Question

I am new to programming so there may be a simple solution to my problem.

What I am trying to accomplish is finding duplicate fields entries in a MySQL column the entries are going to be dates example 01/20/13

I need to search through the date field and determine how many duplicate entries there are for a date then upon finding out the number I need to create a loop that will take the date change it into the name of the day the date is then add a . and a number after the .

So for example 01/20/13 is Sunday so in my database example below there are 4 duplicate entries so the output needs to be Sunday.1, Sunday.2, Sunday.3, Sunday.4

I have this functionality working in the PHP program below but I need it to search for duplicates in a 7 day range example 01/20/13 - 01/26/13, starting with Sunday and ending with Saturday. I don’t want to hard code a weeks worth of date values in. That is where my problem lies I don’t know where to go from here I am stuck. I can only think of hard coding the weeks worth of values in to get the desired results but I know there is an easier way.

My PHP code:

<?php

    require '../TimeCard/DB.php';

    $i = 1;

    try{
    $stmt = $conn->prepare('SELECT `date` FROM `timeRecords` WHERE `employeeID`= 1 ORDER BY date ');
    $stmt->execute();
    } catch(PDOException $e){
        echo'ERROR: ' . $e->getMessage();
    }

    while($row = $stmt->fetch())
  {
    if ($row['date'] == "01/20/13"){

        $date = date("l", strtotime($row['date']));
        $$date = "Hello today is " . $date . "." . $i . "<br>";
        echo $Sunday;
        $i++;
    }
  }

?>

Database:

+----+------------+----------+--------+-------------------+-----------+------------+-----------+---------+
| id | employeeID | date     | timeIn | jobDescription    | equipType | unitNumber | unitHours | timeOut |
+----+------------+----------+--------+-------------------+-----------+------------+-----------+---------+
| 10 |          1 | 01/20/13 | 20:30  | Blah1             |           | 01E        |      NULL | NULL    |
|  2 |          1 | 01/20/13 | 1:00   | Test              |           | 02S        |      NULL | NULL    |
|  9 |          1 | 01/20/13 | 13:00  | Blah12            |           | 02E        |      NULL | NULL    |
|  8 |          1 | 01/20/13 | 6:00   | Blah              |           | 02D        |      NULL | NULL    |
|  1 |          1 | 01/21/13 | 09:12  | Worked in hubbard | Excavator | 01E        |      8375 | 09:14   |
|  3 |          1 | 01/22/13 |        |                   |           | NULL       |      NULL | NULL    |
|  4 |          1 | 01/23/13 |        |                   |           | NULL       |      NULL | NULL    |
|  5 |          1 | 01/24/13 |        |                   |           | NULL       |      NULL | NULL    |
|  6 |          1 | 01/25/13 |        |                   |           | NULL       |      NULL | NULL    |
|  7 |          1 | 01/26/13 |        |                   |           | NULL       |      NULL | NULL    |
+----+------------+----------+--------+-------------------+-----------+------------+-----------+---------+

Answer 1

Your approach seems to be too complicated. Find count for today’s date and given employee instead.

SELECT date, COUNT(*) FROM timeRecords 
  WHERE employeeID = 1 AND date BETWEEN DATE_SUB(NOW(),INTERVAL 1 WEEK) AND NOW() 
  GROUP BY date
  ORDER BY date DESC