PHP If 语句无法正常工作，not null

发布于 2024-11-17 23:44:31 字数 580 浏览 0 评论 0原文

我很难理解为什么下面的 if 语句总是导致错误。我正在创建一个函数，该函数将测试与脚本的传入连接，该脚本将拒绝某些机器人建立的连接。

在下面的测试中，在应用 if 逻辑时，我期望一个 TRUE，因为数组 $value 和 $test 值都应该匹配...导致 NOT无效的？

$bots = array(0 => "PaperLiBot", 1 => "TweetmemeBot", 2 => "Appsfirebot", 3 => "PycURL", 4 => "JS-Kit", 5 => "Python-urllib");

$test = strtolower("PaperLiBot");


foreach($bots as $value)
{   
    $i = strtolower(strpos($value, $test));

    if ($i != NULL)
    {
        echo "Bot is found";
        exit;
    }else
    {
        echo "not found";
    }

}

原文

I'm struggling to understand why my if statement below always results in false. I am creating a function which will test incoming connections to a script which will reject connections made by certain bots.

In my test below, on applying the if logic, I'm expecting a TRUE as both the array $value and $test value should match... resulting in a NOT NULL?

$bots = array(0 => "PaperLiBot", 1 => "TweetmemeBot", 2 => "Appsfirebot", 3 => "PycURL", 4 => "JS-Kit", 5 => "Python-urllib");

$test = strtolower("PaperLiBot");


foreach($bots as $value)
{   
    $i = strtolower(strpos($value, $test));

    if ($i != NULL)
    {
        echo "Bot is found";
        exit;
    }else
    {
        echo "not found";
    }

}

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羁拥 2024-11-24 23:44:31

我认为你正在努力实现这个目标

foreach($bots as $value)
{   
    $i = strpos(strtolower($value), $test);

    if ($i !== false){
        echo "Bot is found";
        exit;
    }else{
        echo "not found";
    }

}

I think you are trying to accomplish this

foreach($bots as $value)
{   
    $i = strpos(strtolower($value), $test);

    if ($i !== false){
        echo "Bot is found";
        exit;
    }else{
        echo "not found";
    }

}

回复收藏 0 原文

梦断已成空 2024-11-24 23:44:31

你想写：

if(stristr($value, $test)){
 // found
}else{
 // not found

}

you want to write:

if(stristr($value, $test)){
 // found
}else{
 // not found

}

回复收藏 0 原文

套路撩心 2024-11-24 23:44:31

PHP 中的 null 可以相互类型转换为 '0'、''、' ' 等...您需要使用严格比较来检查基于 0 的数组索引：

if ($i !== NULL) { 
 ...
}

注意额外的 = 在不等式运算符中。它强制 PHP 比较类型和值，而不仅仅是值。当比较值时，PHP 会根据需要进行类型转换以使测试正常进行，这意味着 null == 0 为 true，但 null === 0 为 false 。

null in PHP is mutually type-castable to '0', '', ' ', etc... You need to use the strict comparisons to check for a 0-based array index:

if ($i !== NULL) { 
 ...
}

note the extra = in the inequality operator. It forces PHP to compare type AND value, not just value. When comparing value, PHP will typecast however it wants to in order to make the test work, which means null == 0 is true, but null === 0 is false.

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