Php 如何在全文检索SQL中获取与列值匹配的字符串

就像下面的

query

，它是show

score

。我想要与匹配的字符串或单词相同

选择意见，将（意见）与布尔模式下的（'nice language'）匹配作为分数
根据意见
其中，匹配（意见）与布尔模式中的（'nice language'）；

产出：

+---------------------------------------+-------+
| Opinion                               | Score |
+---------------------------------------+-------+
| PHP is a very nice language           |     2 |
| Spain is a nice country to visit      |     1 |
| Perl is not as nice a language as PHP |     2 |
+---------------------------------------+-------+

但我们需要以下输出：

+---------------------------------------+--------------+
| Opinion                               | String       |
+---------------------------------------+--------------+
| PHP is a very nice language           | nice language|
| Spain is a nice country to visit      | nice         |
| Perl is not as nice a language as PHP | nice language|
+---------------------------------------+-------+------+

---

如果您正在寻找如下所示的结果

然后使用下面的查询

select Opinion, 
      case 
        when Opinion like '%nice language%' or (Opinion like '% nice %' and Opinion like '% language %') then 'nice language' 
        when Opinion like '% nice %' then 'nice' 
        when Opinion like '% language %' then 'language' 
      end string  
from table1
where 
    Opinion like '%nice language%' 
      OR (Opinion like '% nice %' and Opinion like '% language %') 
      OR Opinion like '% nice %' 
      OR Opinion like '% language %'

---

希望这对您有所帮助。

编辑完成，请检查。有什么建议吗？@DurgeshPandey搜索字符串是静态的还是动态的？@Viki888搜索字符串是静态的。1:-在全文搜索中不可能吗？然后搜索字符串是动态的？@DurgeshPandey如果是动态的，然后，您必须拆分字符串并相应地替换查询中的内容。那么全文搜索呢？您所说的

全文搜索

到底是什么意思？