Question

我有一个字典术语表，现在我想创建另一个表格，其中包含“主要”术语的ID以及在其定义中包含“主要”术语名称的相关术语的ID -

table: terms 
| id | name | definition |


Table: related 
| term_id | related_term_id |

在PHP中创建INSERT语句非常简单：

$result = mysql_query("SELECT * FROM terms");
while ($row = mysql_fetch_array($result)) {
    $name = $row['name'];
    $result2 = mysql_query("SELECT id,name FROM terms WHERE description LIKE \"%$name%\" AND id!=".$row['id']);

    while ($row2 = mysql_fetch_array($result2)) {
        echo "INSERT INTO related(term_id,related_term_id) VALUES(".$row['id'].",".$row2['id'].");";
    }
}

但是，由于我没有太多使用MySQL的经验，我想知道是否可以纯粹在SQL中完成 - 假设使用INSERT-SELECT语句？

谢谢！

Answer 1

INSERT INTO related (term_id, related_term_id)
SELECT a.id as term_id, b.id AS related_term_id
FROM terms a, terms b
WHERE a.id != b.id
AND b.description LIKE CONCAT('%', b.name, '%')

Answer 2

如果terms表很大（超过1000-10000行），那将会非常慢：

INSERT INTO related (term_id, related_term_id)
  SELECT describing.id, described.id
  FROM terms describing
    JOIN terms described 
      ON CONCAT(' ', described.description, ' ') LIKE CONCAT('% ', describing.name, ' %')

假设所有描述都有空格分隔单词而没有其他标记，例如,或?或!或.如果不是这样，则代码会更复杂

Answer 3

更改您的INSERT查询：

echo "INSERT INTO related(term_id,related_term_id) VALUES('". $row['id'] ."','". $row2['id'] ."');";

mysql选择相关术语

3 个答案: