Date is displaying wrong, but it's ok in database - php

Question

I've researched a lot, but nothing helped me. I have 2 views: one for sign up and another one to edit your infos, if you forgot something.
Let's say I wrote in the date label the following month and year: 98/9889 (yeah, i'm retarded). It will shows correctly, but when I get my infos to edit it, it shows 31/1219, but in my database it's still 98/9889. Any idea?
SIGN UP VIEW:
<label for="antitetanica">Data da última dose antitetânica:</label>
<input type="text" id="antitetanica" name="antitetanica" value="echo date("m/Y", strtotime('antitetanica'))"/><br /><br>
EDIT VIEW:
<label for="antitetanica">Data da última dose antitetânica:</label>
<input type="input" name="antitetanica" id="antitetanica" value="<?php echo date("m/Y", strtotime('antitetanica')) ?>"/><br /><br>
Btw i was using DATE type in my database, but now i'm using varchar 255.
EDIT: now using the DATE type in my db
EDIT 2: actual code looks like (sorry, first time working with DateTime):
SIGN UP VIEW:
<input type="input" id="antitetanica" name="antitetanica" value="
<?php $date = date_create_from_format( 'm/Y', $antitetanica )->format('m/Y'); ?>" /><br /><br>
EDIT VIEW:
<label for="antitetanica">Data da última dose antitetânica:</label>
<input type="input" name="antitetanica" id="antitetanica" value="<?php $date = new DateTime( $antitetanica ); echo $date->format('m/Y'); ?>" /><br /><br>

Answer 1

Because strtotime doesn't recognise 'antitetanica' ('tetanus') as a valid date:
PHP strtotime
Even assuming that you've missed a '$' off the start of that string and are actually attempting to feed strtotime a string value which has come from your database - this is not what it is for. You can't feed it 98/9889 and hope to get a valid date.
EDIT
You had the right idea - but strtotime was the wrong way to create your date object.
If your input is "03/2013", you need to convert that into a valid date object that PHP can work with. You do that with DateTime::createFromFormat();
$date = DateTime::createFromFormat( 'm/Y', '03/2013' ); //<< Object-Oriented Style
or
$date = date_create_from_format( 'm/Y', '03/2013' ); //<< Procedural Style
Once you have a date object, you can then output it in whatever format you like:
$date->format('m/Y'); //<< Object
or
date_format( $date, 'm/Y' ); //<< Procedural
So - in your original question (storing the date as a string in the database), it would look like:
echo date_create_from_format( 'm/Y', $antitetanica )->format('m/Y');
But now that you have an actual DATE object in your database, that statement looks something like:
echo date_create_from_format( 'Y-m-d', $antitetanica )->format('m/Y');
Which is NOT the ideal way to do it ... but I'm painting a picture for you here.
A better way to work with dates now that you have actual DATES in your database, is with DateTime
When you retrieve your data from the database, you can now do something like:
$date = new DateTime( $antitetanica ); //<< (where $antitetanica = "2013-03-01")
echo $date->format('m/Y'); //<< "03/2013"
SO - How to get the date from a user in the first place?
There are several ways you could do this. You could get the user to enter the date using multiple 'select' form elements, or multiple 'text' form elements, or a single 'text' form element - and each of these ways would need to be handled differently in PHP.
Let's say you decide to have the user enter the date "03/2013" into a single text field. Over on the server side of things, you'd need to follow exactly the same steps as above - that is:
Convert user input to a valid DateTime object
Format and insert the date into your database
So:
$date = DateTime::createFromFormat( 'm/Y', $userInput );
INSERT INTO myTable ( antitetanica ) VALUES ( '$date->format('Y-m-d')' ) //<< Pseudocode
As 2 text input fields, that might look like:
$date = DateTime::createFromFormat( 'm/Y', $userMonth . '/' . $userYear );
INSERT INTO myTable ( antitetanica ) VALUES ( '$date->format('Y-m-d')' ) //<< Pseudocode