获取关键字出现的次数答案

【问题标题】：Get number of times keyword is present获取关键字出现的次数
【发布时间】：2013-12-29 19:57:24
【问题描述】：

假设我有以下数据，最后一列关键字不过是其他 4 列的组合。

╔════╦══════╦════════════╦═════════╦════════════╦════════════════════════════════╗
║ ID ║ Name ║    Add1    ║  Add2   ║    Add3    ║            Keyword             ║
╠════╬══════╬════════════╬═════════╬════════════╬════════════════════════════════╣
║  1 ║ John ║ W Brown St ║ Edison  ║ Washington ║ JohnW Brown StEdisonWashington ║
║  2 ║ Paul ║ E High Rd  ║ Peapack ║ New Jersey ║ PaulE High RdPeapackNew Jersey ║
║  3 ║ John ║ Greams Rd  ║ Peapack ║ Washington ║ JohnGreams RdPeapackWashington ║
╚════╩══════╩════════════╩═════════╩════════════╩════════════════════════════════╝

现在用户可以输入任何文本字段 Name、Add1、Add2、Add3 并单击搜索，它应该搜索所有字段。

为此，我正在尝试，

SELECT *, COUNT(ID) AS FREQUENCY FROM TABA WHERE ID IN
(
 (SELECT ID FROM TABA WHERE KEYWORD LIKE '%WASHINGTON%')
 UNION ALL
 (SELECT ID FROM TABA WHERE KEYWORD LIKE '%JOHN%')
 UNION ALL
 (SELECT ID FROM TABA WHERE KEYWORD LIKE '%PEAPACK%')
)
GROUP BY ID
ORDER BY FREQUENCY

预期输出：

╔════╦══════╦════════════╦═════════╦════════════╦═══════════╗
║ ID ║ Name ║    Add1    ║  Add2   ║    Add3    ║ Frequency ║
╠════╬══════╬════════════╬═════════╬════════════╬═══════════╣
║  3 ║ John ║ Greams Rd  ║ Peapack ║ Washington ║         3 ║
║  1 ║ John ║ W Brown St ║ Edison  ║ Washington ║         2 ║
║  2 ║ Paul ║ E High Rd  ║ Peapack ║ New Jersey ║         1 ║
╚════╩══════╩════════════╩═════════╩════════════╩═══════════╝

但我得到的所有频率值都是 1。我做错了什么？

【问题讨论】：

+1 用于酷炫的 ASCII 艺术表
那是因为你的内部查询丢失了计数，只选择了IDs。

标签： sql sql-server

【解决方案1】：

... WHERE ID IN ( ...) ... 部分只会进行“是或否”检查，但您需要计数。这必须通过 JOIN 来完成。

试试这个：

SELECT TABA.*, COUNT(sub.ID) AS FREQUENCY FROM TABA JOIN 
(
 (SELECT ID FROM TABA WHERE KEYWORD LIKE '%WASHINGTON%')
 UNION ALL
 (SELECT ID FROM TABA WHERE KEYWORD LIKE '%JOHN%')
 UNION ALL
 (SELECT ID FROM TABA WHERE KEYWORD LIKE '%PEAPACK%')
) sub ON sub.ID=TABA.ID
GROUP BY TABA.ID
ORDER BY FREQUENCY

【讨论】：

【解决方案2】：

为什么不做一个没有子查询的简化版本：

SELECT  ID, Name, Add1, Add2, Add3, COUNT(*) AS FREQUENCY 
  FROM TABA 
 WHERE ID IN
     (SELECT ID 
        FROM TABA 
        WHERE KEYWORD LIKE '%WASHINGTON%'
           OR KEYWORD LIKE '%JOHN%'
           OR KEYWORD LIKE '%PEAPACK%')
GROUP BY ID, Name, Add1, Add2, Add3
ORDER BY 6

'Order by 6'表示选择范围内的第六个元素，即频率

【讨论】：

@DanielE。你是对的，我再次阅读了 OP 的问题并更好地理解了它。我会改正的。

【解决方案3】：

SELECT TABA.*, sub.FREQUENCY  
FROM
(
  SELECT ID, COUNT(*) as FREQUENCY
  FROM 
  (
   (SELECT ID FROM TABA WHERE KEYWORD LIKE '%WASHINGTON%')
   UNION ALL
   (SELECT ID FROM TABA WHERE KEYWORD LIKE '%JOHN%')
   UNION ALL
   (SELECT ID FROM TABA WHERE KEYWORD LIKE '%PEAPACK%')
  ) a
  GROUP BY ID
) sub 
INNER JOIN TABA ON sub.ID=TABA.ID
ORDER BY FREQUENCY DESC

sql fiddle

【讨论】：