我使用GET来获取结果的ID。PHP中的SQL返回错误
$id = $_GET['id'];
然后我用下面的代码:
<?
$q = $database->friendlyDetails($id);
while($row=mysql_fetch_assoc($q))
{
$hu = $row['home_user'];
$ht = $row['home_team'];
$hs = $row['home_score'];
$au = $row['away_user'];
$at = $row['away_team'];
$as = $row['away_score'];
$game = $row['game'];
$name = $row['name'];
$match = $row['match_report1'];
$compid = $row['compid'];
$date = $row['date_submitted'];
$sub = $row['user_submitted'];
}
?>
而且friendDetails-
function friendlyDetails($i)
{
$q = "SELECT *
FROM ".TBL_SUB_RESULTS."
INNER JOIN ".TBL_FRIENDLY."
ON ".TBL_FRIENDLY.".id = ".TBL_SUB_RESULTS.".compid
WHERE ".TBL_SUB_RESULTS.".id = '$i'";
return mysql_query($q, $this->connection);
}
出于某种原因,该代码将只返回的是在ID = 1。 任何人都可以看到任何明显的我做错了吗?
编辑 SUB成绩表
ID | compid | home_user | home_team | away_user | away_team | home_score | away_score | report1 |日期|提交
FRIENDLY TABLE
ID |名称|游戏
你的数据库看起来像什么?你能发几行吗? – jordanstephens 2010-05-20 21:16:22
什么是错误? – 2010-05-20 21:17:08
您可能还想清理输入。使用'$ _GET ['id']'的原始内容运行查询是危险的。 – thetaiko 2010-05-20 21:20:32