我想总结一个查询的输出,但我正在挣扎。下面的查询检查fax_authorized表,并根据他们的电子邮件地址生成扩展摘要。总结来自MySQL查询的输出
我有以下查询:
SELECT extended_number, IF(COUNT(extension_id)<5,1,0) AS 'fax5',
IF(COUNT(extension_id)>5 AND COUNT(extension_id)<11,1,0) AS 'fax10',
IF(COUNT(extension_id)>10,1,0) AS 'fax10+', COUNT(extension_id) AS 'fax_total'
FROM fax_authorized fa, extension e
WHERE fa.extension_id = e.id
GROUP BY extended_number;
输出看起来是这样的:
+-----------------+------+-------+--------+-----------+
| extended_number | fax5 | fax10 | fax10+ | fax_total |
+-----------------+------+-------+--------+-----------+
| 0009*004 | 1 | 0 | 0 | 1 |
| 0139*601 | 0 | 1 | 0 | 6 |
| 0139*743 | 1 | 0 | 0 | 2 |
| 0139*996 | 1 | 0 | 0 | 1 |
+-----------------+------+-------+--------+-----------+
我想什么是总结基础上离开了这个数据(extended_number,4),以便为上面的示例0139将具有fax5 = 2,fax10 = 1,fax_total = 9。
我曾经尝试这样做,不工作:
SELECT extended_number, SUM(IF(COUNT(extension_id)<5,1,0)) AS 'fax5',
SUM(IF(COUNT(extension_id)>5 AND COUNT(extension_id)<11,1,0)) AS 'fax10',
SUM(IF(COUNT(extension_id)>10,1,0)) AS 'fax10+',
COUNT(LEFT(extension_id,4)) AS 'fax_total'
FROM fax_authorized fa, extension e
WHERE fa.extension_id = e.id
GROUP BY LEFT(extended_number,4);
编辑:我用Saharsh沙阿回答下面来解决这个问题,可是撞到了另外一个,我认为是值得加入。我试图从他的嵌套子查询解决方案创建一个视图,但无法执行,因为视图中不允许使用子查询。所以,我创建了两个相关的观点如下:
CREATE VIEW vw_xoom_faxdetail AS
SELECT LEFT(extended_number,4) client_id,
IF(COUNT(extension_id)<5,1,0) AS 'fax5',
IF(COUNT(extension_id)>5 AND COUNT(extension_id)<11,1,0) AS 'fax10',
IF(COUNT(extension_id)>10,1,0) AS 'fax10plus',
COUNT(extension_id) AS 'fax_total'
FROM fax_authorized fa, extension e WHERE fa.extension_id = e.id
GROUP BY client_id;
CREATE VIEW vw_xoom_fax AS
SELECT client_id, SUM(fax5) fax5, SUM(fax10) fax10,
SUM(fax10plus) 'fax10+', SUM(fax_total) fax_total
FROM vw_xoom_faxdetail
GROUP BY client_id;
你说的错误是什么意思? –
@Jack - 对不起,我的意思是它不起作用,只是给我一个语法错误 – btongeorge