PHP - How Do I Convert This Time Stamp To Just The Date?

hello, im trying to figure out the difference between 2 numbers in minutes.  here is what i have:

Code: [Select]
$tsignin="09-05 10:30:00";
$tsignina="09-06 11:30:00";

$log_in_time_stringsa = strtotime($tsignin);
$log_out_time_stringsa = strtotime($tsignina);
$difference_in_secondssa = ($log_out_time_stringsa - $log_in_time_stringsa);
$tsigna = ($difference_in_secondssa / 60);
 
  $tsigna2 = number_format(round($tsigna));
the 09-05 is sept 5, and 09-06 is sept 6.

this works if the month and date are not included.  any ideas for me so i can figure out the total difference between date/times?

Hi,   I'm new to php,   I'm trying to get the month value from a date stamp (ie I have a variable with the following sort of value: 2014-08-20 07:32:20 )   how do I create a new variable with just the month values (ie 08) of the date stamp??   any help would be appreciated.   Thanks !

I have this code to change the validity date of an account.
I do a check against timestamp and current validity which works  fine.
Then I have radio buttons you can check to choose to extend with one, three, six or 12 months through a form and a post variable - they work too.
When I write the variables out they look fine but when I want it in my database - nothing happens!
I mean the original data is never changed.
Can you see what is wrong?

I tried changing the data type from date to datetime to timestamp with the same result.
I suppose I will learn something new from phpfreaks today again.

Code is here for the 12 month radio button:
Code: [Select]
$newdate = strtotime ( '+12 months' , strtotime ($fromDate) ) ;
$newdate = date ('Y-m-d H:i:s', $newdate );
echo "$newdate";
$query = "UPDATE TABLE user_data SET paid_days='$newdate' WHERE user='$user'";
mysql_query($query);When I write the variable $newdate it shows up as 2013-01-10 21:55 (right now) which seems like a valid format to me.
Do I need to convert it somehow?

I am trying to enter a new time stamp into mysql database when a user logs in at my site.
I read on the update syntax but can't seem to get it to do it.. It reads the values just doesnt write the values into mysql.

$update_query = mysql_query UPDATE users SET (last_access, ip_address, browser, hostname) values ($date, $IP, $browser, $hostname) WHERE username = '$login'") or die("QUERY FAILED: " . mysql_error()); 
$update = ($update_query) or die ("QUERY FAILED: " . mysql_error()); 

This topic has been moved to MySQL Help.

http://www.phpfreaks.com/forums/index.php?topic=315056.0

I use this on my sites and it works fine on one and not the others and I can't figure out why. The sites are all hosted at the some place and even share an IP but for some reason only one of them is working. Here's the stamp: Copyright &copy; <?php echo date(Y); ?>.
So can someone tell me what the problem is?
Thanks

I'm having a problem with getting the timestamp to attach to the filename.  Currently when it's uploaded in the database it says $timestamp in front of the filename instead of the time.

Scroll down towards the bottom of the code to see it.  Thanks for your help!


$username = $_GET['username'];
$item_id = $_GET['id'];

  define( 'DESIRED_IMAGE_WIDTH', 150 );
  define( 'DESIRED_IMAGE_HEIGHT', 150 );

  $source_path = $_FILES[ 'thumb' ][ 'tmp_name' ];
  $timestamp = time();

  //
  // Add file validation code here
  //

  list( $source_width, $source_height, $source_type ) = getimagesize( $source_path );

  switch ( $source_type )
  {
    case IMAGETYPE_GIF:
      $source_gdim = imagecreatefromgif( $source_path );
      break;

    case IMAGETYPE_JPEG:
      $source_gdim = imagecreatefromjpeg( $source_path );
      break;

    case IMAGETYPE_PNG:
      $source_gdim = imagecreatefrompng( $source_path );
      break;
  }

  $source_aspect_ratio = $source_width / $source_height;
  $desired_aspect_ratio = DESIRED_IMAGE_WIDTH / DESIRED_IMAGE_HEIGHT;

  if ( $source_aspect_ratio > $desired_aspect_ratio )
  {
    //
    // Triggered when source image is wider
    //
    $temp_height = DESIRED_IMAGE_HEIGHT;
    $temp_width = ( int ) ( DESIRED_IMAGE_HEIGHT * $source_aspect_ratio );
  }
  else
  {
    //
    // Triggered otherwise (i.e. source image is similar or taller)
    //
    $temp_width = DESIRED_IMAGE_WIDTH;
    $temp_height = ( int ) ( DESIRED_IMAGE_WIDTH / $source_aspect_ratio );
  }

  //
  // Resize the image into a temporary GD image
  //

  $temp_gdim = imagecreatetruecolor( $temp_width, $temp_height );
  imagecopyresampled(
    $temp_gdim,
    $source_gdim,
    0, 0,
    0, 0,
    $temp_width, $temp_height,
    $source_width, $source_height
  );

