Php 检查字符串中是否存在数组元素

我原以为用原生php函数实现这一点很简单，但我发现了一些不同的、都相当复杂的方法，人们试图实现它。检查字符串是否包含数组中的一个或多个元素的最有效方法是什么？i、 e，下面-其中$data['description']是一个字符串。Obv下面的in_数组检查中断，因为它希望param 2是一个数组

$keywords = array(
            'bus',
            'buses',
            'train',
    );

    if (!in_array($keywords, $data['description']))
            continue;

---

您可以使用正则表达式来实现这一点-

if( !preg_match( '/(\b' . implode( '\b|\b', $keywords ) . '\b)/i', $data['description'] )) continue;

结果regexp将是

/（\bbus\b |\bbuses\b |\b train\b）/

，假设字符串是折叠/分隔的值列表 例1：

arrayInString( array( 'red' , 'blue' ) , 'red,white,orange' , ',' );
// Would return true
// When 'red,white,orange' are split by ',',
// the 'red' element matched the array

arrayInString( array( 'apple' , 'banana' ) , 'I ate an apple' );
// Would return true
// As 'I ate an apple' contains 'apple'

例2：

arrayInString( array( 'mouse' , 'cat' ) , 'mouse' );
// Would return true
// When 'mouse' is split by ',' (the default deliminator),
// the 'mouse' element matches the array which contains only 'mouse'

arrayInString( array( 'car' , 'bus' ) , 'I was busy' );
// Would return true
// As 'bus' is present in the string, even though it is part of 'busy'

假设字符串是纯文本，您只需在其中查找指定单词的实例 例1：

arrayInString( array( 'red' , 'blue' ) , 'red,white,orange' , ',' );
// Would return true
// When 'red,white,orange' are split by ',',
// the 'red' element matched the array

arrayInString( array( 'apple' , 'banana' ) , 'I ate an apple' );
// Would return true
// As 'I ate an apple' contains 'apple'

例2：

arrayInString( array( 'mouse' , 'cat' ) , 'mouse' );
// Would return true
// When 'mouse' is split by ',' (the default deliminator),
// the 'mouse' element matches the array which contains only 'mouse'

arrayInString( array( 'car' , 'bus' ) , 'I was busy' );
// Would return true
// As 'bus' is present in the string, even though it is part of 'busy'

假设正在搜索整个单词和短语 此函数将从字符串中的短语数组中查找（不区分大小写）短语。如果找到，将重新组合短语，$position返回字符串中的索引。如果未找到，则返回FALSE

function findStringFromArray($phrases, $string, &$position) {
    // Reverse sort phrases according to length.
    // This ensures that 'taxi' isn't found when 'taxi cab' exists in the string.
    usort($phrases, create_function('$a,$b',
                                    '$diff=strlen($b)-strlen($a);
                                     return $diff<0?-1:($diff>0?1:0);'));

    // Pad-out the string and convert it to lower-case
    $string = ' '.strtolower($string).' ';

    // Find the phrase
    foreach ($phrases as $key => $value) {
        if (($position = strpos($string, ' '.strtolower($value).' ')) !== FALSE) {
            return $phrases[$key];
        }
    }

    // Not found
    return FALSE;
}

有趣的是，该函数演示了一种查找整个单词和短语的技术，这样就可以找到“bus”而不是“滥用”。只需在干草堆和针周围留出空间：

$pos = strpos(" $haystack ", " $needle ")

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我无法让strpos（$inString，$inArray）工作。第二个参数不能是strpos中的数组。你真的测试过这个吗？嗨，奥拉夫，谢谢你的评论。我不记得我做了什么测试，因为这个答案是两年前提交的。但是，我已经对代码进行了调整，以便它现在应该遍历数组并执行预期的检查。干净且简单。正是我需要的。