Two days ago I had an interview as a php developer and they gave me a task to perform.
I completed 90% of the task, but I failed while trying to add date, month and year together.
I have 3 dropdowns for date, month and year:
Date Of Birth : <br />
Date : <select name="date">
<?php $i=1; for($i=1; $i<=31; $i++){ ?> <option> <?php echo $i?> </option> <?php } ?>
</select>
Month : <select name="month">
<option value="January">January</option>
<option>Febuary</option>
<option>March</option>
<option>April</option>
<option>May</option>
<option>June</option>
<option>July</option>
<option>August</option>
<option>September</option>
<option>October</option>
<option>November</option>
<option>December</option>
</select>
Year :
<?php
$currently_selected = date('Y');
$earliest_year = 1950;
$latest_year = date('Y');
print '<select name="year">';
foreach ( range( $latest_year, $earliest_year ) as $i ) {
print '<option value="'.$i.'"'.($i === $currently_selected ? ' selected="selected"' : '').'>'.$i.'</option>';
}
print '</select>';
?>
I have only one column in the database for the date of birth.
I want to add the date, month and year together, which should result in the following: date/month/year.
How could I accomplish this?
if(isset($_POST['submit'])){
$name = trim($_POST['username']);
$date = $_POST['date'];
$month = $_POST['month'];
$year = $_POST['year'];
if($name == ''){
$error = "Add Name";
}
$DateOfBirth = $date.'/'$month;
if(!$error){
echo $DateOfBirth;
}
}
Assuming your date of birth field is a MySQL DATE field, you need to make sure that you are creating a date in the correct format i.e. YYYY-MM-DD. You can do that by using strtotime to convert your input data into a timestamp, and then date to convert that into the appropriate format:
$DateOfBirth = date('Y-m-d', strtotime("$date $month $year"));
For example:
$year = 2015;
$month = 'August';
$date = 5;
$DateOfBirth = date('Y-m-d', strtotime("$date $month $year"));
echo $DateOfBirth;
Output:
2015-08-05
Demo on 3v4l.org
Note
If you really want the result to be in dd/mm/yyyy format, just change the first input to date to 'd/m/Y'.
The reason is your are not concatenating the variables in php. Check below code. You are almost there but with small coding mistake.
What you are trying to do is concatenate a variable with a string. So you have to concatenate variable with a '.' and then add your string. In here / works as the string. When adding a string it should be with in quotations '/' . And the final output will be something like this. $variable . '/'. Now concatenate another variable. Then it would be like this. $variable01 . '/' . $variable02.
$DateOfBirth = $date.'/'.$month.'/'.$date ;
Output
12/02/2020
Check the added example with respect to your requirement
example
I am new to PHP. I have a selection box with a range of years starting at current year - 5 (2014) through the year 2050. My selection box contains the correct years but I need it to default to the current year. Right now, it defaults to 2014. I've been working on this for hours. I thought the $current_year variable was needed in the loop in some way - in order to set the default - but that doesn't work. After researching and seeing various examples, I'm now thinking I need to build an if statement within the loop but I really don't know.
<form id="calendarForm" name="calendarForm" method="post" action="">
<label for="select_year">Select the year: </label>
<select id="select_year" name="select_year">
<?php
$date = new DateTime();
// Sets the current year
$current_year = $date->format('Y');
// Year to start available options.
$earliest_year = ($current_year - 5);
$year = $earliest_year;
for ($year = $earliest_year; $year <= LATEST_YEAR; $year++) {
echo "<option value='$year'>$year</option>";
}
?>
</select>
Modify your for loop as follows:
for ($year = $earliest_year; $year <= LATEST_YEAR; $year++) {
if($year==$current_year)
{
echo "<option value='$year' selected>$year</option>";
}
else{
echo "<option value='$year'>$year</option>";
}
}
Try the following. You need to adjust your php tags but neat.
<option value='$year' <?php if($year == date("Y")) {echo "selected";}?> >$year</option>";
It will check against the current year and will be the selected option if matched.
does anyone know how I would echo out the selected date as text with the date month and year separated outside of the form? I tried echoing out $date $month and $year outside of the form however this doesn't give me the correct date thankyou for the help
<?
