PHP如果不等于(！=)和或(||)问题.为什么不起作用? [英] PHP if not equal(!=) and or (||) issue. Why doesnt this work?

问题描述

我知道这是简单的PHP逻辑，但这根本行不通...

I know this is simple PHP logic but it just won't work...

 $str = "dan";
 if(($str != "joe") 
   || ($str != "danielle")
   || ($str != "heather")
   || ($str != "laurie")
   || ($str != "dan")){         

 echo "<a href='/about/".$str.".php'>Get to know ".get_the_author_meta('first_name')." &rarr;</a>";
                  }

我在做什么错了?

推荐答案

我不确定您想要什么，但是该逻辑将始终为true.您可能要使用AND(&&)，而不是OR(||)

I am not exactly sure what you want, but that logic will always evaluate to true. You might want to use AND (&&), instead of OR (||)

曾经测试过的最远的语句是($str != "danielle")，并且当一条语句产生真值时，PHP进入该块只会有两种可能的结果.

The furthest statement that is ever tested is ($str != "danielle") and there are only two possible outcomes as PHP enters the block as soon as a statement yields true.

这是第一个:

$str = "dan";

$str != "joe" # true - enter block
$str != "danielle" #ignored
$str != "heather" #ignored
$str != "laurie" #ignored
$str != "dan" #ignored

这是第二个:

$str = "joe";

$str != "joe" # false - continue evaluating
$str != "danielle" # true - enter block
$str != "heather" #ignored
$str != "laurie" #ignored
$str != "dan" #ignored

如果将OR更改为AND，则它将继续求值，直到返回false:

If the OR was changed to AND then it keeps evaluating until a false is returned:

$str = "dan";

$str != "joe" # true - keep evaluating
$str != "danielle" # true - keep evaluating
$str != "heather"  # true - keep evaluating
$str != "laurie" # true - keep evaluating
$str != "dan"  # false - do not enter block

尽管该解决方案无法很好地扩展，但您应保留一个排除列表数组，并对照该列表进行检查:

The solution doesn't scale well though, you should keep an array of the exclude list and check against that do:

$str = "dan";
$exclude_list = array("joe","danielle","heather","laurie","dan")
if(!in_array($str, $exclude_list)){          
    echo " <a href='/about/".$str.".php'>Get to know ".get_the_author_meta('first_name')." &rarr;</a>";
}

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