獲取多行並將其存儲到變量。 MySQL和PHP

Question

我正在為系統創建一個平均公里數運行計算器。 因此，我試圖在數據庫上獲取最后兩個日期和最后兩個日期。

這是我當前的代碼：

<?php

$conn = mysqli_connect('localhost','root','','mypms');

 $sqldates = "SELECT dateinput FROM sched ORDER BY dateinput DESC LIMIT 2";
 $sqlrun = "SELECT reading FROM sched ORDER BY reading DESC LIMIT 2"; 

 $resultdate = mysql_query($conn,$sqldates);
 $resultrun = mysql_query($conn,$sqlrun);


 while($rowdate = mysql_fetch_assoc($resultdate))
 {
     this ->
 }

 $date_difference = $date2 - $date1;

 while($rowrun = mysql_fetch_assoc($resultrun))
 {
     this ->
 }

 $run_difference = $run2 - $run1;
 $averagerun = $run_difference/$date_difference

 echo $averagerun;
?>

我應該寫上我的什么this ，所以我可以存儲我的$ resultrun和$ resultdate到DATE1，$ DATE2和$ RUN1，$ RUN2。

注意：dateinput的datetype是我的mysqldatabase上的日期。

Answer 1

請嘗試使用此查詢-它會返回您的結果

SELECT 
run_difference/date_difference AS averagerun
FROM
    (
        SELECT
        DATEDIFF(date2,date1) AS date_difference,
        reading2 - reading1 AS run_difference
        FROM
        (
            SELECT 
            (SELECT dateinput FROM sched ORDER BY dateinput LIMIT 0, 1) AS date1, 
            (SELECT dateinput FROM sched ORDER BY dateinput LIMIT 1, 1) AS date2,
            (SELECT reading FROM sched ORDER BY reading DESC LIMIT 0, 1) AS reading1,
            (SELECT reading FROM sched ORDER BY reading DESC LIMIT 1, 1) AS reading2 
            FROM DUAL
        ) t1
    )t2

或“一個班輪”：

SELECT 
    (SELECT reading FROM sched ORDER BY reading DESC LIMIT 1, 1)-(SELECT reading FROM sched ORDER BY reading DESC LIMIT 0, 1)   /   DATEDIFF((SELECT dateinput FROM sched ORDER BY dateinput LIMIT 1, 1), (SELECT dateinput FROM sched ORDER BY dateinput LIMIT 0, 1))  AS averagerun
FROM DUAL