PHP if not equal(!=) and or (||) 问题。为什么这不起作用？
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PHP if not equal(!=) and or (||) issue. Why doesnt this work?
提问by Greg Thompson
I know this is simple PHP logic but it just won't work...
我知道这是简单的 PHP 逻辑，但它不起作用......
$str = "dan";
if(($str != "joe")
|| ($str != "danielle")
|| ($str != "heather")
|| ($str != "laurie")
|| ($str != "dan")){
echo "<a href='/about/".$str.".php'>Get to know ".get_the_author_meta('first_name')." →</a>";
}
What am I doing wrong?
我究竟做错了什么？
回答by Gazler
I am not exactly sure what you want, but that logic will always evaluate to true. You might want to use AND (&&), instead of OR (||)
我不确定您想要什么，但该逻辑将始终评估为true. 您可能想要使用 AND (&&)，而不是 OR (||)
The furthest statement that is ever tested is ($str != "danielle") and there are only two possible outcomes as PHP enters the block as soon as a statement yields true.
测试过的最远的语句是 ( $str != "danielle") 并且只有两种可能的结果，因为一旦语句产生真，PHP 就会进入块。
This is the first:
这是第一个：
$str = "dan";
$str != "joe" # true - enter block
$str != "danielle" #ignored
$str != "heather" #ignored
$str != "laurie" #ignored
$str != "dan" #ignored
This is the second:
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
If the OR was changed to AND then it keeps evaluating until a false is returned:
如果 OR 更改为 AND，则它会继续评估，直到返回 false：
$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
The solution doesn't scale well though, you should keep an array of the exclude list and check against that do:
但是，该解决方案不能很好地扩展，您应该保留一个排除列表的数组并检查它：
$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')." →</a>";
}
回答by joakimdahlstrom
Another approach is
另一种方法是
$name = 'dan';
$names = array('joe', 'danielle', 'heather', 'laurie', 'dan');
if(in_array($name,$names)){
//the magic
}
回答by Marc B
Welcome to boolean logic:
欢迎使用布尔逻辑：
$str = 'dan'
$str != "joe" -> TRUE, dan is not joe
$str != "danielle" -> TRUE, danielle is not dan
$str != "heather") -> TRUE, heather is not dan
$str != "laurie" -> TRUE, laurie is not dan
$str != "dan" -> FALSE, dan is dan
Boolean logic truth tables look like this:
布尔逻辑真值表如下所示：
and:
和：
TRUE && TRUE -> TRUE
TRUE && FALSE -> FALSE
FALSE && FALSE -> FALSE
FALSE && TRUE -> FALSE
or:
或者：
TRUE || TRUE -> TRUE
TRUE || FALSE -> TRUE
FALSE || TRUE -> TRUE
FALSE || FALSE -> FALSE
Your statement boiled down to:
你的陈述归结为：
TRUE || TRUE || TRUE || TRUE || FALSE -> TRUE
回答by trutheality
Based on your comment in Glazer's answer, it looks like you want to enter the if block when $stris not one of the listed names.
根据您在 Glazer 的回答中的评论，您似乎想在$str不是列出的名称之一时输入 if 块。
In that case it would be more readable if you write it as
在这种情况下，如果您将其写为
if( !( ($str == "joe") || ($str == "danielle") || ($str == "heather") || ($str == "laurie") || ($str == "dan") ) )
This actually readsas "if it's not one of these people..." to someone looking at your code. Which is equivalent to the slightly less obvious
对于查看您的代码的人来说，这实际上读作“如果它不是这些人中的一个……”。这相当于稍微不那么明显的
if( ($str != "joe") && ($str != "danielle") && ($str != "heather") && ($str != "laurie") && ($str != "dan") )
The fact that they're equivalent is called DeMorgan's law in logic.
它们等价的事实在逻辑上被称为德摩根定律。
回答by Patrick R
try this
尝试这个
$str = "dan";
if($str == "joe" || $str == "daniella" || $str == "heather" || $str == "laurine" || $str == "dan"){ ... }
回答by Patrick R
When you compare two strings you have to use strcmp().
当您比较两个字符串时，您必须使用 strcmp()。
