php - 检查句子是否有相同的单词

Question

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包含每个 content_id 中其他句子所拥有的单词。

例如:

检查 content_id = 1 它们是 sentence1 和 sentence2。从tb_word可以看出sentence1和sentence2由同一个词a组成。如果两句中a的个数为>=2，则结果为a。所以如果我打印结果，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，我让 functionTotal 来计算每个 content_id 拥有多少 sentence :

$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;

从那个函数我得到 $total 的值，并且我需要检查一些单词(来自 tb_word)在 2 的所有可能性之间的相似性句子

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的作用是:

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。结果也必须按 content_id 分组。对不起我糟糕的英语和糟糕的解释。 cmiiw，请帮帮我..谢谢:)

Answer 1

这个实际上可以由DBMS自己完成，一次查询两步。首先，为了在相同的内容中准备句子组合，您进行自连接:

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，您可以将上述结果与单词结合并计算出现次数。像这样:

SELECT s.content_id,
       s.sentence_id_1,
       s.sentence_id_2,
       c.word,
       Count(*) AS jml
FROM   (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 )) 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,
          s.sentence_id_1,
          s.sentence_id_2,
          c.word
HAVING Count(*) >= 2; 

上述查询的结果将为您提供容器、句子 1 和 2、单词以及出现次数(2 次或更多)。您现在需要的只是将结果收集到数组中，正如我所见，您已经知道该怎么做。

如果我误解了你的目标，请告诉我。