How can I check if a constant is defined in a PHP class?
class Foo {
const BAR = 1;
}
Is there something like property_exists() or method_exists() for class constants? Or can I just use defined("Foo::BAR")?
检查是否存在类常量
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4条回答 默认 最新
- dsfhe34889789708 2014-06-11 09:26关注
Yes, just use the class name in front of the constant name.
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- 回答 4 已采纳 Yes, just use the class name in front of the constant name. http://www.php.net/manual/en/function
- 2013-01-20 21:05回答 6 已采纳 First, these are not variables, but constants. And you can check their existence by using the def
- 2017-02-19 23:14回答 1 已采纳 This is a good example of how late static binding works. I won't rewrite the documentation for it
- 2021-04-28 04:46心妍妈妈的博客 在php中变量类型我们有常用变量与常量，下面...defined() 函数检查某常量是否存在。若常量存在，则返回 true，否则返回 false。 代码如下 复制代码 if (defined('MYCONSTANT')) {echo "常量MYCONSTANT存在";}else{e...
- 2016-11-15 17:42回答 1 已采纳 You need to include namespace in the constant() call - see code @ 3v4l.org EDIT To correct the pr
- 2014-10-27 11:20回答 1 已采纳 PHP 5,4 does not allow for expressions in the class properties declaration. Such feature has been
- 2012-03-22 19:17回答 1 已采纳 Do I need to create an internal function to iterate the constants and return an array? Nope!
- 2021-04-16 13:52克莱登大学门卫的博客 php判断常量、变量和函数是否为函数判断常量是否被定义：if (defined('CONST_NAME')) {//do something}判断变量是否存在：isset() ，注意变量未声明或声明时赋值为NULL，isset均返回FALSE，如：if (isset($var_name)...
- 2014-07-13 17:21回答 2 已采纳 In fact this works just fine: http://3v4l.org/pkNXs class Client { const QUERY_SELECT = 'foo'
- 2018-02-04 15:20回答 7 已采纳 例如 ``` public enum AuditStatus{ /**审核中*/ AUDITING(0), /**审核通过*/
- 2015-05-16 13:40回答 2 已采纳 I am not exactly sure what you expect, but note that !== has a higher precedence than & so you are
- 2021-04-16 13:53编程之翼的博客 defined() 函数检查某常量是否存在。若常量存在，则返回 true，否则返回 false。代码如下if (defined('MYCONSTANT')) {echo "常量MYCONSTANT存在";}else{echo "常量MYCONSTANT不存在";}echo "";isset函数是检测变量...
- 回答 1 已采纳 change: defined(HELLO) to: defined('HELLO') http://php.net/manual/en/function.defined.php
- 2021-04-11 13:41Miss豆子的博客 请注意，此函数并不关心常量的值，它仅关心常量是否存在。即使常量的值为null或false函数仍将返回true。define("GOOD",false);if(defined("GOOD")){print"GOODisdefined";//prints"GOODisdefined"if(GOOD){print"...
- 2021-04-20 01:08Being漠的博客 defined() 函数检查某常量是否存在。若常量存在，则返回 true，否则返回 false。代码如下复制代码if (defined('MYCONSTANT')) {echo "常量MYCONSTANT存在";}else{echo "常量MYCONSTANT不存在"...
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