为什么在 PHP 中 true 大于 3
我想知道为什么 PHP 中的以下语句返回 true?
I am wondering why following statement in PHP is returning true?
true>=4
例如这样的行会回显 1
echo true>=4;
谁能解释一下这背后的逻辑?
Can anyone explain me the logic behind this?
推荐答案
除了 Davids 的回答之外，我还想添加一些东西来增加深度.
In addition to Davids answer, I thought to add something to give a little more depth.
PHP 与其他编程语言不同，如果您不小心操作符/语法，您可能会陷入像您遇到的那种棘手的坑洞.
PHP unlike other programming languages, if your not careful with your operators/syntax you can fall into tricky pot holes like the one you experience.
正如大卫所说，
4 也是真(因为它非零)，真等于真，所以它也大于或等于 true.
4 is also true (because it's non-zero), and true is equal to true, so it's also greater than or equal to true.
考虑到这一点真大于假.
真 = 1
假=0
举个例子:
$test = 1;
if ($test == true){
echo "This is true";
}else{
echo "This is false";
}
上面会输出
这是真的
但是如果你接受这个:
$test = 1;
if ($test === true){
echo "This is true";
}else{
echo "This is false";
}
上面会输出:
这是假的
添加的等号查找完全匹配，从而查找 integer 1 而不是 PHP 读取 1 为真.
The added equals sign, looks for an exact match, thus looking for the integer 1 instead of PHP reading 1 as true.
我知道这有点离题，但只是想解释一下 PHP 包含的一些坑.
希望对您有所帮助
回答你的问题:
回声真>=4;
您看到 1 作为输出的原因是因为 true/false 被解释为数字(见上文)
Reason you are seeing 1 as output, is because true/false is interpreted as numbers (see above)
无论你是在做 echo true>=4 还是只是 echo true; php 都将 true 设为 1，将 false 设为 0
Regardless if your doing echo true>=4 or just echo true; php puts true as 1 and false as 0
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