掌握给定关键字前后的x个单词? [英] Grab x number of words before and after a given keyword?

问题描述

如何在PHP字符串中的给定关键字前后捕获[x]个单词?我正在尝试将针对关键字的mysql查询的结果标记为片段.

How can I go about grabbing [x] number of words before and after a given keyword in a string in PHP? I am trying to tokenize results from a mysql query tailored to the keyword as a snippet.

推荐答案

$string = 'This is a test string to see how to grab words from an arbitrary sentence. It\'s a little hacky (as you can see from the results) - but generally speaking, it works.';

echo $string,'<br />';

function getWords($string,$word,$before=0,$after=0) {
    $stringWords = str_word_count($string,1);
    $myWordPos = array_search($word,$stringWords);

    if (($myWordPos-$before) < 0)
        $before = $myWordPos;
    return array_slice($stringWords,$myWordPos-$before,$before+$after+1);
}

var_dump(getWords($string,'test',2,1));
echo '<br />';
var_dump(getWords($string,'this',2,1));
echo '<br />';
var_dump(getWords($string,'sentence',1,3));
echo '<br />';
var_dump(getWords($string,'little',2,2));
echo '<br />';
var_dump(getWords($string,'you',2,2));
echo '<br />';
var_dump(getWords($string,'results',2,2));
echo '<br />';
var_dump(getWords($string,'works',2,2));

echo '<hr />';


function getWords2($string,$word,$before=0,$after=0) {
    $stringWords = str_word_count($string,1);
    $myWordPos = array_search($word,$stringWords);
    $stringWordsPos = array_keys(str_word_count($string,2));

    if (($myWordPos+$after) >= count($stringWords))
        $after = count($stringWords) - $myWordPos - 1;
    $startPos = $stringWordsPos[$myWordPos-$before];
    $endPos = $stringWordsPos[$myWordPos+$after] + strlen($stringWords[$myWordPos+$after]);

    return substr($string,$startPos,$endPos-$startPos);
}

echo '[',getWords2($string,'test',2,1),']<br />';
echo '[',getWords2($string,'this',2,1),']<br />';
echo '[',getWords2($string,'sentence',1,3),']<br />';
echo '[',getWords2($string,'little',2,2),']<br />';
echo '[',getWords2($string,'you',2,2),']<br />';
echo '[',getWords2($string,'results',2,2),']<br />';
echo '[',getWords2($string,'works',1,3),']<br />';

但是，如果单词多次出现，您想怎么办?还是如果单词没有出现在字符串中?

But what do you want to happen if the word appears multiple times? Or if the word doesn't appear in the string?

编辑

getWords2的扩展版本，以最多返回指定次数的关键字出现

Extended version of getWords2 to return up to a set number of occurrences of the keyword

$string = 'PHP is a widely-used general-purpose scripting language that is especially suited for Web development. The current version of PHP is 5.3.3, released on July 22, 2010. The online manual for PHP is an excellent resource for the language syntax and has an extensive list of the built-in and extension functions. Most extensions can be found in PECL. PEAR contains a plethora of community supplied classes. PHP is often paired with the MySQL relational database.';

echo $string,'<br />';

function getWords3($string,$word,$before=0,$after=0,$maxFoundCount=1) {
    $stringWords = str_word_count($string,1);
    $stringWordsPos = array_keys(str_word_count($string,2));

    $foundCount = 0;
    $foundInstances = array();
    while ($foundCount < $maxFoundCount) {
        if (($myWordPos = array_search($word,$stringWords)) === false)
            break;
        ++$foundCount;
        if (($myWordPos+$after) >= count($stringWords))
            $after = count($stringWords) - $myWordPos - 1;
        $startPos = $stringWordsPos[$myWordPos-$before];
        $endPos = $stringWordsPos[$myWordPos+$after] + strlen($stringWords[$myWordPos+$after]);

        $stringWords = array_slice($stringWords,$myWordPos+1);
        $stringWordsPos = array_slice($stringWordsPos,$myWordPos+1);

        $foundInstances[] = substr($string,$startPos,$endPos-$startPos);
    }
    return $foundInstances;
}

var_dump(getWords3($string,'PHP',2,2,3));

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