如果被称为用户的SQL表在被称为gangs的结构中有一个结果，那么如果该结果是我要查找的结果，则在列表中显示所有找到的结果。这是我到目前为止的代码，witch不工作，请帮忙，谢谢
$sql = "SELECT * FROM users WHERE id='".mysql_real_escape_string($_SESSION['user_id'])."'";
$query = mysql_query($sql) or die(mysql_error());
$row = mysql_fetch_object($query);
$id = htmlspecialchars($row->id);
$userip = htmlspecialchars($row->userip);
$name = htmlspecialchars($row->name);
$sitestate = htmlspecialchars($row->sitestate);
$password = htmlspecialchars($row->password);
$mail = htmlspecialchars($row->mail);
$money = htmlspecialchars($row->money);
$exp = htmlspecialchars($row->exp);
$rank = htmlspecialchars($row->rank);
$health = htmlspecialchars($row->health);
$points = htmlspecialchars($row->points);
$profile = htmlspecialchars($row->profile);
$gang = htmlspecialchars($row->gang);
<?php
$sql = "SELECT * FROM Gangs WHERE name='".mysql_real_escape_string($_GET['name'])."'";
$query = mysql_query($sql) or die(mysql_error());
$row = mysql_fetch_object($query);
$Gang_name = htmlspecialchars($row->name);
$Gang_owner = htmlspecialchars($row->owner);
$Gang_money = htmlspecialchars($row->money);
$Gang_exp = htmlspecialchars($row->exp);
$Gang_level = htmlspecialchars($row->level);
$Gang_profile = htmlspecialchars($row->profile);
?>
<?php
$result = mysql_query("SELECT * FROM users WHERE gang = '".$gang_name."'");
if ($result) {
while($row = mysql_fetch_assoc($result)) {
$members = $row['name'];
}
}
?>
<?php echo $members; ?>
看起来应该只使用带有联接的SELECT查询
"SELECT * FROM users as u JOIN gangs as g on u.gang = g.name WHERE g.name = '".mysql_real_escape_string($_GET['name'])."'";
如果你试图构建一个结果行数组，这个：
$members = $row['name'];
应为：
$members[] = $row['name'];
您还应该像一样在循环之前声明$memberes变量
$members = array();