从变量变量调用定义的常量-解网

问：

我试图在一个单独的函数中引用一个定义的常量。我得到的错误是指未定义的变量以及定义为每个 FOO 和 BAR 的常量的变量。

class Boo {
public function runScare() {
    $this->nowScaring('FOO');
    $this->nowScaring('BAR');
}
private function nowScaring($personRequest) {
    define ('FOO', "test foo");
    define ('BAR', "test bar");
    $person = ${$personRequest};
    echo "<br/>Query selected is: " . $person . "<br/>";
}
}
$scare = new Boo;
$scare->runScare();

php oop 变量

class Boo { 
  public function runScare() { 
      $this->nowScaring(FOO); // no quotes
      $this->nowScaring(BAR); // no quotes
  } 
  private function nowScaring($person) {
      // And no need to "grab their values" -- this has already happened
      echo "<br/>Query selected is: " . $person . "<br/>"; 
  } 
}

如果出于某种原因，你想得到常量的值，而你所拥有的只是它在变量中的名称，你可以使用常量函数来实现：

define ('FOO', "test foo"); 

$name = 'FOO';
$value = constant($name);

// You would get the same effect with
// $value = FOO;

在这种特殊情况下，类常量可能更适合：

class Boo { 
  const FOO = "test foo";
  const BAR = "test bar";


  public function runScare() { 
      $this->nowScaring(self::FOO); // change of syntax
      $this->nowScaring(self::BAR); // no quotes
  } 
  private function nowScaring($person) {
      echo "<br/>Query selected is: " . $person . "<br/>"; 
  } 
}

0赞 Timur 12/16/2011 #3

class Boo {
public function runScare() {
    $this->nowScaring('FOO');
    $this->nowScaring('BAR');
}
private function nowScaring($personRequest) {
    if( !defined('FOO') ){
        define ('FOO', "test foo");
    }
    if( !defined('BAR') ){
        define ('BAR', "test bar");
    }
    $person = constant($personRequest);
    echo "<br/>Query selected is: " . $person . "<br/>";
}
}
$scare = new Boo;
$scare->runScare();

但我认为在某个类的方法中定义常量不是一个好主意。当然，在大多数情况下，您不需要按变量检索它们的值。

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从变量变量调用定义的常量

Call defined constant from variable variable

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