Mysql查询以计算链接数并按状态划分资源
我有一个查询打印出来的链接列表(1列)和他们的计数(第二)。我想做的是将结果除以它们的状态 范例 我的代码:(Mysql、Php、Html)Mysql查询以计算链接数并按状态划分资源,mysql,count,Mysql,Count,我有一个查询打印出来的链接列表(1列)和他们的计数(第二)。我想做的是将结果除以它们的状态 范例 我的代码:(Mysql、Php、Html) $sql=“选择'bookinglink',count('bookinglink`)作为'count' 从'quotedb'开始 其中`bookinglink`!=''和(状态='quote'或状态='booking') 按“bookinglink”分组,按计数说明排序; $result=$mysqli->query($sql); 而($rs=$resu
$sql=“选择'bookinglink',count('bookinglink`)作为'count'
从'quotedb'开始
其中`bookinglink`!=''和(状态='quote'或状态='booking')
按“bookinglink”分组,按计数说明排序;
$result=$mysqli->query($sql);
而($rs=$result->fetch_object()){
回显“$rs->bookinglink.”“$rs->count.”;
}
所以我需要确保查询统计所有链接,然后统计状态为“报价”和“预订”的链接。可以在一个查询中使用它吗
你能帮我解决这个问题吗,因为我不太确定该怎么做。查询可以写如下:
SELECT bookinglink, COUNT(bookinglink) AS bookinglinkcount,
SUM(status = 'quote') AS quotes,
CAST(100 * SUM(status = 'quote') / COUNT(bookinglink) AS DECIMAL(5, 2))
AS quotes_percentage,
SUM(status = 'booking') AS bookings,
CAST(100 * SUM(status = 'booking') / COUNT(bookinglink) AS DECIMAL(5, 2))
AS bookings_percentage
FROM quotedb
WHERE bookinglink != ''
GROUP BY bookinglink;
Sum(当…然后…结束时的情况)x
SELECT bookinglink, COUNT(bookinglink) AS bookinglinkcount,
SUM(status = 'quote') AS quotes,
CAST(100 * SUM(status = 'quote') / COUNT(bookinglink) AS DECIMAL(5, 2))
AS quotes_percentage,
SUM(status = 'booking') AS bookings,
CAST(100 * SUM(status = 'booking') / COUNT(bookinglink) AS DECIMAL(5, 2))
AS bookings_percentage
FROM quotedb
WHERE bookinglink != ''
GROUP BY bookinglink;