在MySQL中使用带有变量的LIKE

Question

OK so I would like to get a match for a query of the database using a LIKE operator in my SQL statement.

I'm looking at a field that has a date and time in it, like this

2012-04-27 12:00:00

I only want to compare the first part of the string, just the date bit, so I was using LIKE but I don't get any matches.

Here's the code

$colname_cruises = "-1";
if (isset($_GET['id'])) {
$colname_cruises = str_replace('-', ' ', $_GET['id']);
}

$colname_date = "-1";
if (isset($_GET['date'])) {
$colname_date = $_GET['date'];
}

$query_cruises = sprintf("SELECT * FROM cruises WHERE title = %s AND departs LIKE %s", GetSQLValueString($colname_cruises, "text"), GetSQLValueString($colname_date, "text"));
$cruises = mysql_query($query_cruises, $connection) or die(mysql_error());
$row_cruises = mysql_fetch_assoc($cruises);
$totalRows_cruises = mysql_num_rows($cruises);

I'm pretty certain just doing a LIKE %s isn't enough... what else can I add in here ?

Answer 1

if you only want to compare the first part of the string, just the date then why dont you use MySQL DATE() which

Extracts the date part of the date or datetime expression expr

 SELECT DATE('2003-12-31 01:02:03');
   -> '2003-12-31'

Answer 2

For date comparison you can use between. Try to use something like that:

departs between <date> and DATE_ADD(<date>,INTERVAL 1 DAY)

Answer 3

Please try

 "SELECT * FROM cruises WHERE title = '".$colname_cruises."' AND departs LIKE '".$colname_date."%'";

If you want to fetch record where departs value starting string containg this date then you have to use Like $str% .

thanks