我对数据库执行两次查询,它们的功能相同,但具有其他值。
CODE1:
$sql= "SELECT sum(amount) as Samount, sum(add25) as Sadd25, sum(add50) as Sadd50, sum(deplacement) as Sdeplacement FROM workhours WHERE reportID = '$reportID'";
$sumWorkhoursData=mysqli_query($link,$sql);
unset($sql);
while($sWorhoursInfo=mysqli_fetch_array($sumWorkhoursData)){
$totalWork = $sumWorkhoursInfo['Samount'];
print_r(var_dump($sWorkhoursInfo));
}
var_dump给出NULL
如果$sql
手动复制到数据库,那么它会得到一个好的结果,即一行。
CODE2:
$sql= "select sum(time) as Stime, sum(amount) as Samount FROM trip WHERE reportID = '$reportID'";
$sumTripData=mysqli_query($link,$sql);
unset($sql);
while ($sTripInfo=mysqli_fetch_array($sumTripData)){
$totalTrip = $sTripInfo['Stime'];
print_r(var_dump($sTripInfo));
}
在这种情况下,var_dump给出array(4) { [0]=> string(2) "20" ["Stime"]=> string(2) "20" [1]=> string(3) "151" ["Samount"]=> string(3) "151" }
这里有什么问题?
答案 0 :(得分:5)
您在$sWorhoursInfo
中输入了错误while($sWorhoursInfo
,$sWorkhoursInfo
错过了k
while($sWorkhoursInfo...