PHP - Updating A Last Access Time Stamp

This topic has been moved to MySQL Help.

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

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

The following code works. The user checks off what foods they want to be added. Then the foods and their nutrient values get added into the database. However, if the user refreshes, it updates the database with the values they selected before - thus doubling the amount of foods stored in the database. How do I stop this from happening?

Right now i'm taking values from the POST array and running an INSERT SQL query to put them into the database. (look for the following comment below: //Look here PHPFREAKS.COM)

Code: [Select]
<?php include 'top.php'; ?>
<?php

if (!loggedin())
{//1 if start
header('Location: need-to-log-in-mmp.php');
}//1 if end
?>
    <title>My Meal Planner - Vitamin K Tracker</title>
</head>

<div id="container">

<?php include 'header.php'; ?>

<?php include 'nav.php'; ?>



<div id="content-container">

<div id="content_for_site">
<h2>My Meal Planner</h2> <br />
                     
     


     
       
                        My Meal Que:
<br />
<ul>
<form action="my-meal-planner.php" method="POST">
<?php
//do sql query to return the foods and nutrients that a person added to their que

$query = "SELECT DISTINCT
foods.id, `name`, `source`,`vit_k`, `cal`,`protein`, `fiber`, `carbs`, `sugar`, `sodium`, `chol`, `total_servings`
 FROM `foods`
 LEFT JOIN users_foods
 ON foods.id=users_foods.food_id

WHERE users_foods.user_id =". $_SESSION['user_id']."
 ORDER BY users_foods.id";



$query_run = mysql_query($query, $link);
//echo mysql_errno($link) . ": " . mysql_error($link). "\n";
while ($row = mysql_fetch_array($query_run)){
?> <input type="checkbox" name="<?php
echo $row["id"];
?>"<?php
    echo ' value="'.$row["id"].'" />';
echo ' Food Name: '. $row["name"]." | Vitamin K: ".
    $row["vit_k"]." | Fiber: ".
    $row["fiber"]." | Calories: ".
    $row["cal"]." | Protein: ".
    $row["protein"]." | Carbs:".
    $row["carbs"]." | Sugar: ".
    $row["sugar"]." | Sodium: ".
    $row["sodium"]." | Cholesterol: ".
    $row["chol"]." | Source: ".
    $row["source"]." | Total Servings: ".
    $row["total_servings"]. "<br />How many servings will you have? "?><input type="text" name="<?php
echo $row["id"]."serving";
?>"<?php
    echo ' />';
   echo '<br /><br />';
}
     

?>
</ul>
Select a date: <input id="date" type="text" name="add_to_calendar"><br />
<input type="submit" value="add to calendar" ><br />





<input type="button" value="delete" name="delete">

</form>

<?php
//updates the date of the food 
if (isset($_POST['add_to_calendar']) && !empty($_POST['add_to_calendar'])){

$add_to_calendar = $_POST['add_to_calendar'];
$date_array = explode("/", $add_to_calendar);
                   

//Look here PHPFREAKS.COM
for ($i=1; $i<1000; $i++){
if (!empty($_POST[$i])){
$id = $_POST[$i];
$serv = $id. "serving";
$serving = $_POST[$serv];
$month = $date_array[0];
$day = $date_array[1];
$year = $date_array[2];
//echo 'month: '. $month . ' day: ' . $day . ' year: ' . $year . "<br />"; 

if (substr($month,0,1) == '0'){
//takes out a zero in front of month if it exists
$month = substr($month,1,1);}

if (substr($day,0,1) == '0'){
//takes out a zero in front of day if it exists
$day = substr($day,1,1);}

//echo 'POST ARRAY READS AS: <br />';
//print_r($_POST);
//echo '<br />';
$query = "INSERT INTO `users_calendar` VALUES ('','".$_SESSION['user_id']."','".$id."','".$month."','".$day."','".$year."','".$serving."')";
//echo '<br /> The query reads as: ' . $query;
$query_run = mysql_query($query) or die(mysql_error());
//columns in users_calendar:
//DONEid    DONEuser__id   DONEfood_id    DONEcalendar_month   DONElendar_day    DONEcalendar_year    servings
  }
}




}
?>

<br />


</div>



<script type="text/javascript" src="js/datepicker.js"></script>
<div id="clear"></div>
<?php //include 'footer.php'; ?>


</div>
</div>
</body>
</html>

Hi guys,

How can I compare

Code: [Select]
2011-12-17T12:00:06+0000
with a current date('c')?

cheers in advance

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 !