  //
  // Copy cropped region from temporary image into the desired GD image
  //

  $x0 = ( $temp_width - DESIRED_IMAGE_WIDTH ) / 2;
  $y0 = ( $temp_height - DESIRED_IMAGE_HEIGHT ) / 2;

  $desired_gdim = imagecreatetruecolor( DESIRED_IMAGE_WIDTH, DESIRED_IMAGE_HEIGHT );
  imagecopy(
    $desired_gdim,
    $temp_gdim,
    0, 0,
    $x0, $y0,
    DESIRED_IMAGE_WIDTH, DESIRED_IMAGE_HEIGHT
  );

  //
  // Render the image
  // Alternatively, you can save the image in file-system or database
  //

  header( 'Content-type: image/jpeg' );
  imagejpeg( $desired_gdim, "image_files/".'$timestamp'.$_FILES["thumb"]["name"] );
  
  
  
mysql_connect("localhost", "root", "", "") or die(mysql_error());
mysql_select_db("") or die(mysql_error());

$timestamp = time();
$item_image = '"$timestamp"'.$_FILES['thumb']['name'];
  
$sql="UPDATE product SET thumb='$item_image' WHERE id = '$item_id'";

 if (!mysql_query($sql))

  {

  die('Error: ' . mysql_error());

This topic has been moved to JavaScript Help.

http://www.phpfreaks.com/forums/index.php?topic=306905.0

Code: [Select]
for ($z=1;$z<=30*365; $z++)
{
$bal[date('Y-m-d', strtotime("+ ". $z ." days"))] = $balance;//the balance of loan here.;
}

As you can see Im working on loading the daily balance of a 30 year loan in an array.

The code goes to hell in the year 2038.  I have scoured the internet for help.  I have gone to the wikipedia link and have looked at other posts but there isn't quite something that can help me.  How can I load dates beyond 01-18-2038 into my array?

I have heard of ways to do it with MySQL and the DATETIME() function, but that doesn't seem to make sense for what I am trying to accomplish.

Please point me in the right direction.

Thanks.

Hi everyone...

I would like to implement a questionnaire/survey system that has only two Answers (Yes /No).

Basically, this questionnaire system will be widely used on Mobile Phones.

It will be looking something like this below: http://awardwinningfjords.com/2009/06/16/iphone-style-checkboxes.html

Lets say a user selects a response as No (Using the above slider button for a question). That response should be recorded in the database with User Details + Time Stamp.

Also, I would like to generate a report for each question (Responses of Multiple Users) and a report at a User Level for the entire questionnaire/survey (s).

Can some one guide me the fastest & easiest way to achieve this? I'm a new learner of PHP.

Regards
Sandeep

Hey guys,
How would I go about subtracting Today from a previous day to find the difference?

For example, I want to subtract TODAY from a previous date in my database, to determine if the difference is greater than 1 day.

Any ideas? I tried doing the subraction in TIMESTAMPS, but when I convert the date back to Y-m-d H:i:s, I got some weird year and time.
 

I have tried a large number of "solutions" to  this but everytime I use them I see 0000-00-00 in my date field instead of  the date even though I echoed and can see that the date looks correct.

Here's where I'm at:
I have a drop down for the month (1-12) and date fields (1-31) as well as a text input field for the year.  Using the POST array, I  have combined them into the xxxx-xx-xx format that I am using in my field as a date field in mysql. 

<code>
$date_value =$_POST['year'].'-'.$_POST['month'].'-'.$_POST['day'];
echo $date_value;
</code>

This outputs 2012-5-7 in my test echo but 0000-00-00 in the database.

I have tried unsuccessfully to use in a numberof suggested versions of:
strtotime()
mktime

Any help would be extremely appreciated.
I am aware that I need to validate this data and insure  that it is a valid date. That I'm okay with. I would like some help on getting it into the database.

PHP 5.2.6

We rent PHP server space on a server in California. Several of my pages generated by PHP show the current date and time, so when the page is printed, the user knows when the data was printed. Our users could be anywhere in the US, I am in Michigan. How do I convert the server time (PDT) into local time, regardless of where the user is?

Thanks.

Hi,

I am trying to convert a String date into numeric date using PHP function's, but haven't found such function.
Had a look at date(), strtotime(), getdate();

e.g. Apr 1 2011  ->    04-01-2011

Could someone please shed some light on this?

Regards,
Abhishek

I am getting syntax error, unexpected T_CONSTANT_ENCAPSED_STRING
on the date stamp line below but I can't see what is causing the problem.
when I comment it out the problem goes away so I know I it is there.
please help.
 
$email_message = "Form details below.\n\n";
 $email_message = date("m/d/Y")"\n";