php - If语句错误且变量

我已经有一段时间这个问题了，最后我将其范围缩小了（我认为）。
我试图对收入设置一个限制，该限制是$ storagecap每个$仓库=到500 $ storagecap。在我向您展示代码之前，我将告诉您我的问题。
它显示了您目前拥有2个仓库的限额的示例，它说我的限额为1000，但它使我的收入停止在499（我将其设置为等于0个仓库），因此显示499/1000。 Cron_turn将您的收入添加到您的帐户中。

我不知道为什么知道$ storagecap为1000时为什么会在499停下来。

storagecap.php

<?php

$warehouse = $building['warehouse'];

if ($warehouse == 1) {
$storagecap = 500;
}elseif($warehouse == 2) {
$storagecap = 1000;
}elseif($warehouse == 3) {
$storagecap = 1500;
}elseif($warehouse == 4) {
$storagecap = 2000;
}elseif($warehouse == 5) {
$storagecap = 2500;
}elseif($warehouse == 6) {
$storagecap = 3000;
}elseif($warehouse == 7) {
$storagecap = 3500;
}elseif($warehouse == 8) {
$storagecap = 4000;
}elseif($warehouse == 9) {
$storagecap = 4500;
}elseif($warehouse == 10) {
$storagecap = 5000;
}elseif($warehouse == 11) {
$storagecap = 5500;
}elseif($warehouse == 12) {
$storagecap = 6000;
}elseif($warehouse == 13) {
$storagecap = 6500;
}elseif($warehouse == 14) {
$storagecap = 7000;
}elseif($warehouse == 15) {
$storagecap = 7500;
}elseif($warehouse == 16) {
$storagecap = 8000;
}elseif($warehouse == 17) {
$storagecap = 8500;
}elseif($warehouse == 18) {
$storagecap = 9000;
}elseif($warehouse == 19) {
$storagecap = 9500;
}elseif($warehouse == 20) {
$storagecap = 10000;
}elseif($warehouse == 0) {
$storagecap = 499;
}else{
$storagecap = 10;
}
?>

* cron_turn.php *

<?php
include("functions.php");
connect();
include("safe.php");
include("storagecap.php");
$get_users = mysql_query("SELECT * FROM `stats`") or die(mysql_error());

while($user = mysql_fetch_assoc($get_users)) {
  //Get
  $gold = $user["gold"];
  $food = $user["food"];
  $energy = $user["energy"];

  //Increment
  $gold += $user["income"];
  $food += $user["farming"];
  $energy += 5;

  //Verify and correct
  if ($energy > 100) {
     $energy = 100;
  };

  if ($gold > $storagecap) {
     $gold = $storagecap;
  };

  if ($food > $storagecap) {
     $food = $storagecap;
  };

  //Submit
  $update = mysql_query("UPDATE `stats` SET
                    `gold`= '".$gold."',
                    `food`= '".$food."',
                    `energy`= '".$energy."' WHERE `id`='".$user['id']."'") or     die(mysql_error());
}
?>

最佳答案

在将值分配给$building['warehouse']之前，要先包含storagecap.php文件。由于要在if语句中将null与int进行比较，因此0表示为true。将包含向下移动到您具有$building['warehouse']值的点