Code: [Select]
<?php
$myServer = "202.138.125.155";
$myUser = "micro_fms";
$myPass = "micro_fms*#$";
$myDB = "micro_fms"; 

//create an instance of the  ADO connection object
$conn = new COM("ADODB.Connection")or die("Cannot start ADO");

//define connection string, specify database driver
$connStr = "PROVIDER='SQLOLEDB'; SERVER=".$myServer.";UID=".$myUser.";PWD=".$myPass.";DATABASE=".$myDB; 

//Open the connection to the database
$conn->open($connStr);

//$table_name=$_GET['Device'];
$table_data=$_GET['Data'];
$data=(explode(",",$table_data));

$table_name=$data[0];
$speed=$data[1];
$time=$data[2];
$date=$data[3];
$lat=$data[4];
$long=$data[5];
//$newdate = substr($date,6,2) . "-" . substr($date,0,2) . "-" substr($date,3,2);
$datetime=$date.$time;


$new_lat = substr($lat,0,2).'.'.substr($lat,2);
$new_long = substr($long,0,2).'.'.substr($long,2);

echo $table_name."<br>";
echo $speed."<br>";
echo $new_lat."<br>";
echo $new_long."<br>";
echo $datetime."<br>";

$query="insert into $table_name (DeviceNumber,Speed,dtDateTime,Latitude,Longitude)values('$table_name','$speed','$datetime','$new_lat','$new_long')";


//execute the SQL statement and return records
$rs = $conn->execute($query);
if($rs)
{
echo 'Values Inserted';
}
else
{
echo "faile";
}
?>


Hi friends...
I have two variables ($date and $time). I concat these two strings and make a new variable called "$datetime". Now in above code I want to insert it into table but I got the following error. Code: [Select]
Fatal error: Uncaught exception 'com_exception' with message '<b>Source:</b> Microsoft OLE DB Provider for SQL Server<br/><b>Description:</b> Conversion failed when converting datetime from character string.' in C:\wamp\www\FMS\fms.php:44 Stack trace: #0 C:\wamp\www\FMS\fms.php(44): com->execute('insert into A00...') #1 {main} thrown in C:\wamp\www\FMS\fms.php on line 44
In my database, data type of column in which I want to insert is Date time stamp though I have string type ($datetime) variable.
I think I have to convert this $datetime variable string to "DateTime type first but don't have any idea about it.
Please help??/
Thanks in advance....

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 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";

PHP date and time function is not showing correct time on my local system

I have the following php code

date_default_timezone_set("Africa/Lagos");

$date = date('d-m-y h:i:s');

echo "Server Time ".$date ."<br>";

echo "The time is " . date("h:i:sa")."<br>";

$current_datetime = date("Y-m-d") . ' ' . date("H:i:s", STRTOTIME(date('h:i:sa')));

echo "Current time1: ".$current_datetime . "<br>";

 

Output

Server Time 21-05-21 09:55:39
The time is 09:55:39am
Current time1: 2021-05-21 09:55:39

 

Expected Output

Server Time 21-05-21 10:55:39
The time is 10:55:39am
Current time1: 2021-05-21 10:55:3

 

Any help would be appreciated.

 

Edited May 21 by Ponel

I am trying to simulate an ad expiration and carry out an action if the ad is expired. And I cannot get the if/else to work properly... I've tried many variations and I cannot see what I am doing wrong here. Any tips please 3 hours and counting of no solution!

$ad_start = time()-14 . "<br />"; // 14 days from today in the past (negative)
echo $ad_start;

$current_time = time() . "<br />"; // current epoch time
echo $current_time;

$days_past = $ad_start - $current_time; // days past

echo "<br />$days_past days have past since the ad started!<br />";

if($days_past <= 14)
{
echo "<br />Ad is less than 14 days.  Not expired.";
}
else
{
echo "<br />Ad is over 14 days. Expired.";
}

OK So I've got a datepicker that sends a date in d/m/y format. My DB stores the data in Unix Timestamp Which I can convert the date to with strtotime however this does the exact date & time.

All I want is the actual day. I've spent hours trying to convert this with just the day with mixed results...

Thanks.

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.