将MySQL表关联到特定值

Question

What I want to do is quite simple: I want to recover data from a database, only where the ID of the first table (stored in session) is similar to the ID of the second table. 我想做的事情很简单：我想从数据库中恢复数据，只有第一个表（存储在会话中）的ID与第二个表的ID相似。 My code is this one below. 我的代码如下。 The only problem: the present code gives me the following error: 唯一的问题：当前代码给我以下错误：

Query is invalid: You have an error in your SQL syntax; 查询无效：您的SQL语法有误； check the manual that corresponds to your MySQL server version for the right syntax to use near '.' 请查看与您的MySQL服务器版本相对应的手册，以获取在'。'附近使用的正确语法。 at line 1 在第1行

What am I doing wrong? 我究竟做错了什么？ Thanks for your help! 谢谢你的帮助！

<?php include "base.php"; ?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-    strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">  
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />  

<?php
//We check if the user is logged
if(isset($_SESSION['Username']))
{
?>

<?php

//query
$query = mysql_query("select TitreEvent, DescriptionEvent from users_event where ID = .$_SESSION    [id].") or die ('Query is invalid: ' . mysql_error());

//write the results

while ($row = mysql_fetch_array($query)) {
echo $row['TitreEvent'] . " " . $row['DescriptionEvent'] . "";

// close the loop
}

?>


<div class="message">To access this page, you must be logged.<br />
<a href="http://www.groupe90.webege.com/index.php">Log in</a></div>
<?php
}
?>
             <div class="foot"><a href="<?php echo $url_home; ?>">Go Home</a> - <a    href="http://www.webestools.com/">Webestools</a></div>
     </body>
 </html>

Answer 1

Your PHP is broken where you build the query string: 您的PHP在构建查询字符串的位置已损坏：

$query = mysql_query("select [...snip...] where ID = .$_SESSION    [id].") or die ('Query is invalid: ' . mysql_error());
                                                    ^-- here

you never ended the string, so the $_SESSION portion is actually part of the query string, producing 您永远不会结束字符串，因此$ _SESSION部分实际上是查询字符串的一部分，产生了

where ID = Array   [id].

It should be 它应该是

 where ID = " . $_SESSION['id']) ...

Answer 2

$query = mysql_query("select TitreEvent, DescriptionEvent from users_event where ID = '$_SESSION[id]'") or die ('Query is invalid: ' . mysql_error());

Try the above code 试试上面的代码

You have to put your condition in single quote 您必须将条件放在单引号中

将MySQL表关联到特定值

问题描述

2 个解决方案

解决方案1
2 已采纳 2013-07-25 18:10:34

解决方案2
0

将MySQL表关联到特定值

问题描述

2 个解决方案

解决方案1 2 已采纳 2013-07-25 18:10:34

解决方案2 0

解决方案1
2 已采纳 2013-07-25 18:10:34

解决方案2
0