¿Tomar x número de palabras antes y después de una palabra clave determinada?

Question

¿Cómo puedo obtener el [x] número de palabras antes y después de una palabra clave dada en una cadena en PHP? Estoy tratando de mostrar los resultados de una consulta de MySQL adaptada a la palabra clave como un fragmento.

Answer 1

$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 />'; 

Pero, ¿qué quieres que suceda si la palabra aparece varias veces? O si la palabra no aparece en la cadena?

EDITAR

Versión ampliada del getWords2 para regresar hasta un número determinado de veces que aparece la palabra clave

$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));