在带有OR的PHP中的IF语句中,如何跟踪哪个条件评估为false？

Question

例如,在这样的PHP代码中:

if($this->function1() && $this->function2())
{
  //everything is alright
}
else
{
  //function1 or function2 returned false
  //whodunnit?
}

是否有语言结构或某些东西可以帮助我找出else块中的罪魁祸首？我不想再次运行条件,只是为了找出问题所在.

请忍受我的伪代码.我的实际代码在这里看起来完全脱离了上下文...顺便说一句,我正在使用CodeIgniter.

Answer 1

if($this->function1()) {
    if($this->function2()) {
        // both worked
    } else {
        // function1 returned a true value
        // function2 returned a false value
    }
} else {
    // function1 returned a false value
    // (you could optionally call $this->function2() here)
}

或者,如果您总是想要运行这两个功能:

$result1 = $this->function1();
$result2 = $this->function2();
if($result1 && $result2) {
    // everything is okay
} else {
    // look at the results to figure out what to do
}

或者,如果你想做第一个但没有2个级别的缩进:

if(!$this->function1()) {
    // function1 returned false
    // (you can call $this->function2() here if you want)
} elseif (!$this->function2()) {
    // function2 returned false
} else {
    // everything is okay
}