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我正在尝试在 MySQL 数据库中搜索用户特定的网站活动，为此我创建了一个表单和流程来搜索统计表并返回每个记录与查询匹配的用户 ID。但是我不断收到此消息，不知道为什么：

You have an error in your SQL syntax; check the manual that corresponds to your MySQL     server version for the right syntax to use near '@hotmail.com' at line 1

这是我正在使用的表格

<form method='get' action='searchuser.php'  id='searchuser'>
<input type='text' input name='txtSearch' id='txtSearch'>
    <input type='submit' name='submit' value='Search'>
</form>";

这是进程文件

<?php
require_once( "Functions.php" );
$header = makeHeader();

$con= connect();


$user = $_GET['txtSearch'];

$query = "SELECT * FROM statistics WHERE userID = $user";
$result=mysql_query($query) or die (mysql_error());

echo"<table border='1'><th>User</th><th>IP</th>
    <th>Date</th><th>Page visited</th><th>Page from</th>";

while($row = mysql_fetch_assoc($result))
{
$username = $row['userID'];
$ip =  $row['ipAddress'];
$date = $row['dateOfVisit'];
$pagev = $row['pageVisited'];
$pagef = $row['pageFrom'];

echo "<tr><td>".$row->UserID."</td><td>".$row->ipAddress."</td><td>".$row->dateOfVisit."</td><td>".
    $row->pageVisited."</td><td>".$row->pageFrom . "<br/>\n"."</td></tr>";
    }
IF (mysql_num_rows($queryresult) == "") 
            {
                Echo "<p>Sorry there were no results for your         search<p> <br /><br /> <p><A HREF='javascript:javascript:history.go(-1)'>Click here to go back to previous page</A></p>";
            } 


$footer = makeFooter();

?>
4

2 回答 2

3

SQL问题出在

userID = $user

将字符串放入sql查询的正确方法是

  • 用引号分隔它
  • 并使用 mysql_real_escape_string 来转义内部可能发生的分隔符。

所以。正确的代码是

  $user =  mysql_real_escape_string($user);
  $query = "SELECT * FROM statistics WHERE userID = '$user'";
于 2012-04-20T12:38:00.910 回答
-2

试试这个

$user = $_GET['txtSearch'];
$query = "SELECT * FROM statistics WHERE userID = ".$user." ";

这对我有用

于 2012-04-20T12:52:49.917 回答