问题描述
是否有可能在不使用$的情况下引用了变量?
is it possible that a variable is referenced without using the $?
例如:
if ($a != 0 && a == true) {
...
}
我不这么认为，但是代码(不是我写的)没有显示错误，我认为这很奇怪.我忽略了代码，并且a也不是常量.
I don't think so, but the code (not written by me) doesn't show an error and I think it's weird. I've overlooked the code and a is not a constant either.
推荐答案
在PHP中，可以定义一个常量，该常量将不具有$，但是一个变量必须具有一个.但是，这不是变量，也不能替代变量.常量只应定义一次，并且在脚本的整个生命周期中都不得更改.
In PHP, a constant can be defined, which would then not have a $, but a variable must have one. However, this is NOT a variable, and is not a substitute for a variable. Constants are intended to be defined exactly once and not changed throughout the lifetime of the script.
define('a', 'some value for a');
此外，您不能在双引号或HEREDOC字符串内插值常量的值:
Additionally, you cannot interpolate the value of a constant inside a double-quoted or HEREDOC string:
$a = "variable a"
define('a', 'constant a');
echo "A string containing $a";
// "A string containing variable a";
// Can't do it with the constant
echo "A string containing a";
// "A string containing a";
最后，PHP可能发布Use of undefined constant a - assumed 'a'的通知，并将其解释为错误引用的字符串"a".查看您的错误日志以查看是否正在发生.在这种情况下，"a" == TRUE是有效的，因为字符串"a"是非空的，并且将其与布尔TRUE进行松散比较.
Finally, PHP may issue a notice for an Use of undefined constant a - assumed 'a' and interpret it as a mistakenly unquoted string "a". Look in your error log to see if that is happening. In that case, "a" == TRUE is valid, since the string "a" is non-empty and it is compared loosely to the boolean TRUE.
echo a == TRUE ? 'true' : 'false';
// Prints true
// PHP Notice: Use of undefined constant a - assumed 'a'
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