Question

我的代码在do while中使用7个变量，每次do while循环时都会更改我想替换7个变量而不创建全新的变量，因此每个变量仅显示一次且没有更多变量，我希望变量在每次while循环后都会更改，并且在显示新变量之前会有一个小的延迟

<html>
<body>    
<?php
include('connection.php');//connection to databse
$sql = "select COUNT(*)from dbo.arrivals";//getting number of rows from database

$max;
$stmt = sqlsrv_query( $conn, $sql);
sqlsrv_fetch($stmt);
$max= sqlsrv_get_field( $stmt,0 );

$var= array();
$counter=0;

for($i=0;$i<=$max;$i++)
{

$sql = "WITH NthRowCTE AS(SELECT ROW_NUMBER() OVER (ORDER BY ArrivalTime)AS RNum, * FROM dbo.arrivals) SELECT * FROM NthRowCTE WHERE RNum =".$i;

$stmt = sqlsrv_query( $conn, $sql);
sqlsrv_fetch($stmt);
$flight = sqlsrv_get_field( $stmt,3 );
array_push($var,"$flight");
}

$f= array("","","","","","","");

$counter=0;
$range=7;
do{
$f[0]="empty";
$f[1]="empty";
$f[2]="empty";
$f[3]="empty";
$f[4]="empty";
$f[5]="empty";
$f[6]="empty";
if($counter==$range)
{
$range=$range+7;

}
for($i=0;$i<7;$i++)
{
$counter++; 
//$i=$counter-1;
if($counter<=$max)
{
$f[$i]= $var[$counter];
}
else
{
$counter=$max;
break;
}
}
echo "<br />";
echo "<br />";
echo  $f[0];
echo "<br />";
echo $f[1];
echo "<br />";
echo $f[2];
echo "<br />";
echo $f[3];
echo "<br />";
echo $f[4];
echo "<br />";
echo $f[5];
echo "<br />";
echo $f[6];
echo "<br />";
}
while($counter!=$max);       
?>             
</body>
</html>

如何在不创建新表的情况下替换表数据

0 个答案: