Question

我正在尝试显示此页面上mysql行的信息。我正在使用$ _GET，因为该ID包含在该页面的链接中：www.example.com/page.php？id = 1但它返回此错误：

错误：您的SQL语法出错;检查与MySQL服务器版本对应的手册，以便在第1行的'='1''附近使用正确的语法

有谁知道如何解决这个问题？

以下代码：

<?php

    $username="xxx";
    $password="xxx";
    $database="xxx";
    mysql_connect(localhost,$username,$password);
    @mysql_select_db($database) or die( "Unable to select database");


include 'library/config.php';
include 'library/opendb.php';

if(isset($_GET['id']))
{
   $query  = "SELECT id, title, content, contactname, contactemail, contactnumber ".
             "FROM vacancies".
             "WHERE id = '{$_GET['id']}'";
   $result = mysql_query($query) or die('Error : ' . mysql_error());
   list($id, $title, $content, $contactname, $contactemail, $contactnumber) = mysql_fetch_array($result, MYSQL_NUM);

   $content = htmlspecialchars($content);
}

if(isset($_POST['update']))
{
   $id = $_POST['id'];
   $title   = $_POST['title'];
   $content = $_POST['content'];
   $contactname = $_POST['contactname'];
   $contactemail = $_POST['contactemail'];
   $contactnumber = $_POST['contactnumber'];

   if(!get_magic_quotes_gpc())
   {
      $title   = addslashes($title);
      $content = addslashes($content);
      $contactname = addslashes($contactname);
      $contactemail = addslashes($contactemail);
      $contactnumber = addslashes($contactnumber);
   }

   // update the article in the database
   $query = "UPDATE vacancies
            SET title = '$title', content = '$content', contactname = '$contactname', contactemail = '$contactemail', contactnumber = '$contactnumber'".
        "WHERE id = '$id'";
   mysql_query($query) or die('Error : ' . mysql_error());

   // then remove the cached file
   $cacheDir = dirname(__FILE__) . '/cache/';
   $cacheFile = $cacheDir . '_' . $_GET['id'] . '.html';

   @unlink($cacheFile);

   // and remove the index.html too because the file list
   // is changed
   @unlink($cacheDir . 'index.html');

   echo "<b>Job Entry: '$title' updated</b>";

   // now we will display $title & content
   // so strip out any slashes
      $title   = stripslashes($title);
      $content = stripslashes($content);
      $contactname = stripslashes($contactname);
      $contactemail = stripslashes($contactemail);
      $contactnumber = stripslashes($contactnumber);

}

include 'library/closedb.php';
?>

Answer 1

查看http://us2.php.net/manual/en/function.mysql-query.php

问题是你在这里使用了太多的单引号：

“WHERE id ='{$ _GET ['id']}'”;

并且您的查询未按预期执行。改为使用mysql_real_escape_string（）。

Answer 2

一个问题：

   $query = "UPDATE vacancies
            SET title = '$title', content = '$content', contactname = '$contactname', contactemail = '$contactemail', contactnumber = '$contactnumber'".
        "WHERE id = '$id'";

导致最后一列和WHERE子句之间没有空格。将其更改为：

   $query = "UPDATE vacancies
            SET title = '$title', content = '$content', contactname = '$contactname', contactemail = '$contactemail', contactnumber = '$contactnumber' ".
        "WHERE id = '$id'";

或我的首选格式：

$query = <<<END
UPDATE vacancies
SET title = '$title',
    content = '$content',
    contactname = '$contactname',
    contactemail = '$contactemail',
    contactnumber = '$contactnumber'
WHERE id = '$id'
END;

注意：你应该使用mysql_real_escape_string（）来逃避字段。

Answer 3

删除

周围的引号

 {$_GET['id']}

和

$id

在您的所有查询中。

您的id是我假设的整数类型，它不能使用带引号的版本，或者尝试将整数键与字符串“1”匹配

- 改变这一行

$result = mysql_query($query) or die('Error : ' . mysql_error());

到

$result = mysql_query($query) or die('Error : ' . mysql_error() . "\n\n" . $query);

然后，您可以确切地看到进入数据库的查询。然后你可以在这里发帖给我们看。

也请发帖

describe <tablename>;

Answer 4

试试这个：

$query  = "SELECT id, title, content, contactname, contactemail, contactnumber ".
         "FROM vacancies ".
         "WHERE id = '".$_GET['id']."'";

我总是试图将变量从我的字符串中删除，只需将它们添加到句点中，我发现它消除了很多混淆。

mysql语法问题

4 个答案: