer 3

好吧，

警告：mysqli_query（）要求参数1为mysqli，在第182行的/home5/virtua15/public_html/gatewayaviation/pages/register.inc.php中给出null

是因为你不能使用mysqli_query（$ dbc，$ qf）;当$ dbc不是从mysql_connect或mysql_init返回时，因为它期望链接作为第一个参数，并返回它所需的链接。

这也是您收到第二个错误的原因。 mysqli没有向$ rf返回任何内容。

我建议重构整个页面。首先给出变量有意义的名称，转义SQL字符串参数以帮助避免注入。

清除这些警告并再次拍摄。

PHP Dropdown框没有填充

3 个答案: