select count(*) as c from `销售表` as s left join `成员表` as u on u.`姓名`=s.`姓名` group by concat(`年`,`月`,`日`) where u.id=1 ;
SELECT count( * ) AS c FROM sell AS s LEFT JOIN user AS u ON u.ln_name = s.nl_name WHERE u.ln_id =1 GROUP BY concat( s.nl_year, s.nl_month, s.nl_day )这个查出来的数据只是统计了该id数出现的次数?如何应用到前面的问题呢?我的具体需求是: 输入2008年3月 则显示三月的天数1§2§3§4§5....30§31 而后根据天数显示某个销售人员(销售人员不再重复出现)的销售数量如1号1个;2号3个......姓名 1(日期) §2 § 3§ ... §31 张三 2(数量) §3 § 4§ ... §12 不知道表述的是否清楚。请各位帮忙了!谢谢了先。
SELECT count( * ) AS c ,u.ln_name,s.nl_day FROM sell AS s LEFT JOIN user AS u ON u.ln_name = s.nl_name WHERE s.nl_year='2008' and s.nl_month='3' GROUP BY concat( s.ln_name,'|', s.nl_year,'|', s.nl_month,'|', s.nl_day ) //得到结果后,保存为2维数组:$sellstats = array(); while($row = mysql_fetch_array($result)){ $name = $row[ln_name]; $day = $row[ln_day]; if(!$sellstats["$name"]) $sellstats["$name"] = array(); //名字作为第一维 $sellstats["$name"][$day] = $row[c]; //日期作为第二维 }//遍历2维数组: foreach($sellstats as $name=>$days){ echo "$name | $days[1] | $days[2] |... | $days[31] <br />"; }
select count(*) as c from `销售表` as s
left join `成员表` as u on u.`姓名`=s.`姓名`
group by concat(`年`,`月`,`日`) where u.id=1 ;
FROM sell AS s
LEFT JOIN user AS u ON u.ln_name = s.nl_name
WHERE u.ln_id =1
GROUP BY concat( s.nl_year, s.nl_month, s.nl_day )这个查出来的数据只是统计了该id数出现的次数?如何应用到前面的问题呢?我的具体需求是:
输入2008年3月
则显示三月的天数1§2§3§4§5....30§31
而后根据天数显示某个销售人员(销售人员不再重复出现)的销售数量如1号1个;2号3个......姓名 1(日期) §2 § 3§ ... §31
张三 2(数量) §3 § 4§ ... §12 不知道表述的是否清楚。请各位帮忙了!谢谢了先。
FROM sell AS s
LEFT JOIN user AS u ON u.ln_name = s.nl_name
WHERE s.nl_year='2008' and s.nl_month='3'
GROUP BY concat( s.ln_name,'|', s.nl_year,'|', s.nl_month,'|', s.nl_day ) //得到结果后,保存为2维数组:$sellstats = array();
while($row = mysql_fetch_array($result)){
$name = $row[ln_name];
$day = $row[ln_day];
if(!$sellstats["$name"]) $sellstats["$name"] = array(); //名字作为第一维
$sellstats["$name"][$day] = $row[c]; //日期作为第二维
}//遍历2维数组:
foreach($sellstats as $name=>$days){
echo "$name | $days[1] | $days[2] |... | $days[31] <br />";
}