Question

我一直在修改我的PHP代码以将数据插入到SQL表中，而且我总是得到ERROR QUERYING DATABASE。这些值来自普通的HTML表单，然后当我点击提交（action = memberadd.php）时，我从下面的代码中收到错误消息。我错过了什么，却看不出它是什么????

    <?php 
    $first_name = $_POST['first_name'];
    $last_name = $_POST['last_name'];
    $city = $_POST['city'];
    $state = $_POST['state'];
    $country = $_POST['country'];
    $industry = $_POST['industry'];//only showing one - need to fix
    $profile_visibility = $_POST['profile_visibility'];
    $position = $_POST['position'];
    $status = $_POST['status'];
    $profile_link = $_POST['profile_link'];
    $skills = $_POST['skills'];


    //connects and sends information to the database
    $dbc = mysqli_connect('localhost', 'root', 'root', 'main') or die('Error connecting to       MySQL server.');

   //inserts data into the member_details table main db
   $query = "INSERT INTO 'member_details' (first_name, last_name, city, state, country, industry, profile_visibility, position, status, profile_link, skills) 
   VALUES ('$first_name', '$last_name', '$city', '$state', '$country', '$industry', '$profile_visibility', '$position', '$status', '$profile_link', '$skills')";

   $result = mysqli_query($dbc, $query) or die('Error querying database.');
   mysqli_close($dbc);

   echo '<h2>Here are your details.</h2>';
   echo '<h3>First Name: '.$first_name.'<br />';
   echo '<h3>Last Name: '.$last_name.'<br />';
   echo 'City: '.$city.'<br />';
   echo 'State: '.$state.'<br />';
   echo 'Country: '.$country.'<br />';
   echo 'Industry: '.$industry.'<br />';//only showing one - need to fix
   echo 'Profile: '.$profile_visibility.'<br />';
   echo 'Position: '.$position.'<br />';
   echo 'Status: '.$status.'<br />';
   echo 'Link: '.$profile_link.'<br />';
   echo 'Skills: '.$skills.'<br /></h3>';

   ?>

Answer 1

问题

由于@Fred -ii-有noted，问题在于INSERT语句中的表名围绕引号。

$query = "INSERT INTO 'member_details' (first_name, ...
                      ^              ^

解决方案

如果你想引用＆＃39;表格或列名称，您应该使用反引号，您可以在MySQL documentation page上阅读更多相关信息。

$query = "INSERT INTO `member_details` (first_name, ...

检测错误

要检查MySQLi数据库错误请求，可以使用一些方法来获取错误信息。可能最有用的是mysqli_error() will give you an error string。

$result = mysqli_query($dbc, $query);
if(!$result)
{
  printf("Errormessage: %s\n", mysqli_error($dbc));
}

作为@Fred -ii-还mentioned，您应该在开发新代码时正确使用错误报告。理想情况下，您应该在php.ini中配置它，但也可以通过将以下内容添加到页面顶部来轻松完成。

error_reporting(E_ALL); 
ini_set('display_errors', 1);

最后，您对SQL Injection Attacks 持开放态度。您应该考虑使用prepared statements with MySQLi来帮助防止这种情况发生。

Answer 2

你可以使用像...这样的mysql查询。

$query = "INSERT INTO `member_details` SET first_name = '".$first_name."', last_name = '".$last_name."', city = '".$city."', state = '".$state."', country = '".$country."', industry = '".$industry."', profile_visibility = '".$profile_visibility."', position = '".$position."', status = '".$status."', profile_link = '".$profile_link."', skills = '".$skills."'";

执行插入PHP代码时查询数据库时出错

2 个答案: