PHP - Set Default Date.

Hi,

I have a job listing website which displays the closing date of applications using:

$expired_date

(This displays a date such as 31st December 2019)

I am trying to show a countdown/number of days left until the closing date.

I have put this together, but I can't get it to show the number of days.

                    <?php 
                    $expired_date = get_post_meta( $post->ID, '_job_expires', true );
                    $hide_expiration = get_post_meta( $post->ID, '_hide_expiration', true );
                    if(empty($hide_expiration )) {
                        if(!empty($expired_date)) { ?>
                                <span><?php echo date_i18n( get_option( 'date_format' ), strtotime( get_post_meta( $post->ID, '_job_expires', true ) ) ) ?></span>
                    
                    <?php
                                        $datetime1 = new DateTime($expired_date);
                                        $datetime2 = date('d');
                                        $interval = $datetime1->diff($datetime2);
                                        echo $interval->d;

                     ?>
                        <?php }
                    } 
                    ?>

Can anyone help me with what I have wrong?

Many thanks

(continuing from topic title)
So if I set a date of July 7 2011 into my script, hard coded in, I would like the current date to be checked against the hard coded date, and return true if the current date is within a week leading up to the hard coded date.

How could I go about doing this easily?

I've been researching dates in php but I can't seem to work out the best way to achieve what I'm after.

Cheers
Denno

Hi Guys..

How can I change a date on the fly ?
Everything is UTC on my server.

How can I change a date to something else on the fly?

Ie:

$timezone = "cet";
$datetime = "2011-09-04 19:53:00";

echo $datetime($timezone);

So I can give it a datetime and have it echo the datetime as if it were in the other timezone?

Thanks
Graham

I'm getting this Time Zone error.
Perhaps it's a compatibility issue with PHP 5.3. Looked all over for an answer without finding one.
Here is the error message


Warning: date() [function.date]: It is not safe to rely on the system's timezone settings.
You are *required* to use the date.timezone setting or the date_default_timezone_set() function.
In case you used any of those methods and you are still getting this warning,
you most likely misspelled the timezone identifier. We selected 'America/New_York' for 'EST/-5.0/no DST'
instead in /blocked.php on line 41
12/02/12

Here is the code. Line 41 is near the bottom, the one with the d,m,y.  Perhaps the echo date (d/m/y") needs to be changed. Appreciate any help!

Code: [Select]
   <table border="3" width="16%" align="center" cellspacing="0" bgcolor="#FF6600" bordercolor="red" bordercolordark="red" bordercolorlight="red">
                            <tr>
                                    <td width="176">

                                            <p align="center"><?php // shows IP Number on Page
            echo $ip; ?> </p>

                                    </td>
                            </tr>
                    </table>
                    <p align="center"><?php // Show the user agent
        echo 'Your user agent is: <b>'.$_SERVER['HTTP_USER_AGENT'].'</b><br />';?></p>
                    [b]<h1 align="center"><?php echo date("d/m/y");?></h1>[/b]
            </td>
    </tr>
</table
[,code]

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

Hi there,

I have a string '12/04/1990', that's in the format dd/mm/yyyy.

I'm attempting to convert that string to a Date, and then insert that date into a MySQL DATE field.

The problem is, every time I try to do so, I keep getting values like this in the database: 1970-01-01.


Any ideas?


Much appreciated.

Hi guys, I'm putting together a small event system where I want the user to add his own date and time into a textfield (I'll probably make this a series of drop-downs/a date picker later).
This is then stored as a timestamp - "0000-00-00 00:00:00" which displays fine until I try to echo it out as a UK date in this format - jS F Y, which just gives today's date but not the inputted date.



Here's the code I have right now:

Code: [Select]
$result = mysql_query("SELECT * FROM stuff.events ORDER BY eventdate ASC");

echo "<br />";
echo mysql_result($result, $i, 'eventvenue');
echo ", ";

$dt = new DateTime($eventdate);

echo $dt->format("jS F Y");


In my mysql table eventdate is set up as follows:
 
field - eventdate
type - timestamp
length/values - blank
default - current_timestamp
collation - blank
attributes - on update CURRENT_TIMESTAMP
null - blank
auto_increment - blank


Any help as to why this could be happening would be much appreciated, thanks.

i have a table that shows payments made but  want to the payments only showing from a set date(06/12/14) and before this date i dont want to show   this is my sql that doesnt seem to work and is showing dates before the specified date. .

"SELECT * FROM payments2014, signup2014, editprop2014 WHERE signup2014.userid = payments2014.payment_userid AND editprop2014.prop_id = signup2014.prop_id AND signup2014.userid !='page1' AND signup2014.userid !='page6' AND signup2014.userid !='page4' AND payments2014.payment_transaction_status !='none' AND payments2014.payment_transaction_status !='CANCELLEDa' AND payments2014.payment_type !='deposit' AND payments2014.payment_paid_timestamp NOT LIKE '%2012%' AND payments2014.payment_paid_timestamp NOT LIKE '%2011%' AND payments2014.payment_paid_timestamp >= '06/12/14' ORDER BY payments2014.payment_id DESC"

i have some other parts in the statment but this one that should be filtering is 

host_payments2014.payment_paid_timestamp >= '06/12/14' 

thanks in advance

Hi, Currently I am making a module for joomla. every article has an publish date, if the article was published in 7 days ago, it will displayed as "article in last week", My idea is to use today's date - publish date, if the result is greater than 7 and smaller than 14, the article will be displayed as "article in last week.
Any one know how to write this code?
Here is what I have got, but not working.

