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PHP - Calculate Age From Date Of Birth
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; Similar TutorialsHi 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. Is 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 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 $year_diff; } ?> ..but I'm having a syntax error (unexpected T_LINE), most probably down to my novice ability, I bet I've missed something simple. I'm still learning guys and I'd really appreciate any help at all. Hi there, I'm new to PHP so sorry if this is a really basic question. How do i post date of birth collected from a form, into a database? I have the fields in the form set up as 'day' 'month' 'year' all of which are drop-down boxes. I tried doing it one way which i saw on a different website, but it didn't work. Here is what i tried: Code: [Select] '$_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 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 Can you please help how to validate the date of birth in code igniter including leap years
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
$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? 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>"; } ?> Hi there, i am using a form with 2 inputs which are equipped with a datepicker: Date 1 & Date 2, is it possible to calculate how many days are there from Date 1 to Date 2 (including the selected ones) ? On my form the dates are in this format: September 08, 2011 (i guess i can change that to numeric only, if that helps) Tamper data shows them getting posted like this: September+14%2C+2011 Any help / hints will be appreciated ! I have date stored in database in any of the given forms 2020-06-01, 2020-05-01 or 2019-04-01 I want to compare the old date with current date 2020-06-14 And the result should be in days. Any help please? PS: I want to do it on php side. but if its possible to do on database side (I am using myslq) please share both ways🙂 Edited June 14, 2020 by 684425Hi all, I am trying to figure out how to calculate 5 working days prior to a given date. I have done some googling but can only see examples of how to add 5 working days onto a date, such as this: Code: [Select] $holidayList = array(); $j = $i = 1; while($i <= 5) { $day = strftime("%A",strtotime("+$j day")); $tmp = strftime("%d-%m-%Y",strtotime("+$j day")); if($day != "Sunday" and $day != "Saturday" and !in_array($tmp, $holidayList)) { $i = $i + 1; $j = $j + 1; } else $j = $j + 1; } $j = $j -1; echo strftime("%A, %d-%m-%Y",strtotime("+$j day")); Does anyone know how to calculate 5 working days prior to a date? Many thanks, Greens85 I have PHP code that generates a long list of Birth Years in a Drop-down list, and I want to make the control "sticky". I know how to make something like this sticky - thanks to help from you guys... Code: [Select] <!-- Gender --> <label for="gender">Gender:</label> <select id="gender" name="gender"> <option value="">--</option> <option value="F"<?php echo (isset($gender) && $gender == "F") ? 'selected="selected"' : ''; ?>>Female</option> <option value="M"<?php echo (isset($gender) && $gender == "M") ? 'selected="selected"' : ''; ?>>Male</option> </select> But my Birth Year code is a little more complex and I'm racking me brain how to do it... Code: [Select] <!-- Birth Year --> <label for="birthYear">Year Born:</label> <select id="birthYear" name="birthYear"> <option value="">--</option> <?php // Display dates for Users between 18 and 100. for($i = $newestYear; $i >= $oldestYear; $i--){ echo '<option value="' . $i . '">' . $i . '</option>'; } ?> </select> How do I make this second set of code "sticky"?? Thanks, Debbie Hey, I'm using a script which allows you to click on a calendar to select the date to submit to the database. The date is submitted like this: 2014-02-08 Is there a really simple way to prevent rows showing if the date is in the past? Something like this: if($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. 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. 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? (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, 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.
               <?phpÂ
               $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>
              Â
               <?php
                              $datetime1 = new DateTime($expired_date);
                              $datetime2 = date('d');
                              $interval = $datetime1->diff($datetime2);
                              echo $interval->d;
               ?>
                  <?php }
               }Â
               ?>
Can anyone help me with what I have wrong? Many thanks Hey guys im stuck in a bind and need some input. What im trying to do is display a number based from community levels. ex: This user need another 2 tasks to get next level. here are the levels 1- 0-2 task new 2- 2-5 tasks beginer 3- 5-7 tasks pro 4- 7+ tasks advanced The way it works is by two different tasks. basicly i wanna know which is the higher number in the tasks either how many posts or comment and figure out based on our levels how many tasks he need to complete in order for him to reach the next level. the first query sees how many post this user has the second check how many comment he has I wanna take the highest number and that belongs to whicheber catgory, do a calculation. and have this kind of output This user is level 2 and need 1 more task to reach the next level. Like i know if he need one more that he has 4 tasks done. I know im all over the place with this one im just trying to get someone who might have an idea of my issue and point me in the right direction. Either if this is a known ?? does this procedure has a specific name anything will help me. |
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