Question

What could I add to the following code to filter the returned results using keywords in an array?

发出关键词并发出单调呼的法典位于贾瓦斯里。

下面的代码开立档案,使用请求使用一个点子,开立档案查找点,并从点子到档案末检索所有记录线。然后,它形成一个JSON物体,然后送回Javacript。

function tail(){
    $keywords = json_decode($_REQUEST[ keywords ]);

    $file = "/path/to/the/log.log";
    $handle = fopen($file, "r");
    clearstatcache();       

    if ($_REQUEST[ pointer ] ==   ) {
        fseek($handle, -1024, SEEK_END);
    } else {
        fseek($handle, $_REQUEST[ pointer ]);
    }

    while ($buffer = fgets($handle)) { 
        $log .= $buffer . "<br />
";
    } 



    $output = array("pointer" => ftell($handle), "log" => $log);
    fclose($handle);

    echo json_encode($output);
}

请告诉我,我如何能够利用关键词阵列中的关键词对检索的数据进行过滤。

Answer 1

您可按每一空间分界线,并对照所提供关键词核对每一字:

while ($buffer = fgets($handle)) { 
  $words = explode(   , $buffer);
  foreach ($words as $word) {
    if (in_array($word, $keywords)) {
      $log .= $buffer . "<br />
";
      break;
    }
  }
}

或对照行文核对每一关键词:

while ($buffer = fgets($handle)) { 
  foreach ($keywords as $keyword) {
    if (strstr($buffer, $keyword)) {
      $log .= $buffer . "<br />
";
      break;
    }
  }
}

Answer 2

改写:

while ($buffer = fgets($handle)) { 
    foreach ($keywords as $kw) { // Loop keywords
        if (strpos($buffer, $kw) !== FALSE) { // Search for this keyword
            // If we get here, we found a keyword
            $log .= $buffer . "<br />
";
            break;
        }
    }
}

如果你想以对个案敏感的方式加以匹配,你可以使用<代码>stripos(。

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