mysql php - help needed

[amkma]

I have a Mysql table with 4 fields.
id, name, date_of_birth and email.
I want to display some record from Mysql, where date_of_birth is todays date, But I don't want to display those records whose email field is null or empty.

I don't know how to do this.

[activeco]

SELECT * from MyTable where date_of_birth = CURDATE() and where EXISTS (SELECT email FROM mytable);

should work.
Also you could try ...and where email <>"";

or something in that line.

[DaveSawers]

Presumably you want to identify a birthday and thus you will only want the part of the date of birth that excludes the year part and compare it against the current date without the current year added.

If so, then activeco's SQL won't help you.

For a more specific answer, let us know the format of the date_of_birth field in your MySQL database.

[activeco]

You are right Dave. Amkma probably needs month and date only in which case something like (RIGHT(CURDATE(),5)=RIGHT(date_of_birth,5)) could be used, under condition that date is in yyyy-mm-dd format.

[wige]

This should work, I believe:

SELECT * FROM table WHERE MONTH(date_of_birth) = MONTH(CURDATE()) AND DAY(date_of_birth) = DAY(CURDATE()) AND email NOT NULL AND email !='';

Edit for less processing overhead:

SELECT * FROM table WHERE DATEFORMAT(date_of_birth, '%c %e') = DATEFORMAT(CURDATE(), '%c %e') AND email NOT NULL AND email !='';

You could go even easier, if there is a large database, and replace the second DATEFORMAT() with a PHP variable containing the current date and month, separated by a space.

[amkma]

Many thanks to all of you.

The following solution helped me out:

$curdate=date("YmdHis");

SELECT * FROM table WHERE date_of_birth = $curdate AND email !=''


Cheers.