CODEWITHSUNDEEP
mysqldate
anonymous
anonymous

Reputation: 1

Compare two dates in a database table are equal

I'm new here. So, be easy on me. I have a table called 'data' in which one of the columns is 'date' stored in YYYY-MM-DD format. Here is the code that I have been working on. Basically, what I want to do is compare if two dates stored in my table are equal. But, each time I run the code, I keep getting error: Undefined offset right where the code says $lisdate[$i+1]. How do I compare the dates stored in a table? Thank you.

My code

$sth2 = mysql_query("SELECT * FROM data WHERE dest_name='www.myren.net.my'");

while($rowstemp = mysql_fetch_assoc($sth2))
{
   $lisdate[] = $rowstemp['date'];  
   $lisaverage[] = $rowstemp['avg_rtt'];
}

$rows = mysql_num_rows($sth2);
$addAverage[0] = $lisaverage[0];
$numbers = 0;
$j = 0;


for($i = 0; $i<$rows; $i++)
{
   if($lisdate[$i] == $lisdate[$i+1])
   {
      $addAverage[$j] = $addAverage[$j] + $lisaverage[$i+1];
   }
   else 
   {
      $j++;
      $numbers = $numbers +1;
   }
}

Upvotes: 0

Views: 184

Answers (2)

spencer7593
spencer7593

Reputation: 108450

It's not entirely clear what that code in the question is attempting to achieve.

(As Ollie pointed out in his answer, it bears repeating: the order that MySQL returns the rows is guaranteed ONLY if there's an ORDER BY clause on the query. The result from a "duplicate date" check that's being performed in the code is only going to detect a "duplicate" date value if it appears on contiguous rows.)

We see the code loading a couple of arrays, using the date and avg_rtt column values returned by the query. Then there's some manipulation on the array... the end result will be another it's not clear what the actual intent of that rigmarole is, what it's actually trying to achieve.

If there are any rows It looks like there's going to be another array... if there are no "duplicate" date values found, it's going to have a single element (with the value of rtt_avg from the first row), and $numbers will be set to the number of elements in the original array.

If there is a "duplicate" date values found, the results from the code seem very odd to me, a sparsely populated array. Why?

Personally, I'd be looking to get an actual statement of the specification, and have the database do that processing for me, rather than mucking through two arrays.

If what we want is set of values with no duplicated date values, I'd use a GROUP BY and some aggregation, for example: 

 SELECT d.dest_name
      , d.date
      , AVG(d.avg_rtt)            AS avg_avg_rtt
      , MAX(d.avg_rtt)            AS max_avg_rtt
      , MIN(d.avg_rtt)            AS min_avg_rtt
      , SUM(d.avg_rtt)            AS sum_avg_rtt
      , COUNT(1)                  AS cnt
      , COUNT(d.avg_rtt)          AS cnt_avg_rtt
      , COUNT(DISTINCT d.avg_rtt) AS cnt_distinct_avg_rtt
   FROM data d
  WHERE d.dest_name = 'www.myren.net.my'
  GROUP
     BY d.dest_name
      , d.date
  ORDER
     BY d.dest_name
      , d.date

If I was looking for just a count of distinct date values, like what $numbers is going to contain, then just:

 SELECT d.dest_name
      , COUNT(DISTINCT d.date) AS cnt_distinct_date
   FROM data d
  WHERE d.dest_name = 'www.myren.net.my'
  GROUP BY d.dest_name
  ORDER BY d.dest_name

The "looping through array" that the original query just looks like an odd way to achieve a result, whatever that result is supposed to be.

Upvotes: 0

O. Jones
O. Jones

Reputation: 108746

Gordon has it right. Your for loop is overrunning the end of your array because you're working with consecutive pairs of rows. Change it to read ; $i < $rows - 1; to correct this problem.

But, you have another problem. SQL rows in result sets have an unpredictable order unless your query includes an ORDER BY clause. If these rows, without that clause, appear in ascending order by date, it's dumb luck. Put ORDER BY date in your query.

Upvotes: 2

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