php SQL语法错误

Question

我尝试了几种方式来解决这个问题，但错误消息几乎保持不变：

15检索分数时出错 您的 SQL 语法有错误；检查与您的 MySQL 服务器版本相对应的手册，以在第 1 行的“table WHERE id>15 1 ORDER BY id ASC LIMIT 0,100”附近使用正确的语法

我打的电话是

http://myserver.com/get_dbupdates2.php?theid=15

    $type   = isset($_GET['type']) ? $_GET['type'] : "global";
$offset = isset($_GET['offset']) ? $_GET['offset'] : "0";
$count  = isset($_GET['count']) ? $_GET['count'] : "100";
$sort   = isset($_GET['sort']) ? $_GET['sort'] : "id ASC";

// Localize the GET variables
$udid  = isset($_GET['udid']) ? $_GET['udid'] : "";
$name  = isset($_GET['name']) ? $_GET['name']  : "";
$clubname  = isset($_GET['clubname']) ? $_GET['clubname']  : "";
$theid  = isset($_GET['theid']) ? $_GET['theid']  : ""; 


// Protect against sql injections
$type   = mysql_real_escape_string($type);
$offset = mysql_real_escape_string($offset);
$count  = mysql_real_escape_string($count);
$sort   = mysql_real_escape_string($sort);
$udid   = mysql_real_escape_string($udid);
$name   = mysql_real_escape_string($name);
$clubname   = mysql_real_escape_string($clubname);
$theid   = mysql_real_escape_string($theid);

    echo $theid;

// Build the sql query
//$sql = "SELECT * FROM $table WHERE ";
$sql = "SELECT * FROM $table WHERE id>$theid ";

switch($type) {
    case "global":
        $sql .= "1 ";
        break;
    case "device":
        $sql .= "udid = '$udid' ";
        break;
    case "name":
        $sql .= "name = '$name' ";
        break;
    case "clubname":
        $sql .= "clubname = '$clubname' ";
        break;
    case "theid":
        $sql .= "theid = '$theid' ";
        break;
}

$sql .= "ORDER BY $sort ";
$sql .= "LIMIT $offset,$count ";

$result = mysql_query($sql,$conn);

谁能看出我哪里出错了？

最诚挚的问候， -代码

编辑

看到这两行

//$sql = "SELECT * FROM $table WHERE ";
$sql = "SELECT * FROM $table WHERE id>$theid ";

如果我注释掉最后一行，取消注释第一行，脚本运行正常并返回数据。但是保持原样会产生错误。

所以这让我相信问题与此有关

id>$theid ";  

问候 -代码

Answer 1

命名表'table'必须这样引用

SELECT * FROM `table`

并且您必须在条件之间定义 AND 或 OR，例如$sql .= "AND clubname = '$clubname' ";