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MySQL Problem

Posted on 2010-09-11
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Last Modified: 2012-05-10
In a solution provided in a previous question I was given this:

$rd = $_REQUEST['start_date'];
$sql = "SELECT * FROM `listings` WHERE `date` >= date_add(str_to_date('$rd', '%d/%M/%Y'), interval -3 day) and `date` < date_add(str_to_date('$rd', '%d/%M/%Y'), interval 4 day) ";

I need to know if this would work:


$rd = $_REQUEST['start_date'];
$sql = "SELECT * FROM `listings` WHERE str_to_date(`date`, '%d/%M/%Y') >= date_add(str_to_date('$rd', '%d/%M/%Y'), interval -3 day) and str_to_date(`date`, '%d/%M/%Y') < date_add(str_to_date('$rd', '%d/%M/%Y'), interval 4 day) ";

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Question by:grantunwin
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2 Comments
 
LVL 58

Accepted Solution

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cyberkiwi earned 2000 total points
ID: 33653363
Only if the column `date`contains not date data, but VARCHAR data in d/M/Y format.
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LVL 58

Expert Comment

by:cyberkiwi
ID: 33653370
Using any function on the `date` column also causes an index on that column to be unusable.  Consider making a date range that will cover the date+time involved.
If `date` contains time data and you want to include the 7th day, then try this

$rd = $_REQUEST['start_date'];
$sql = "SELECT * FROM `listings` WHERE `date` >= date_add(str_to_date('$rd', '%d/%M/%Y'), interval -3 day) and `date` < date_add(str_to_date('$rd', '%d/%M/%Y'), interval 5 day) ";

< ".. interval 5 days "

will include any hour on the day "$rd +4 days"
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