输出日期与原始日期不同

Question

Here's my problem. output date different from the string i cut. I've tried the code below and the day not match with the original date. it should "26" but output "02", give me solution.

CODE :

$line = 'File=D:\Shared\print\2017\september\26\printing\baryun - cetak transferpaper.tif';
     $r = "/^File=D:\\\\[a-zA-Z]+\\\\[a-z]+\\\\([0-9]{4})\\\\([a-z]+)\\\\([0-9]{1,2})/";

       if (preg_match($r, $line, $match)) {
            $date = $match[1]."-".$match[2]."-".$match[3];
            echo $date."<br>";
            $xtime = strtotime($date);
            $tes = date("Y-m-d",$xtime);

            echo $tes;
          }

OUTPUT :

2017-september-26
2017-09-02

And when i try this code

 if (preg_match($r, $line, $match)) {
            $date = $match[1]."-".$match[2]."-".$match[3];
            echo $date."<br>";
            $xtime = date_create($date);
            $tes = date_format($xtime,"Y-m-d");

            echo $tes;

OUTPUT, the day 01. please help me

2017-september-26
2017-09-01

Answer 1

It's because the date string you are trying to pass is not-correctly formatted.

Do like below:-

$r = "/([0-9]{4})\\\\([a-z]+)\\\\([0-9]{1,2})/"; // curtale your regex
if (preg_match($r, $line, $match)) {
    $date = $match[3]."-".$match[2]."-".$match[1]; //change the pattern
    echo $date."<br>";
    $xtime = strtotime($date);
    $tes = date("Y-m-d",$xtime);

    echo $tes;
}

https://eval.in/872478