PHP - Date Convert From Sql
Hi
In sql we have dates like this 134238 In php it shows - 19:10:11 . we can make it by using date('d.m.y', $date); where $date=134238 ; my question is how can we turn 19:10:11 in to 134238 using PHP Similar TutorialsI 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. 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 while I'm reading the cell value '20/09/1980' from an excel file using phpexcel class method $student[$i]['d_o_b'] = $objWorksheet->getCell('G' . $intRow)->getValue(); getting the value as '29484' pls help me to convert it into the date format again. Hi all, I'm having a bit of trouble a script running on a site where it converts a date of birth in a database shown like this '30/04/1993' to an actual age, for instance 18 in this case. Only the script I'm using below shows this age as 17, not 18 as it should be. Code: [Select] <?php $birthday = $row_getdets['dob']; function birthday ($birthday){ list($day,$month,$year) = explode("/",$birthday); $day_diff = date("d") - $day; $month_diff = date("m") - $month; $year_diff = date("Y") - $year; if ($day_diff < 0 || $month_diff < 0) $year_diff--; return $year_diff; } ?> So i've tried to remedy this myself with the following: Code: [Select] <?php $birthday = $row_getdets['dob']; function birthday ($birthday){ list($year,$month,$day) = explode("/",$birthday); $year_diff = date("Y") - $year; $month_diff = date("m") - $month; $day_diff = date("d") - $day; if ($month_diff < 0) $year_diff--; else if (($month_diff==0) && ($day_diff < 0)) $year_diff--; return $year_diff; } ?> ..but I'm having a syntax error (unexpected T_LINE), most probably down to my novice ability, I bet I've missed something simple. I'm still learning guys and I'd really appreciate any help at all. Is it possible to convert time(); to date();? 2011-02-05T18:07:57.00+0000 This is a string. sorry, I know this is simple, but stuck on it... how would one go about converting 1/21/11 to a UNIX timestamp, something like 1293035229? and while we are at it, how to convert 1293035229 to 1/21/11 ? (i know the timestamp is wrong, just for example) thanks!!!!! Hi, Sorry, me again! Ok, so i've got my url: http://www.mydomain.com/?ec3_after=2010-08-01&ec3_before=2010-08-07 I've then got the following code to get the values: $afterDateParts = split("-", $_GET['ec3_after']); $after = $afterDateParts[2] . " " . $afterDateParts[1] . " " . $afterDateParts[0]; echo $after; which returns: 01 08 2010 (correct for this demo) I've then got this to convert the above into a nicer format: $convertMe = strtotime($after); echo date('d-M-Y', $convertMe); But it always returns: 01 Jan 1970 For the love of god, I cannot work out why. Is it really not that simple? Please someone put me out my misery. I'm loosing hair by the minute! Lol TIA Hello, i need a fast convertor that converts the date format 03-03-2008, 08:26 PM into a timestamp fast. Like, you paste 03-03-2008, 08:26 PM into a post form, submit and it gives you the time stamp. I made a convertor a whille ago, but you need to insert each numbers into a separate input box (day, months, year etc) I cant remember how exactly it functions etc asi haventtouched PHP In a while..... hello, i have the following code and it is ok. what im trying to do is convert a 2010-09-20 post from a form and have it read out as Monday September 20, 2010. is this possible? Code: [Select] $dateselected="$_POST[Y]-$_POST[M]-$_POST[D]"; echo "<table border='1'>"; echo "<tr><td width='100%' colspan='4' align='center'>$_POST[M]-$_POST[D]-$_POST[Y]</td></tr>"; Hi. I have an input field where a date is entered, format dd-mm-yy. I need to query the database to see if this date exists. How can I convert the date to yyyymmdd before the query? Thanks So this is a simple code that finds out the difference between two dates and displays it in number of days. $date1=date_create("2013-03-15"); $date2=date_create("2013-12-12"); $diff=date_diff($date1,$date2); echo $diff->format("%R%a days"); // RESULT +272 days
My first question. Is it possible to remove the + sign in the result above? Second question. Is it possible to show "months" if it's greater than 30 days? And years if the days are greater than 365? How would I do this? I wish to know how i can convert this date 24-09-2010 to this format 2010-09-24 This is my first real jump into PHP, I created a small script a few years ago but have not touched it since (or any other programming for that matter), so I'm not sure how to start this. I need a script that I can run once a day through cron and take the date from one table/filed and insert it into a different table/field, converting the human readable date to a Unix date. Table Name: Ads Field: endtime_value (human readable date) to Table Name: Node Field: auto_expire (Converted to Unix time) Both use a field named "nid" as the key field, so the fields should match each nid field from one table to the next. Following a tutorial I have been able to insert into a field certain data, but I don't know how to do it so the nid's match and how to convert the human readable date to Unix time. Thanks in advance!. Hey, I'm using a script which allows you to click on a calendar to select the date to submit to the database. The date is submitted like this: 2014-02-08 Is there a really simple way to prevent rows showing if the date is in the past? Something like this: if($currentdate < 2014-02-08 || $currentdate == 2014-02-08) { } Thanks very much, Jack Hello. I'm new to pHp and I would like to know how to get my $date_posted to read as March 12, 2012, instead of 2012-12-03. Here is the code: Code: [Select] <?php $sql = " SELECT id, title, date_posted, summary FROM blog_posts ORDER BY date_posted ASC LIMIT 10 "; $result = mysql_query($sql); while($row = mysql_fetch_assoc($result)) { $id = $row['id']; $title = $row['title']; $date_posted = $row['date_posted']; $summary = $row['summary']; echo "<h3>$title</h3>\n"; echo "<p>$date_posted</p>\n"; echo "<p>$summary</p>\n"; echo "<p><a href=\"post.php?id=$id\" title=\"Read More\">Read More...</a></p>\n"; } ?> I have tried the date() function but it always updates with the current time & date so I'm a little confused on how I get this to work. Alright, I have a Datetime field in my database which I'm trying to store information in. Here is my code to get my Datetime, however it's returning to me the wrong date. It's returning: 1969-12-31 19:00:00 $mysqldate = date( 'Y-m-d H:i:s', $phpdate ); $phpdate = strtotime( $mysqldate ); echo $mysqldate; Is there something wrong with it? 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 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. |