Php 数组的重复输出

Php 数组的重复输出,php,codeigniter,Php,Codeigniter,问题:为什么输出重复?我的代码中有什么问题?:(谢谢。可能会尝试使用$total\u diff如下: echo json_encode($hours); // [9,9,9,] echo json_encode($mins); //[20,36,46] foreach($timein\u out作为$timetest){ $total_diff[]=strotime($timetest[“Time\u out”])-strotime($timetest[“Time\u in”]); 对于(

问题:为什么输出重复?我的代码中有什么问题?:(谢谢。

可能会尝试使用
$total\u diff
如下:

 echo json_encode($hours); // [9,9,9,]
 echo json_encode($mins); //[20,36,46]
foreach($timein\u out作为$timetest){
$total_diff[]=strotime($timetest[“Time\u out”])-strotime($timetest[“Time\u in”]);
对于($key=0;$key可以尝试使用
$total\u diff
如下:

 echo json_encode($hours); // [9,9,9,]
 echo json_encode($mins); //[20,36,46]
foreach($timein\u out作为$timetest){
$total_diff[]=strotime($timetest[“Time\u out”])-strotime($timetest[“Time\u in”]);

对于($key=0;$key可能是,您希望内部循环位于外部。请尝试此操作

foreach($timein_out as $timetest) {
    $total_diff[] = strtotime($timetest["Time_out"]) - strtotime($timetest["Time_in"]);

    for($key=0;$key<count($total_diff);$key++) {
        $hours[] = intval(floor($total_diff[$key]/3600));
        $mins[] =  intval(($total_diff[$key]-$hours[$key]*3600)/60); 
    }
    unset($total_diff);
}
foreach($timein\u out作为$timetest)
{
$total_diff[]=strotime($timetest[“Time\u out”])-strotime($timetest[“Time\u in”]);
}

对于($key=0;$key可能是,您希望内部循环位于外部。请尝试此操作

foreach($timein_out as $timetest) {
    $total_diff[] = strtotime($timetest["Time_out"]) - strtotime($timetest["Time_in"]);

    for($key=0;$key<count($total_diff);$key++) {
        $hours[] = intval(floor($total_diff[$key]/3600));
        $mins[] =  intval(($total_diff[$key]-$hours[$key]*3600)/60); 
    }
    unset($total_diff);
}
foreach($timein\u out作为$timetest)
{
$total_diff[]=strotime($timetest[“Time\u out”])-strotime($timetest[“Time\u in”]);
}

对于($key=0;$key,
$timein\u out
的值是多少?
$timein\u out
的值是多少?+1比接受的答案更接近。它运行2n次,接受的答案运行n*n次……是的……我认为这一个比另一个更快……)+1比被接受的答案更好。它运行了2n次,被接受的答案运行了n*n次……是的……我认为这一个比另一个快……)对不起,老兄,我认为另一个快得多。对不起,老兄,我认为另一个快得多。