Php 从二维数组中求时间和
您好,我将查找总和,但是出现了错误,如为foreach()提供的Php 从二维数组中求时间和,php,Php,您好,我将查找总和,但是出现了错误,如为foreach()提供的无效参数如何检查错误 我的代码是 我有这样的时间 $SQL = "SELECT o_duration FROM total WHERE location=('".$location_c[$k]."') AND sub_cat=('".$id."') AND dateid BETWEEN '".$trgD."' AND '".$date2."'"; $result = mysql_query($SQL); $s="0"; $dura
无效参数如何检查错误
我的代码是
我有这样的时间
$SQL = "SELECT o_duration FROM total WHERE location=('".$location_c[$k]."') AND sub_cat=('".$id."') AND dateid BETWEEN '".$trgD."' AND '".$date2."'";
$result = mysql_query($SQL);
$s="0";
$duration_c[$k]=array();
while ($db_field = mysql_fetch_array($result)) {
$duration_c[$k][$s]= $db_field['o_duration'];
$s++;
}
然后我用for循环在表中打印它
<?php
for($i=0;$i<$l;$i++)
{
?>
<td width="154" class="alt" align="center"><?php echo sum_the_time($duration_c[$i])
?></td> <?php } ?>
function sum_the_time($times) {
$seconds = 0;
foreach ($times as $time1)
{
list($hour,$minute,$second) = explode(':', $time);
$seconds += $hour*3600;
$seconds += $minute*60;
$seconds += $second;
}
$hours = floor($seconds/3600);
$seconds -= $hours*3600;
$minutes = floor($seconds/60);
$seconds -= $minutes*60;
// return "{$hours}:{$minutes}:{$seconds}";
return sprintf('%02d:%02d:%02d', $hours, $minutes, $seconds);
}
函数和时间($times){
$seconds=0;
foreach($time1的倍)
{
列表($hour,$minute,$second)=分解(“:”,$time);
$seconds+=$hour*3600;
$seconds+=$minute*60;
$seconds+=$second;
}
$hours=地板($seconds/3600);
$seconds-=$hours*3600;
$分钟=地板($秒/60);
$seconds-=$minutes*60;
//返回“{$hours}:{$minutes}:{$seconds}”;
返回sprintf(“%02d:%02d:%02d',$hours,$minutes,$seconds);
}
为foreach()提供的参数无效。
表示传递给函数的数组为空(或不是数组)
添加一个条件以检查它是否有效
if(is_array($times) and count($times)>0){
foreach ($times as $time1){
(...)
}
}
问题是什么?显示了为foreach()@ganna提供的错误无效参数,这是由于未设置$duration\u c[$i]
处的值所致。通过回显检查是否有任何值。您可以在sql查询本身中使用sum(o_duration)
来获取时间之和。lenght
在PHP中不是函数,请使用count($times)