在PHP中,如何更容易地将整数值映射为字符串?
我如何重新考虑以下代码,使其更简洁、更易于维护在PHP中,如何更容易地将整数值映射为字符串?,php,Php,我如何重新考虑以下代码,使其更简洁、更易于维护 if ($row['vocation'] == 1) $vocation = "sorcerer"; if ($row['vocation'] == 2) $vocation = "druid"; if ($row['vocation'] == 3) $vocation = "paladin"; if ($row['vocation'] == 4) $vocation = "knight"; if ($row['vocation'] == 5)
if ($row['vocation'] == 1) $vocation = "sorcerer";
if ($row['vocation'] == 2) $vocation = "druid";
if ($row['vocation'] == 3) $vocation = "paladin";
if ($row['vocation'] == 4) $vocation = "knight";
if ($row['vocation'] == 5) $vocation = "master sorcerer";
if ($row['vocation'] == 6) $vocation = "elder druid";
if ($row['vocation'] == 7) $vocation = "royal paladin";
if ($row['vocation'] == 8) $vocation = "elite knight";
else $vocation = "none";
我建议使用如下数组:
static $vocations = array(
1 => 'sorceror',
2 => 'druid',
3 => 'paladin',
4 => 'knight',
5 => 'master sorceror',
6 => 'elder druid',
7 => 'royal paladin',
8 => 'elite knight',
);
$vocation =
isset($vocations[$row['vocation']]) ? $vocations[$row['vocation']] : 'none';
define('VOCATION_SORCEROR', 1);
define('VOCATION_DRUID', 2);
define('VOCATION_PALADIN', 3);
$vocations = array(
VOCATION_SORCEROR => 'sorceror',
VOCATION_DRUID => 'druid',
VOCATION_PALADIN => 'paladin'
);
正确地开始您的项目,使用const now来表示这些数值常量,并为自己省去一些后顾之忧。(除了像其他人建议的那样使用switch/case之外)下面是一个使用switch来完成此操作的示例:
switch ($row['vocation']) {
case 1:
$vocation = "sorcerer";
break;
case 2:
$vocation = etc..
default:
$vocation = "none";
}
这对于C、Java、C#等许多语言以及其他许多语言来说都很常见。下面的文章可能会更好一些。8个元素是可以的,但是如果列表包含1000个元素呢
$list = array("sorcerer", "druid", ...);
$vocation = "none";
if($row['vocation'] <= count($list)){
$vocation = $list[$row['vocation'] - 1];
}
$list=array(“魔法师”、“德鲁伊”…);
$cavity=“无”;
如果($row['cavity'],我会将建议与数组一起使用,并使用常量来表示整数值,如下所示:
static $vocations = array(
1 => 'sorceror',
2 => 'druid',
3 => 'paladin',
4 => 'knight',
5 => 'master sorceror',
6 => 'elder druid',
7 => 'royal paladin',
8 => 'elite knight',
);
$vocation =
isset($vocations[$row['vocation']]) ? $vocations[$row['vocation']] : 'none';
define('VOCATION_SORCEROR', 1);
define('VOCATION_DRUID', 2);
define('VOCATION_PALADIN', 3);
$vocations = array(
VOCATION_SORCEROR => 'sorceror',
VOCATION_DRUID => 'druid',
VOCATION_PALADIN => 'paladin'
);
还有一个建议:
<?php
class Action
{
const TYPE__ADD = 0;
const TYPE__VIEW = 1;
const TYPE__EDIT = 2;
const TYPE__PATCH = 3;
const TYPE__DELETE = 4;
const TYPE__MAP = [
self::TYPE__ADD => 'add',
self::TYPE__VIEW => 'access',
self::TYPE__EDIT => 'edit',
self::TYPE__PATCH => 'patch',
self::TYPE__DELETE => 'delete'
];
protected $type;
public function setType(int $type)
{
if (!isset(self::TYPE__MAP[$this->type])) throw new \Exception(sprintf('Invalid type. Possible options are: %s.', implode(',', self::TYPE__MAP)));
$this->type = $type;
}
public function getType(): int
{
return $this->type;
}
public function getTypeStr(): string
{
return self::TYPE__MAP[$this->type];
}
}
// Test
$action = new Action();
$action->setType(Action::TYPE__PATCH);
echo 'Action type is: '.$action->getTypeStr().', and its representative int value is: '.$action->getType();
这一点,加上一条注释,如果这些数据来自数据库,您可能会更好地将其存储在一个休假表中并将其连接起来。+1作为示例,但不会影响代码的可读性是一种改进,除非情况变得比设置$vocation复杂得多。噢,我希望原始的asker可以再次找到这个问题,当我更改标题时没有考虑到这一点