好吧,让我们在表users_chars中看起来像这样
pos_zone
255
表二中的
zone_id | name
255 | This_Area
我将如何比较它们并显示名称行而不是id
好的,使用@missdefying发布的方法,我想出来这是好的吗?
////Zone id to name
$zoneid=myQ(myF("SELECT t.zoneid
FROM users_chars c
INNER JOIN zone_settings t ON c.pos_zone = t.zoneid"));
$result=mysql_query("SELECT t.name
FROM users_chars uc
INNER JOIN zone_settings t ON uc.pos_zone = t.zoneid");
while($row = mysql_fetch_array($result))
{
$zonename = "";
if (($row['pos_zone'] == $zoneid)) {
$zonename = "<td>" . $row['name'] . "</td>";}
echo "<td>" . $zonename ."</td>";}
@missdefying谢谢