using a variable outside while loop

Question

I am trying to use a variable outside of a while loop which is in a foreach loop, but it works in the loop and displays the correct date. But when used outide the loop it displays incorrect date 01/01/1970.

I would be grateful if someone could point out my error. Many thanks.

  $Destdate = $_POST['box_rtv'];
  $brtvarray = array();
  $brtvarray = $Destdate;
  $ddate = array();

  foreach ($brtvarray as $destbox) {
    $sql = "SELECT destroydate FROM act WHERE item = '".$destbox."'";
    $result = mysqli_query($conn, $sql) or die('Error selecting item: ' . mysqli_error($conn));
    if (mysqli_num_rows($result) > 0) {
      while ($row_code = mysqli_fetch_array($result)) {
      $ddate[] = date('d/m/Y', strtotime($row_code['destroydate']));
      }
    }
  }

    $company = strtoupper(mysqli_real_escape_string($conn, $_POST['rtvcompany']));
    $activity = "Box Retrieval";
    $address = ucwords(mysqli_real_escape_string($conn, $_POST['address']));
    $service = ucwords(mysqli_real_escape_string($conn, $_POST['service']));
    $success = 'SUCCESS';
    $authorised = ucwords(mysqli_real_escape_string($conn,$_SESSION['ls_name_usr']));
    $dept = strtoupper(mysqli_real_escape_string($conn,$_POST['rtvdept']));
    $boxitems = $_POST['box_rtv'];
    $box = implode(",",$boxitems);
    $array = array();
    $array = $boxitems;

    foreach ($array as $boxes) {
        $outString .= "$box" . "  ";

        $sql = "INSERT INTO `act` (service, activity, department, company, address, `user`, `date`, destroydate, item, new) VALUES ('$service', '$activity', '$dept', '$company', '$address', '$authorised', NOW(), '".$ddate."', '$boxes', 1)";
        $result = mysqli_query($conn, $sql) or die ('Error inserting box:' . mysqli_error($conn));

        $sql = "UPDATE boxes SET status = 1 WHERE customer = '$company' AND custref = '$boxes'";
        $result = mysqli_query($conn, $sql) or die ('Error updating box:' . mysqli_error($conn));

        $sql = "UPDATE files SET boxstatus = 1 WHERE customer = '$company' AND boxref = '$boxes'";
        $result = mysqli_query($conn, $sql) or die ('Error updating filebox:' . mysqli_error($conn));
    }

    $json = array($box);
    $result = json_encode($json);
    echo $result;

share sample value for $row_code['destroydate'] ?

Niklesh Raut
– Niklesh Raut

2018-06-16 11:01:56 +00:00
Commented Jun 16, 2018 at 11:01 — Niklesh Raut
– Niklesh Raut, Commented Jun 16, 2018 at 11:01
@C2486 should be: 29/05/018 Thanks

user1532468
– user1532468

2018-06-16 11:07:25 +00:00
Commented Jun 16, 2018 at 11:07 — user1532468
– user1532468, Commented Jun 16, 2018 at 11:07
Are you sure year is 018 and not 2018 or 18 ?

Niklesh Raut
– Niklesh Raut

2018-06-16 12:49:25 +00:00
Commented Jun 16, 2018 at 12:49 — Niklesh Raut
– Niklesh Raut, Commented Jun 16, 2018 at 12:49

Ivan Vartanyan · Accepted Answer · 2018-06-16 11:00:14Z

3

If your goal is to have array of dates, which is $ddate, then this line

$ddate = date('d/m/Y', strtotime($row_code['destroydate']));

should be:

$ddate[] = date('d/m/Y', strtotime($row_code['destroydate']));

answered Jun 16, 2018 at 11:00

Ivan Vartanyan

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6 Comments

user1532468 Over a year ago

That works but not when I use variable in mysql query. it enters date as 0000-00-00. I am trying to do a lookup on a table and then use that var $ddate in a foreach loop for a mysql insert. I shall update my code to reflect this.

Gabor Over a year ago

If $row['destroydate'] = "29/05/2018", then $ddate[]=$row['destroydate']

Ivan Vartanyan Over a year ago

If in your MySQL table your field format is datetime then you should convert your date to proper format before using it in INSERT query, which is Y-m-d H:i:s. You can read more about in this question stackoverflow.com/questions/2501915/…

user1532468 Over a year ago

It is a date field. Thanks

imabug Over a year ago

you have $ddate as an array, but in your second foreach loop, you're using it as a scalar in your INSERT query. Echo the resulting SQL query string and see what it looks like. I think you'll find it's not what you expected.

|

Niklesh Raut · Accepted Answer · 2018-06-16 13:01:46Z

1

You are not using any standard format for year in date so first you need to make it standard either full year 2018 or short 18

$row_code['destroydate'] = '29/05/018';
$dt_arr = explode("/",$row_code['destroydate']);
$dt_arr[2] = 2000+$dt_arr[2];
$dt = implode("/",$dt_arr);
$date = DateTime::createFromFormat('d/m/Y', $dt);
echo $date->format('Y-m-d');

Live Demo

answered Jun 16, 2018 at 13:01

Niklesh Raut

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