PHP - date field

[wilna]

Hi

I have a date of birth field(dob) in my table, eg 16/05/2011 (dd/mm/yyyy).  I need to run a query but it must not include the year, how do I do that?

 

$sql = "SELECT * FROM detail WHERE dob='$mydate'";

 

The dob is in the format dd/mm/yyyy, how do I change my query to just use dd/mm?

Thank you

[Adam]

The dob is in the format dd/mm/yyyy, how do I change my query to just use dd/mm?

 

You'll need to use the proper 'date' data type first, which will store the date in the format "YYYYMMDD".

[wilna]

Hi, my field is in date format, please see the attached examples.  Can you please be so kind and as to explain how my query must look?

 

[attachment deleted by admin]

[cyberRobot]

You could break apart the date with explode():

 

list($datePart_day, $datePart_month, $datePart_year) = explode('/', $mydate);

 

 

Then just use the new variables in the query:

 

$sql = "SELECT * FROM detail WHERE dob LIKE '%$datePart_month-$datePart_day'";

 

 

Of course you may need to do some checking to make sure month and day are valid. And make sure they are padded with zeros if needed.

[Adam]

@cyberRobot

 

It would be better to use the built-in MySQL functions for this, to avoid splitting the logic and it's a lot more efficient than using a LIKE statement.

 

@wilna

 

Ah okay great - though that wasn't what you said originally. Anyway you want to use the DATE() and MONTH() functions to do this:

 

select * from table where date(dob) = 16 and month(dob) = 5;

[cyberRobot]

@MrAdam - Ah, yes that does look much easier. Of course, they'll still need to use something like explode() to break apart the date contained in $mydate.

[wilna]

Thank you both!  The following solved my problem:

 

$myDate = date('d');
$myDate1 = date('m');
.....
$sql = "SELECT * FROM detail WHERE month(dob)='$myDate1'and day(dob)='$myDate'";

[Adam]

day(dob)='$myDate'

 

Sorry day() yeah, not date().