如何计算mysql数据库中的表水平？

Question

I have a table with 15 Columns ID, USERNAME, PASSWORD, EMAIL, TELEPHONE, FAX, ADDRESS, COUNTRY, CITY, NATIONALITY, EDUCATION, etc in mysql database.

There is some required fields and optional fields. Users will fill the required fields and some optional fields. I need to give every user completeness score (e.g. Completeness Score 40%) to increase there attention to fill all optional fields.

How can I do that in PHP? I've tried to do it by ( SELECT COUNT WHERE ) - but without any result.

Answer 1

Supposed the table is called "users", you do

SQL: 'select * from users where id=X'

(Without adding any of the optional fields in the WHERE clause!)

Then later in PHP, you do something like this:

if (strlen($row['username'])>1)
    $score += 10;
if (is_valid_email($row['email']))
    $score += 10;

... et cetera. There might be better ways of doing this. It is just to give you an Idea! This is a very basic question, you might want to learn some more! Cheers.