$date = array('16-01-14','16-01-28','16-02-14','16-02-28','16-03-14','16-03-28','16-04-14','16-04-28',
'16-05-14','16-05-28','16-06-14','16-06-28','16-07-14','16-07-28','16-08-14','16-08-28','16-09-14','16-09-28','16-10-14','16-10-28',
'16-11-14','16-11-28','16-12-14','16-12-28');
$currentdate = date('y-m-d');
echo $currentdate;
?>
<form>
<select style="width:200px;">
<?php
foreach ($date as $i => $d) {
if ($currentdate >= $d && ($i == count($date)-1 || $currentdate < $date[$i+1])) {
$selected = "selected";
} else {
$selected = "";
}
list($year, $month, $day) = explode('-', $d);
echo "<option $selected>" . date("m/d/Y", strtotime($d)) . "</option>";
echo 'the current billing period is';
}
?>
</select>
</form>
Inside of your loop add a $selected_int variable like so:
foreach ($date as $i => $d) {
if ($currentdate >= $d && ($i == count($date)-1 || $currentdate < $date[$i+1])) {
$selected = "selected";
$selected_int = $i;
} else {
$selected = "";
}
list($year, $month, $day) = explode('-', $d);
echo "<option $selected>" . date("m/d/Y", strtotime($d)) . "</option>";
echo 'the current billing period is';
}
Then, you can reference it like:
echo date('Y-m-d', strtotime($date[$selected_int]));
Addition
I know you've already accepted the answer, but I also wanted to make a suggestion now that I see what you are using the $date for. Since you know the start date, and it is in 14-day periods, it would be easy to write that as part of the loop.
$start_date = date('Y-m-d', strtotime(date('Y').'-01-01'); //First day of the year, for the sake of argument.
$interval = 14;
for ($i = 0; date('Y') == date('Y', strtotime($start_date.' +'.($i * $interval).' days')); $i++) {//While this year is equal to the start date's year with the added interval [If I knew what your logic here was I could offer a better suggestion]
if ($currentdate >= date("Y-m-d", strtotime($start_date.' +'.($i * $interval).' days')) && (date('Y') < date("Y", strtotime($start_date.' +'.(($i + 1) * $interval).' days')) || $currentdate < date("m/d/Y", strtotime($start_date.' +'.(($i + 1) * $interval).' days')))) {
$selected = "selected";
$selected_int = $i;
} else {
$selected = "";
}
echo "<option $selected>" . date("m/d/Y", strtotime($start_date.' +'.($i * $interval).' days')) . "</option>";
}
Basically, this takes the start date, shows it as the first date option, then adds 14 days to it with each pass through. Your if/else statement should still be the same. It checks to see if you are on the last interval of the year, or if the current date is less than the next interval, and also that the current date is greater than the current interval.
After your loop, you can get the date by:
echo date("m/d/Y", strtotime($start_date.' +'.($selected_int * $interval).' days'));
I know it seems like a lot, but it would save you from having to make a date array to begin with.
Use strtotime instead list.
....
// list($year, $month, $day) = explode('-', $d);
echo "<option $selected>" . date("m/d/Y", strtotime($d)) . "</option>";
....
EDIT: Additional information - your code requires a lot modification and likely some structure changes but assuming this is for testing a method and "how to do" instead a final product.
You need to submit the selected date, catch it in the script and use the selected date to do what you need - i.e. retrieve data from database - and this should give you some idea.
<?php
// You need to create these dates by using another method. You cannot hard code these. You can create it with date functions easily.
