我的 table :
id | views | date
1 | 100 | 2017-03-09
2 | 150 | 2017-03-10
3 | 300 | 2017-03-11
4 | 350 | 2017-03-12
我需要像这样计算几天之间的访问次数差异
2017-03-12-->Visitors:350
2017-03-11-->Visitors:300
Difference between days:50
2017-03-11-->Visitors:300
2017-03-10-->Visitors:150
Difference between days:150
2017-03-10-->Visitors:150
2017-03-09-->Visitors:100
Difference between days:50
and so on...
我设法得到了类似的结果,但并不完全是我想要的
$sql = "SELECT * FROM `table` ORDER BY `table`.`id` DESC";
$result = mysql_query($sql) or die(mysql_error());
while($row = mysql_fetch_array($result))
{
$t = $row['views'];
$dat = $row['date'];
$sql1 = "SELECT * FROM `table` ORDER BY `table`.`id` DESC LIMIT 1, 99";
$result1 = mysql_query($sql1) or die(mysql_error());
while($row1 = mysql_fetch_array($result1))
{
$y = $row1['views'];
$dat1 = $row1['date'];
$d = $t-$y;
echo "{$dat}-->Visitors:{$t}";
echo "<br/>";
echo "{$dat1}-->Visitors:{$y}";
echo "<br/>";
echo "Difference between days:{$d}";
echo "<br/><br/><br/>";
}
}
所以我想我需要用一个查询选择同一张表两次。
最佳答案
这里不需要 SQL 技巧。
您正在按日期顺序显示行。只需将最后一行的计数保存在一个变量中,在 php 中进行减法,就可以得到不同的结果。
删除嵌套循环。您只需要一个循环。
$last_views = null;
while($row = mysql_fetch_array($result))
{
$views = $row['views'];
$dat = $row1['date'];
if( $last_views === null )
$delta_views = "";
else
$delta_views = $last_views - $views;
$last_views = $views;
echo "{$dat}-->Visitors:{$views}";
echo "<br/>";
echo "Difference between days:{$delta_views}";
echo "<br/>";
}
关于php - 在单个查询中两次选择同一个表,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/43138417/