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PHP - Convert A Date Of Birth To Age
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. Similar TutorialsIs there a way to add a date of birth into a mysql but display it as the age?... e.g Mysql = 04/06/89 Display = 21 Hi guys, My apologies if this is in the wrong forum but I am not really sure how to go about this. I have not written any code for this but I have four fields in one table one called age and 3 others dobmonth / dobday /dobyear - My question being how would I write some code that automatically fills in the age field based on the date of birth fields? If anyone could point me in the right direction that would be awesome, Appreciated. i have date of birth stored as DATE type in mysql. i tried this so it would show the age but it comes up blank. Code: [Select] $getprof = mysql_query("SELECT * FROM Profile WHERE username='$search'")or die(mysql_error()); while($rowprof = mysql_fetch_assoc($getprof)) { $username1 = $rowprof['username']; $location = $rowprof['location']; $gender = $rowprof['gender']; $dateofbirth = $rowprof['dateofbirth']; $information = $rowprof['information']; } function GetAge($dateofbirth) { // Explode the date into meaningful variables list($BirthYear,$BirthMonth,$BirthDay) = explode("-", $dateofbirth); // Find the differences $YearDiff = date("Y") - $BirthYear; $MonthDiff = date("m") - $BirthMonth; $DayDiff = date("d") - $BirthDay; // If the birthday has not occured this year if ($DayDiff < 0 || $MonthDiff < 0) $YearDiff--; return $YearDiff; } echo $YearDiff; $username = $_POST['username']; $password = $_POST['password']; $month = $_POST['month']; $day = $_POST['day']; $year = $_POST['year']; $query = mysql_query("INSERT INTO users VALUES ('','$username','$password','$month','$day','$year') mysql_query($query); The code above is a sample of what I have but what I want is to store an entire birthdate in ONE SQL cell. More like this... $username = $_POST['username']; $password = $_POST['password']; $month = $_POST['month']; $day = $_POST['day']; $year = $_POST['year']; $query = mysql_query("INSERT INTO users VALUES ('','$username','$password','$birthdate') mysql_query($query); How is this possible? Can I do this and actually use it efficiently in the future? Hi there, I'm new to PHP so sorry if this is a really basic question. How do i post date of birth collected from a form, into a database? I have the fields in the form set up as 'day' 'month' 'year' all of which are drop-down boxes. I tried doing it one way which i saw on a different website, but it didn't work. Here is what i tried: Code: [Select] '$_POST[day] . - . $_POST[month]' . - . $_POST[year]', More info: In the database table this information is going to, the "date of birth" field is set to "DATE" type. Don't know if that makes any difference Can you please help how to validate the date of birth in code igniter including leap years
hello fellas, need some help please if possible. i have created a date of birth section in my form where the user selects his/her date of birth from the dropdown menu. they would first select the day then month then year of their birthday. how would i setup the database to get this to work? i currently have: Code: [Select] day VARCHAR( 2 ) NOT NULL , month VARCHAR( 4 ) NOT NULL , year VARCHAR( 4 ) NOT NULL , is this correct? many thanks I need to add date of birth field to registration form and then save it to databse. I cannot figure out what might be best way of storing the date in the table. I could convert it to unix epoch time, or I could do YYYYMMDD.
Thoughts? What would be the easiest method of saving the DOB?
I am not asking on how to do it, just the format. Thanks
Since I didn't want to type it out myself I wrote a small Date of Birth drop down menu generator. Now I'm wondering how I can make the code copy-able in a text area? The script should be inserting all the code ready and finished into a textarea so you can copy and go. How is it done? Here's the script: <?php echo "<center>"; ?> <form action='' method='POST'> <input type='submit' name='submit' /> </form> <?php $submit = $_POST['submit']; if ($submit) { echo "<form action='' method='POST'>"; echo "<select name='month'>"; for ($m = 01; $m <= 12; $m++) { echo " <option value='" . $m . "'>" . $m . "</option> "; } echo "</select>"; echo "<select name='day'>"; for ($d = 01; $d <= 31; $d++) { echo " <option value='" . $d . "'>" . $d . "</option> "; } echo "</select>"; echo "<select name='year'>"; for ($y = 1900; $y <= 2010; $y++) { echo " <option value='" . $y . "'>" . $y . "</option> "; } echo "</select>"; echo "</form>"; echo "</center>"; } ?> 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. 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 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 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. Is it possible to convert time(); to date();? 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>"; 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..... 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 2011-02-05T18:07:57.00+0000 This is a string. 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? |
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