$date = array('16-01-14','16-01-28','16-02-14','16-02-28','16-03-14','16-03-28','16-04-14','16-04-28','16-05-14','16-05-28','16-06-14','16-06-28','16-07-14','16-07-28','16-08-14','16-08-28','16-09-14','16-09-28','16-10-14','16-10-28','16-11-14','16-11-28','16-12-14','16-12-28');
// Checking if we have a posted form, with the button name user clicked
if (isset($_POST["btnSubmit"])) {
// This is your selected day - use it where you need:
$selectedDate = $_POST["selectedDate"];
// This is where your model start singing and gets necessary info for this date - just printing here as sample
print $selectedDate;
// I need dropDownDate to compare in the SELECT to preselect the appropriate date
$dropDownDate = strtotime($selectedDate);
} else {
// First time visit, preselect the nearest date by using current date
$dropDownDate = time();
}
?>
<form method="post">
<select name="selectedDate" style="width:200px;">
<?php
foreach ($date as $i => $d) {
if ($dropDownDate >= strtotime($d) &&
(!isset($date[$i+1]) || ($dropDownDate < strtotime($date[$i+1])))
) {
$selected = 'selected="selected"';
} else {
$selected = "";
}
list($year, $month, $day) = explode('-', $d);
echo "<option $selected>" . date("m/d/Y", strtotime($d)) . "</option>";
}
?>
</select>
<input type="submit" name="btnSubmit" value="Submit">
</form>
Note that I added a "submit" type input (to submit the form) and changed form method to "post", finally named SELECT as "selectedDate". I also changed your date comparison code line in the loop.
Hope this helps.
Is there a way to convert values to date? I have 3 dropdown lists: day, month and year. If the selected date < today, then the consumers won't be able to click on the buy button til the day i selected in de dropdown.
The problem is, that the date I selected, is a value and not a date. Is there a way to compare my value with the date of today? The code below is the backend of my website, the if statement has to stay on the frontend
$releasedatumdag= get_post_meta( $domeinnaam_extensies->ID, 'releasedatumdag', true );
$releasedatummaand= get_post_meta( $domeinnaam_extensies->ID, 'releasedatummaand', true );
$releasedatumjaar= get_post_meta( $domeinnaam_extensies->ID, 'releasedatumjaar', true );
<tr>
<th>Releasedatum:</th>
<td>
<select name="domeinnaam_extensies_releasedatumdag">
<?php
for($idag = 1; $idag <= 31; $idag++){
echo '<option value="'.$idag.'">'.$idag.'</option>';
}
?>
</select>
<select name="domeinnaam_extensies_releasedatummaand">
<?php
for($imaand = 1; $imaand <= 12; $imaand++){
echo '<option value="'.$imaand.'">'.$imaand.'</option>';
}
?>
</select>
<select name="domeinnaam_extensies_releasedatumjaar">
<?php
for($ijaar = 2000; $ijaar < date("Y")+10; $ijaar++){
echo '<option value="'.$ijaar.'">'.$ijaar.'</option>';
}
?>
</select>
</td>
</tr>
You can use DateTime() to create your date. You then can compare it to another DateTime object representing today since DateTime objects are comparable:
$my_date = DateTime::createFromFormat('d/m/Y', '31/12/2014');
$now = new DateTime();
if ($my_date < $now) {
// do something
}
You have a day, month, and a year so just create a DateTime object and you can get the timestamp for it and compare it to today.
http://www.php.net/manual/en/datetime.construct.php
http://php.net/manual/en/function.checkdate.php
Can string the digits into any format that you want
$dateString = $releasedatumdag . '/' . $releasedatummaand . '/' . $releasedatumjaar;
//Make sure that they didn't try to create Feb. 31
if(checkdate($dateString)) {
$submittedTime = new DateTime($dateString);
$today = new DateTime();
//DO YOUR COMPARISON HERE
}
How do you create a select with the option of 2 days ahead from today picked as the default option (i.e. a 48 hour window) in PHP? This is the code I'm using so far but it doesn't work unfortunately!
<?php
$weekday = array('Sunday','Monday','Tuesday','Wednesday','Thursday','Friday','Saturday');
$days = range (1, 31);
$currentDay = date('F');
echo "<select name='weekday'>";
foreach ($days as $value) {
$default = ($value == $currentDay)?'selected="selected"':'';
echo '<option '.$default.' value="'.$value.'">'.$value."</option>\n";
}
echo '</select> ';
?>
I'm confused as to what your code does.
As far as I can tell $weekday is not used after being instantiated, and you are setting $currentDay to the text representation of the current month (e.g. September).
But if you want to get the day of month of the day 48 hours from now:
$two_days_ahead = date('j', strtotime('+ 48 hours'));