Select record(s) from mysql table where date is greater than or equal to today

问题

I am working on a project that has rows in the database that contains a date. I want to echo the data for the fields that have a date that is equal or greater to today. I have looked at other post and tried a lot of different methods and have yet to succedd. Currently what I have returns an error when I do >= but when I just do = with the statement below, I get all rows from the database.

  $sql = "SELECT * FROM drives WHERE ddest = '{$trimmed}' AND 'leave_date' => DATE_FORMAT(CURDATE(), '%m/%d/%Y') LIKE '%$trimmed%' ORDER BY timeago DESC";

回答1:

Try this

  $sql = "SELECT * FROM drives WHERE ddest = '{$trimmed}' AND 'leave_date' >= NOW() LIKE '%$trimmed%' ORDER BY timeago DESC";

else this,

  $sql = "SELECT * FROM drives WHERE ddest = '{$trimmed}' AND 'leave_date' >= DATE_FORMAT(CURDATE(), '%m/%d/%Y') LIKE '%$trimmed%' ORDER BY timeago DESC";

回答2:

try this query if you want a date that is equal or greater to today.

select * from drives where day(now()) <= day(ddest) and year(ddest) >= year(now());

回答3:

A valid query might look something like this:

SELECT * 
  FROM drives 
 WHERE ddest = '{$trimmed}' 
   AND leave_date >= CURDATE()
 ORDER 
    BY timeago DESC

Any remaining problems are more likely to do with the way you're handling our data rather then with the query itself.

来源：https://stackoverflow.com/questions/41421524/select-records-from-mysql-table-where-date-is-greater-than-or-equal-to-today