在多个if语句中检测哪个条件为假 [英] Detect which condition is false in multiple if statement

问题描述

我尝试缩短代码，因此也缩短了以下类型的if语句:

I try to shorten my code, and so I come along to shorten the following type of if statement:

// a,b,c,d needed to run
if ( empty(a) ) {
    echo 'a is empty';
} elseif ( empty(b) ) {
    echo 'b is empty';
} elseif ( empty(c) ) {
    echo 'c is empty';
} elseif ( empty(d) ) {
    echo 'd is empty';
} else {
  // run code with a,b,c,d
}

有没有一种方法可以检测出哪个条件为假(为空)?

Is there a way to detect which one of the conditions was false (is emtpy)?

if ( empty(a) || empty(b) || empty (c) || empty(d) ) {
     echo *statement n*.' is empty';
} else {
  // run code with a,b,c,d
}

我想到了一个for循环，但是那需要大量的代码更改. 也许有人可以指出我正确的方向.

I thought about a for loop, but that would need massive code changes. Maybe someone can point me to the right direction.

先谢谢您了:)

詹斯

推荐答案

使用compact，array_filter和array_diff:

$arr = compact( 'a', 'b', 'c', 'd' );
if( count( $empty = array_diff( $arr, array_filter( $arr ) ) ) )
{
    echo key( $empty ) . ' is empty';
}
else
{
    echo 'OK';
}

这样，在$empty中，您将拥有所有空值.因此，您可以对所有按键发出警告:

By this way, in $empty you have all empty values. So you can echo a warning for all keys:

echo 'Empty: ' . implode( ', ', array_keys( $empty ) );

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