PHP MySQL查询帮助
我试图使用此查询返回每个实例,其中变量$d['userID']等于单独表中的用户ID,然后回显与该用户ID绑定的用户名 以下是我目前掌握的情况:PHP MySQL查询帮助,php,mysql,Php,Mysql,我试图使用此查询返回每个实例,其中变量$d['userID']等于单独表中的用户ID,然后回显与该用户ID绑定的用户名 以下是我目前掌握的情况: $uid = $d['userID']; $result = mysql_query("SELECT u.username FROM users u LEFT JOIN comments c ON c.userID = u.id WHERE u.id = $uid;")$row = mysql_fetch_assoc($result); echo $
$uid = $d['userID'];
$result = mysql_query("SELECT u.username
FROM users u
LEFT JOIN comments c
ON c.userID = u.id
WHERE u.id = $uid;")$row = mysql_fetch_assoc($result);
echo $row['username'];
试试这个:
$uid = $d['userID'];
$result = mysql_query("SELECT u.username
FROM users u
LEFT JOIN comments c
ON c.userID = u.id
WHERE u.id = $uid");
while($row = mysql_fetch_assoc($result))
{
echo $row['username'];
}
*编辑:*删除分号。如果您希望获得所有结果。将您的
mysql\u fetch\u assoc
包装在while
循环中:
$result = mysql_query("SELECT u.username
FROM users u
LEFT JOIN comments c
ON c.userID = u.id
WHERE u.id = $uid;");
while($row = mysql_fetch_assoc($result)){
echo $row['username'];
}
这应该行得通
$uid = mysql_real_escape_string($d['userID']);
$result = mysql_query("SELECT u.username
FROM users u
LEFT JOIN
comments c
ON c.userID = u.id
WHERE u.id = '$uid'");
while($row = mysql_fetch_assoc($result))
{
//PRINTS ALL INSTANCES OF THE ROW
echo $row['username'];
}
上面的echo打印了什么。echo在这里打印了什么吗?您在$result=…结尾缺少了一个分号,很抱歉被复制了。在页面刷新之前还有其他的回应。你仍然需要在最后加上分号。Thx tsegay,我想我已经喝够了啤酒。该睡觉了:)@Cybernate对你很好,新年快乐。如果是这样,你可以忽略while循环。修复了SQL注入漏洞()