php - 已知条件未被 if 语句捕获

Question

$trial = $row['Trial'];
$level = $row['membership_level'];

print $trial;

$number = count($domains);
if (strlen($domain) == 0) {$number = 0;}
if ($level == 4 and $number >= 10) {exit('You have reached your 10 domain limit and     cannot add anymore');}
if ($level == 3 and $number >= 1) {exit('You have reached your 1 domain limit and cannot add anymore');}
if ($level == 2 and $trail >= 1) {exit('You have already activated the trial, please purchase a license');}

if ($level == 2) {

根据打印的 Trail = 1，根据在下一个 if 语句中失败的级别 = 2。但由于某种原因，它永远不会失败

if ($level == 2 and $trail >= 1)

我已经尝试过 intval、strlen 并不管它。即使我知道 trail = 1 和 level = 2，它也不会在这一点上失败。

那么我错过了什么？

Answer 1

您拼错了 trial/trail：

$trial = $row['Trial'];
...
if ($level == 2 and $trail >= 1)