php - 表单和PHP函数将表单数据保存到mysql在同一个文件中

Question

我一直在尝试解决这个问题，但到目前为止我还无法让它发挥作用。不会引发任何错误，页面会在提交时刷新。我很茫然，但我不完全是专家，对此还很陌生。

这是代码(为了发布而简化):

<?php if (!isset($_POST['submit'])) {
echo              "<!-- Form starts here -->
                  <form id=\"billing\" action=\"\" method=\"post\">
                  <!-- Name -->
                  <div class=\"control-group\">
                  <label class=\"control-label\"><b>Name</b></label>
                  <div class=\"controls\">
                  <input type=\"text\" id=\"name\" name=\"name\" placeholder=\"your name\" class=\"input-large\">
                  </div>
                  </div>
                  <!-- Zip -->
                  <div class=\"control-group\">
                  <label class=\"control-label\"><b>Zip Code</b></label>
                  <div class=\"controls\">
                  <input type=\"text\" id=\"billingzip\" name=\"billingzip\" placeholder=\"5 digit zip\" class=\"input-large\">
                  </div>
                  </div>
                  <!-- Submit -->
                  <div class=\"control-group\">
                  <div class=\"controls\">
                  <button class=\"button save small_green_button\" type=\"submit\">
                  &nbsp;Save&nbsp;
                  </button>
                  </div>
                  </div>
                  </form>";
}
else
{
 $host="localhost";
 $user_name="user";
 $pwd="password";
 $database_name="database";
 $db=mysql_connect($host, $user_name, $pwd) or die(mysql_error());
 $dbsel=mysql_select_db($database_name, $db);

 if (mysql_error() > "") print mysql_error() . "<br>";
               if (mysql_error() > "") print mysql_error() . "<br>";

                   $account_id = users::getAttr('Account', 'account_id'); 
                   $zip = mysql_real_escape_string($_POST['billingzip']);
                   $name = mysql_real_escape_string($_POST['name']);

                           $sql = "INSERT INTO `billing`
                                   SET `account_id` = '{$account_id}',
                                       `zip` = '{$billingzip}',
                                       `name` = '{$name}',
                                       `updated_at` = NOW()";

                           $result = mysql_query($sql, $dbsel) 
                                or  die(mysql_error().$sql);

                 mysql_close($db);


   }
  ?>

Answer 1

首先我看到的是

$result = mysql_query($sql, $dbsel) or  die(mysql_error().$sql);

我认为你应该写:

$result = mysql_query($sql, $db) or  die(mysql_error().$sql);