PHP include() 返回值不一致还是我的无知？
哦，好吧，也许这是我的无知，但是关于 include< 的 PHP 文档/a> 声明说比较返回值时要小心：
<?php
// won't work, evaluated as include(('vars.php') == 'OK'), i.e. include('')
if (include('vars.php') == 'OK') {
echo 'OK';
}
// works
if ((include 'vars.php') == 'OK') {
echo 'OK';
}
?>
但这就是我所经历的； boolean false 如果失败，int 1 如果成功：
$exists = 'Found.php';
var_dump((include $exists)); // Type is: int, value is: 1
$notexists = 'NotFound.php';
var_dump((include $notexists)); // Type is: boolean, value is: false
这是我的错吗？为什么返回值不一致（例如，并不总是布尔值）并且与 PHP 文档不同？
Oh well, maybe it's my ignorance, but PHP documentation about include statement says take care when comparing return value:
<?php
// won't work, evaluated as include(('vars.php') == 'OK'), i.e. include('')
if (include('vars.php') == 'OK') {
echo 'OK';
}
// works
if ((include 'vars.php') == 'OK') {
echo 'OK';
}
?>
But this is what i've experienced; boolean false in case of fail, int 1 if success:
$exists = 'Found.php';
var_dump((include $exists)); // Type is: int, value is: 1
$notexists = 'NotFound.php';
var_dump((include $notexists)); // Type is: boolean, value is: false
Is this my bad? Why returning value is inconsistent (meaning not always boolean, for example) and differs from PHP documentation?
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嗯，这正是文档中所述的内容：
当它变得不一致时，就是当你实际返回包含文件中的某些内容时，
例如在 Found.php 中：
然后
还值得一读 PHP 中布尔值的计算方式：
http://php.net/manual/en/language.types.boolean.php
Well this is exactly what the documentation states:
When it becomes inconsistent is when you actually return something in the included file,
e.g. in Found.php:
And then
It is also worth reading up on how booleans evaluate in PHP:
http://php.net/manual/en/language.types.boolean.php