<?php
$todays_date = date("Y-m-d"); 

$result = mysql_query("select * from jos_content where $test between $todays_date-14 and $todays_date-7");  
while($row = mysql_fetch_array($result))
  {
echo "$todays_date - $row[title]";
}
?>

In a Switch statement, can you give the Default: a specific name, maybe like this...


switch ($resultsCode){
// Missing Primary Key.
case 'COMMENT_MISSING_KEYS_2050':
echo '<h1>System Error</h1>';
echo '<p>A Fatal Error has occurred.  Please contact the System Administrator. (2050)</p>';
break;

default 'DEFAULT_CATCHALL_ERROR_CODE_9999':
echo '<p>You have reached the catch-all error code... (9999)</p>';
break;



Debbie

I have a few dropdown forums and would like it to select the current values that are requested.

Code: [Select]
global $filename,$fileid,$filetype,$filedir,$fileby;
  global $months,$pg;


echo "<h2>Search Files</h2>";
echo "<p>";
  echo "<form name = 'File_Search' action = '#' method = 'POST'>";
echo "<table border=\"0\" width=\"100%\"><tr>";
echo "<td>File Name<br><input type = 'text' name = 'name'></td><td>File Type<br><select name = 'type'>
<option value = '0'>All Types</option>
  <option value = 'o'>Official Files</option>
<option value = '1'>Quests</option>
  <option value = '2'>Graphic/Spritesheets</option>
  <option value = '3'>Entity/Scripts</option>
  <option value = '4'>Sound/Music</option>
  <option value = '5'>QuestPack/Programs</option>
  <option value = '6'>Miscellaneous</option>
  </select></td>";
  echo "<td>Order By<br><select name = 'order'>
  <option value = '0'>Last Updated</option>
  <option value = '1'>ID</option>
  <option value = '2'>Name</option>
  <option value = '3'>Downloads</option>
  <option value = '4'>Rating</option>
<option value = '5'>Points</option>
  <option value = '6'>Random</option>
  </select></td>";
  echo "<td>Acending?<br><select name = 'dir'><option value = '0'>False</option><option selected value = '1'>True</option></select></td>";
echo "</tr><tr><td rowspan = '4'><input type = 'submit' name = 'search' value = 'search'></td></tr></table></form>";
echo "</p>";

So if the $filetype equals 2 I want the dropdown box to show "Graphic/Spritesheets" by default. BTW Official Files is not actually a type so that's why I chose to give it a small letter o and will work the same way as All Types but run a different function.

$fileby is used for the OrderBy field.
$filedir is for the Acending field.
The other data is for the other parts of the script.

If you like you can give advice on the name's default because you may have a better method then what I have in my head.

Hi guys and gals ,

got a minor problem.  I have a table in which i want the "photo" column to have a default value of "noimage.jpg".

I set the default value to "noimage.jpg" and put "as defined" too.

but when i fill the form in and leave the upload field blank it doesnt show the noimage.jpg as it should and in the mysql table it leaves it blank and not with default value.

Here is the inserts.php which adds the data to the mysql table.   

Can you help please.

<CENTER><B>Vehicle Added</B></CENTER>
<BR>

<?php

mysql_connect("localhost", "wormste1_barry", "barry") or die(mysql_error());
mysql_select_db("wormste1_barry") or die(mysql_error());

$CarName = mysql_real_escape_string(trim($_POST['CarName']));
$CarTitle = mysql_real_escape_string(trim($_POST['CarTitle']));
$CarPrice = mysql_real_escape_string(trim($_POST['CarPrice']));
$CarMiles = mysql_real_escape_string(trim($_POST['CarMiles']));
$CarDescription = mysql_real_escape_string(trim($_POST['CarDescription']));
$pic = mysql_real_escape_string(trim($_FILES['uploadedfile']['name'])); 


$target_path = "images/";

$target_path = $target_path . basename( $_FILES['uploadedfile']['name']); 

if(move_uploaded_file($_FILES['uploadedfile']['tmp_name'], $target_path)) {
    echo "The file ".  basename( $_FILES['uploadedfile']['name']). 
    " has been uploaded";
    echo "<br />";
} else{
    echo "There was an error uploading the file, please try again!";
}



mysql_query("INSERT INTO cars 
(CarName, CarTitle, CarPrice, CarMiles, CarDescription, photo) VALUES('$CarName', '$CarTitle', '$CarPrice', '$CarMiles', '$CarDescription', '$pic' ) ") 
or die(mysql_error());  


echo "The vehicle data has been added!";

?>

I am using this php code <?php echo $_GET['CiFrame']; ?> to check the url for the variable CiFrame, this allows me to link to a page through my page containing the iframe.

Here is my iframe code.
<iframe name="CiFrame" width="727" height="805" src="<?php echo $_GET['CiFrame']; ?>" scrolling="auto" frameborder="0"></iframe>

The problem is that if the url does not contain a variable the iframe will not open a page. How can i set a default variable if one is not provided?

Thank You

How to define default folder. For example I have folder images and in that folder specific image. In the root directory I have index.php with
Code: [Select]
<?php require_once("public/includes/header.php"); ?>and in index.php there are links which goes to different folder:
  Code: [Select]
<a href="public/sajt/kategorija.php?id=<?php echo $id;  ?>">which also have header.php, but there is no image. How to make default folder for image, or some similar solution?