PHP私有變量是「未定義」，雖然它被定義

Question

我一直在玩弄一個RUNESCAPE hiscore形象的創造者。人們傾向於在論壇上使用這些簽名來簽名。

而返工的一些代碼，我把它弄壞了。我越來越：

Undefined variable: exp in /assets/user.php on line 8 Fatal error: Cannot access empty property in /assets/user.php on line 8

然而，該變量甚至沒有在第8行（我假設的東西，用它做修剪空格）。

這裏是我的user.php的文件

<?php 

class User { 

public static $names = ["Overall", "Attack", "Defence", "Strength", "Hitpoints", "Ranged", "Prayer", "Magic", "Cooking", 
"Woodcutting", "Fletching", "Fishing", "Firemaking", "Crafting", "Smithing", "Mining","Herblore", "Agility", "Thieving", "Slayer", 
"Farming", "Runecrafting", "Hunter", "Construction", "Summoning","Dungeoneering", "Divination", "Invention"]; 

private $user; 
private $mySkills = []; 
private $exp = []; 

public function __construct($result) { 
    $array = explode(",", $result); 
    $id = 0; 
    $arrId = 1; 
    while($arrId < count($result)) { 
     $this->$mySkills[$id] = $array[$arrId++]; 
     $temp = explode(" ", $array[$arrId++]); 
     $this->$exp[$id++] = $array[0]; 
    } 
} 

public function getTotalXp() { 
    return $this->$exp[0]; 
} 

public function getLevel($skill) { 
    global $names; 
    for ($x = 0; $x <= count($names); $x++) { 
     if(strcasecmp($skill, $names[$x]) == 0) { 
      return $this->$mySkills[$x]; 
     } 
    } 
    return 0; 
} 

public function getExp($skill) { 
    global $names; 
    for ($x = 0; $x <= count($names); $x++) { 
     if(strcasecmp($skill, $names[$x]) == 0) { 
      return $this->$exp[$x]; 
     } 
    } 
    return 0; 
} 
} 
?> 

我正在用$名稱錯誤，但那是因爲我沒有使用在函數全局$名稱。

Answer 1

您使用了錯誤的語法。

使用$this->variableName不$this->$variableName

即改變return $this->$exp[0];到return $this->exp[0];