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php – 检查句子是否有相同的单词

发布时间:2020-12-13 16:18:15 所属栏目:PHP教程 来源:网络整理
导读:tb_content(左)和tb_word(右): ===================================== ================================|id|sentence |sentence_id|content_id| |id|word|sentence_id|content_id|===================================== ===============================
tb_content(左)和tb_word(右):

=====================================    ================================
|id|sentence |sentence_id|content_id|    |id|word|sentence_id|content_id|
=====================================    ================================
| 1|sentence1|    0      |    1     |    | 1|  a |     0     |    1     |
| 2|sentence2|    1      |    1     |    | 2|  b |     0     |    1     |
| 3|sentence5|    0      |    2     |    | 3|  c |     1     |    1     |
| 4|sentence6|    1      |    2     |    | 4|  a |     1     |    1     |
| 5|sentence7|    2      |    2     |    | 5|  e |     1     |    1     |
=====================================    | 6|  f |     0     |    2     |
                                         | 7|  g |     1     |    2     |
                                         | 8|  h |     1     |    2     |
                                         | 9|  i |     1     |    2     |
                                         |10|  f |     2     |    2     |
                                         |11|  h |     2     |    2     |
                                         |12|  f |     2     |    2     |
                                         ================================

我需要检查每个句子是否由每个content_id中的其他句子所拥有的单词组成.

例如 :

检查content_id = 1他们是sentence1和sentence2.从tb_word,我们可以看到sentence1和sentence2由相同的单词a组成.如果两个句子中的a的数量是> = 2,那么a将是结果.因此,如果我打印结果,它必须是:
00Array([0] => a [1] => b)01Array([3] => a)10Array([3] => a)11Array([0] => c [1] = > a [2] => e)其中00表示sentence_id = 0且sentence_id = 0

首先,我使functionTotal计算每个content_id所拥有的句子数:

$total = array();
$sql = mysql_query('select content_id,count(*) as RowAmount 
       from tb_content Group By contente_id') or die(mysql_error());
while ($row = mysql_fetch_array($sql)) {
    $total[] = $row['RowAmount']; 
}
return $total;

从这个函数我得到$total的值,从中我需要检查一些单词(来自tb_word)在2个句子的所有可能性之间的相似性

foreach ($total as $content_id => $totals){
for ($x=0; $x <= ($totals-1); $x++) {
    for ($y=0; $y <= ($totals-1); $y++) {
      $shared = getShared($x,$y);
    }
}

getShared的功能是:

function getShared ($x,$y){
    $token = array();
    $shared = array();
    $i = 0;
    if ($x == $y) {
        $query = mysql_query("SELECT word FROM `tb_word`
                             WHERE sentence_id ='$x' ");
        while ($row = mysql_fetch_array($query)) {
            $shared[$i] = $row['word'];
            $i++;
        }

    } else {
        $query = mysql_query("SELECT word,count(word) as jml 
                             FROM `tb_word` WHERE sentence_id ='$x' 
                             OR sentence_id ='$y' 
                             GROUP BY word ");
        while ($row = mysql_fetch_array($query)) {
            $jml = $row['jml'];
            $token[$i] = $row['word'];
            if ($jml >= 2) {
                $shared[$i] = $token[$i];
            }
            $i++;
        }

但我得到的结果仍然是错误的.结果仍然在不同的content_id之间混合.结果必须也是由content_id分组.抱歉我的英语不好,我的解释也不好. cmiiw,请帮帮我..谢谢:)

解决方法

这个实际上可以由DBMS本身完成,在一个查询中有两个步骤.首先,您进行自我加入以准备相同内容中的句子组合:

SELECT a.content_id,a.sentence_id AS sentence_id_1,b.sentence_id AS sentence_id_2
FROM   tb_content AS a
       JOIN tb_content AS b
         ON ( a.content_id = b.content_id
              AND a.sentence_id <= b.sentence_id )

“< =”将保持相同的句子连接,如“1-1”或“2-2”,但避免双向重复,如“1-2”和“2-1”.接下来,您可以使用单词加入上述结果并计算出现次数.像那样:

SELECT s.content_id,s.sentence_id_1,s.sentence_id_2,c.word,Count(*) AS jml
FROM   (SELECT a.content_id,b.sentence_id AS sentence_id_2
        FROM   tb_content AS a
               JOIN tb_content AS b
                 ON ( a.content_id = b.content_id
                      AND a.sentence_id <= b.sentence_id )) AS s
       JOIN tb_word AS c
         ON ( s.content_id = c.content_id
              AND ( c.sentence_id = s.sentence_id_1
                     OR c.sentence_id = s.sentence_id_2 ) )
GROUP  BY s.content_id,c.word
HAVING Count(*) >= 2;

上述查询的结果将为您提供容器,句子1和2,单词和出现次数(2或更多).您现在需要的只是将结果收集到数组中,正如我所知道的那样.

如果我错过了你的目标,请告诉我.

(编辑:李大同)

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