PHP如果不是空字符串和string ="this"或"that"怎么做？

Question

function test(){
$embedmode = 'normal';
if ( ( $embedmode != '' ) && ( $embedmode != 'normal' || $embedmode != 'popup' || $embedmode != 'popup' ) )
    return "<p>ARVE Error: mode is not set to 'normal', 'popup' or 'special' maybe typo</p>";
elseif ( $embedmode == '')
    $mode = 'default';
else
    $mode = $embedmode;

echo '<pre>';
var_dump($mode);
echo "</pre>";
}
echo test();

这是我的尝试,我现在感到头痛,它发出了回复信息,我不知道为什么

Answer 1

!($var == 'something')会是一样的($var != 'something').

(!$var == 'something')在进行比较之前,Do 会对$ var执行布尔运算.!$ var将返回false,除非$ var为空,因此它基本上会说(false =='something'),这将是假的.